Statistics - User contributions [en]

Problem 2-1, Comparing a single mean to a specified value

2011-11-04T15:40:13Z

Statadmin:

==Problem statement==
''The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that the standard deviation of breaking strength is σ = 3 psi. A random sample of four specimens is tested, and the results are y<sub>1</sub> = 145, y<sub>2</sub> = 153, y<sub>3</sub> = 150, and y<sub>4</sub> = 147.''

<ol style="list-style-type:lower-latin">
<li>''State the hypotheses that you think should be tested in this experiment.''</li>
<li>''Test these hypothesis using α = 0.05. What are your conclusions?''</li>
<li>''Find the ''P''-value for the test in part (b).''</li>
<li>''Construct a 95 percent confidence interval on the mean breaking strength.''</li>
</ol>

==Solution==
[[Image:Gaussian1.png|thumb|left|'''Figure 1:''' Our data compared to a theoretical Gaussian distribution.]]

===Section A: Choosing hypotheses===
In this problem we are given a set of four data points. These data points all come from a distribution of breaking strengths which has an unknown mean μ. We will call this the ''true distribution''. Previous experience indicates that breaking strengths follow a Gaussian ''theoretical distribution'' with a standard deviation of 3 psi, so we assume this for our distribution also. Our task is to determine whether or not the true mean, which is impossible to know exactly, is greater than or equal to 150 psi. We plot this data and the distribution in figure 1.

Since the sample mean is an approximation of the true mean, we define the ''standard error of the mean'' (SEM) to be <math>\sigma/\sqrt{n}=1.5</math>, where n=4 is the number of data points.

<table border=1 cellspacing=0 cellpadding=4>
<tr><th bgcolor="#eeeeee"></th><th bgcolor="#eeeeee">Distribution type</th><th bgcolor="#eeeeee">Mean</th><th bgcolor="#eeeeee">Standard deviation</th></tr>
<tr><th bgcolor="#eeeeee">True distribution</th><td>Normal</td><td><math>\mu\approx\overline{y}=148.25</math></td><td><math>\sigma=3</math></td></tr>
<tr><th bgcolor="#eeeeee">Theoretical distribution</th><td>Normal</td><td><math>\mu_0=150</math></td><td><math>\sigma_0=3</math></td></tr>
</table>

We first state two hypotheses. The null hypothesis is that our data does come from the theoretical distribution: the true mean μ = μ<sub>0</sub>. Our alternative hypothesis states that the data comes from a distribution centered around a different mean.

There are three choices for the alternative hypothesis: μ < 150, μ > 150, and μ ≠ 150. We adopt the convention that the alternative hypothesis will be true if the data does not meet the requirements. In this case, the breaking strength of the fiber is required to be at least 150 psi, so we choose μ < 150 as our alternative hypothesis.

Formally, we state our hypotheses as:<br/>
<center>H<sub>0</sub>: μ = 150<br/>
H<sub>1</sub>: μ < 150</center>

<br clear='all'>
[[Image:Gaussian2.png|thumb|left|'''Figure 2:''' Our plot after normalizing.]]

===Section B: Z-values===
For convenience, we start by standardizing our theoretical distribution to have a mean of zero and a standard deviation of one. To do this, we first center the distribution around zero by subtracting the theoretical mean (150) from each point in the distribution. We then divide each point by the standard deviation (3). The sample mean can be standardized in the same manner. We plot the normalized distribution and sample mean in figure 2.

We now assume that the null hypothesis is true and ask whether or not this assumption makes sense. Given that this assumption is true, the sample mean is most likely to be close to zero. To test this, we define a range over which we consider our sample mean to be unacceptable, the ''rejection region''. If the sample mean is in the rejection region, it is too far from zero and we reject the null hypothesis.

We will define the lower limit to be z<sub>α</sub>, where α=0.05. Graphically, given a standard Gaussian distribution, the area under the curve left of z<sub>0.05</sub> is equal to 5% of the total area. You can either look up z<sub>α</sub> in a table or calculate it using a software package. Using Excel, the appropriate function is <tt>=NORMSINV(alpha)</tt>. The corresponding function in R is <tt>qnorm(alpha)</tt>. Using one of these methods, we find that z<sub>0.05</sub>=−1.64.

We then find the z-value of our data and compare the z-value to z<sub>α</sub>. The formula for the z-value is as follows:

<center><math>z=(\overline{y}-\mu_0)(\frac{1}{\sigma})(\sqrt{n})</math></center>

Because we have already standardized our data, μ<sub>0</sub>=0 and σ=1, so this formula simplifies to <math>\overline{y} \sqrt{n}=-0.417\cdot2=-0.833</math>. Note that the formula above normalizes the data, if it has not already been normalized. The z-value can be interpreted as the distance between the sample mean and μ<sub>0</sub>, scaled by a factor which makes the z-value more extreme with large sample sizes. If we take many samples, our z-value is more likely to fall in the rejection region, because we are more certain of the accuracy of our sample mean.

The rejection region for our z-value is from negative infinity to z<sub>α</sub>. We see that our z-value is greater than z<sub>α</sub>. Therefore, we cannot reject the null hypothesis.

[[Image:Gaussian3.png|thumb|left|'''Figure 3:''' Illustrating the P-value.]]

===Section C: P-values===
Another way to judge how likely it is that our null hypothesis is true is to calculate the P-value. If we were to redo the experiment, taking four new data points, the P-value gives us the probability of our new sample mean being at least as extreme as our original sample mean. Graphically, if we extend the critical region until it reaches our z-value, the P-value is equal to the area of the shaded region (see figure 3).

To calculate the P-value in Excel, use <tt>=NORMSDIST(-ABS(z))</tt>. In R, use <tt>pnorm(-abs(z))</tt>. (We use the negative absolute value because <tt>NORMSDIST</tt> and <tt>pnorm</tt> integrate from negative infinity to the z-value. If the z-value is positive, we instead want to integrate from the z-value to positive infinity, which is mathematically equivalent to integrating from negative infinity to the negative of the z-value.)

For this problem, we find that the P-value is 0.202. Note that a P-value of 0.5 indicates that the sample mean is equal to the mean of the theoretical distribution. You can see this graphically by noting that the z-value will be zero in this case, and integrating the theoretical distribution to zero covers half of the area. (Recall that the total area under a standard Gaussian curve is one.) The further the P-value is from 0.5, the greater the distance between the two means.

<div style="float:left; vertical-align: top; padding-right: 20px; padding-bottom: 20px;">[[Image:Gaussian4.png|thumb|none|'''Figure 4:''' The confidence interval about the sample mean.]]<br>
[[Image:Gaussian5.png|thumb|none|'''Figure 5:''' The confidence interval about the theoretical mean.]]</div>

===Section D: Confidence intervals===
We now return to our original data set and theoretical distribution with the mean of 150 psi; that is, we will no longer use our normalized space.

We will now calculate the range of sample means that would lead us to conclude that the breaking strength of our fiber is at least 150 psi, given an α of 0.05. This range is known as the confidence interval about the sample mean.

To calculate this interval, we ask what sample mean would give us a z-value equal to z<sub>α</sub>. We can determine this by substituting z<sub>α</sub> for z into the formula for z, and solving for <math>\overline{y}</math>:

<center><math>z_\alpha=(\overline{y}-\mu_0)(\frac{1}{\sigma})(\sqrt{n}) \Rightarrow \overline{y} = \mu_0+\frac{z_\alpha \sigma}{\sqrt{n}}=147.53</math></center>

This is the lower limit of our confidence interval. Because any sample mean greater than 150 is acceptable, the upper limit of the confidence interval is infinity. We plot this interval in figure 4. Formally, our confidence interval about the sample mean is

<center><math>147.53 < \overline{y} < \infty</math></center>

We next calculate a confidence interval about the mean of the theoretical distribution, μ<sub>0</sub>. This will give us the range of minimum breaking strengths we could have specified and still found our data acceptable. We can calculate this in much the same way as the previous confidence interval: substitute z<sub>α</sub> for z in the formula for z, but this time solve for μ<sub>0</sub>:

<center><math>\mu_0=\overline{y}-\frac{z_\alpha \sigma}{\sqrt{n}}=151.22</math></center>

This is the upper limit of our confidence interval. The lower limit is zero, because we simply require the theoretical mean to be less than this number. Formally, our confidence interval about the theoretical mean is

<center><math>0 \le \mu_0 < 151.22</math></center>

Problem 5-1, Factorial designs

2010-08-30T23:25:53Z

Statadmin: /* Solution */

==Problem Statement==
''The yield of a chemical process is being studied. The two most important variables are thought to be the pressure and the temperature. Three levels of each factor are selected, and a factorial experiment with two replicates is performed. The yield data follow:''
<table cellpadding=2 cellspacing=0 style="border-top: 1px solid black; border-bottom: 1px solid black"><tr><th rowspan=2 valign=bottom style="border-bottom: 1px solid black;">Temperature (°C)</th><th colspan=3 style="border-bottom: 1px solid black;">Pressure (psig)</th></tr>
<tr><td style="border-bottom: 1px solid black;">200</td><td style="border-bottom: 1px solid black;">215</td><td style="border-bottom: 1px solid black;">230</td></tr>
<tr><td rowspan=2 valign=top>150</td><td>90.4</td><td>90.7</td><td>90.2</td></tr>
<tr><td>90.2</td><td>90.6</td><td>90.4</td></tr>
<tr><td rowspan=2 valign=top>160</td><td>90.1</td><td>90.5</td><td>89.9</td></tr>
<tr><td>90.3</td><td>90.6</td><td>90.1</td></tr>
<tr><td rowspan=2 valign=top>170</td><td>90.5</td><td>90.8</td><td>90.4</td></tr>
<tr><td>90.7</td><td>90.9</td><td>90.1</td></tr></table>

<ol style="list-style-type:lower-latin">
<li>''Analyze the data and draw conclusions. Use α=0.05.''</li>
<li>''Prepare appropriate residual plots and comment on the model's adequacy.''</li>
<li>''Under what conditions would you operate this process?''</li>
</ol>

==Solution==
<div style="float:left; vertical-align: top; padding-right: 20px; padding-bottom: 20px;">
[[Image:5-1yield.png|thumb|180px|none|'''Figure 1:''' This plot of yield vs. pressure, at the different levels of temperature, helps to visualize the different effects in our problem.]]
<br>
[[Image:5-1dataplot.png|thumb|180px|none|'''Figure 2:''' This plot illustrates the various effects on the data in problem 5-1. The hats (^) indicate that these are approximate effects calculated from our sample data.]]</div>

We are given a set of data that describes the yield of a chemical process, but the yield is affected by two factors: temperature and pressure. At each combination of <math>a</math> levels of factor A (temperature) and <math>b</math> levels of factor B (pressure), <math>n</math> data points have been collected (see chart in problem statement). Not only do these two factors influence the yield data directly, but they might also interact to affect the yield in an unexpected manner.

The average of the <math>n</math> data points at a specific temperature and pressure in our problem is <math>\bar{y}_{ij}.</math>. An interaction effect would cause the difference between two <math>\bar{y}_{ij}.</math> values in a column of our chart to vary as pressure is changed. For example, say we set the pressure to 200 psig and observe the average yield increase from 90 to 95 as temperature is increased from 150 °C to 160 °C. But when we set the pressure to 215 psig the yield drops from 90 to 85 as temperature is again increased from 150 °C to 160 °C. This would be an interaction effect, since the effect on the yield cannot be separated into independent temperature and pressure effects. Of course, what appears to be an interaction effect may just be large random error, so we'll have to check for that.

The plot shown in Figure 1 is very useful for visualizing the factor and interaction effects. We can see an apparent temperature effect that indicates 170 °C produces the highest yield. 215 psig also appears to be the best pressure if we want high yield, as the green line is higher than the others. Interaction effects in this type of plot are indicated by non-parallel lines, so there is not much apparent interaction here. Before we draw any conclusions, we must test that this plot provides a true image of the chemical process, and is not just a product of high random error.

Each data point in our problem can be thought of as the sum of several effects:

<center><math>y_{ijk} = \mu + \tau_i + \beta_j + (\tau \beta)_{ij} + \epsilon_{ijk}</math></center>

Where:<br>
<math> i = 1,2,...,a </math><br>
<math> j = 1,2,...,b </math><br>
<math> k = 1,2,...,n </math><br>
and:<br>
<math> y_{ijk} = </math> data point from the <math>i</math>th level of factor A and the <math>j</math>th level of factor B<br>
<math> \mu = </math> the grand mean<br>
<math> \tau_i = </math> the effect of the <math>i</math>th level of factor A<br>
<math> \beta_j = </math> the effect of the <math>j</math>th level of factor B<br>
<math> (\tau \beta)_{ij} = </math> the effect of interaction between factors A and B<br>
<math> \epsilon_{ijk} = </math> a random error affecting the data point from the <math>i</math>th level of factor A and the <math>j</math>th level of factor B<br>

Figure 2 illustrates how these effects sum up to each data point. The blue line indicates the grand mean <math>\bar{y}.. \approx \mu </math>, the red X indicates the sample mean of the <math>i</math>th level of temperature <math>\bar{y}_i.</math>, the green # indicates the #th pressure level effect indicator, the brown triangle indicates the interaction effect indicator, and the black # indicates the data point from the #th level of pressure. The projection of each line onto the x-axis tells you the quantitatively what its estimated effect is, and each point can be thought of as where the data would be if effects closer to the real data point were not present.

=== Hypothesis Testing ===
We would like to know if any of our estimated effects are significant. Therefore, we have three tests to do, one for each type of effect. For factor A, our null hypothesis will be that all the treatment means are equal (they all come from the same distribution). Our alternative hypothesis is that they are not all equal:

<math> H_0\!:~\tau_1 = \tau_2 = ... = \tau_a = 0 </math><br>
<math> H_1\!:~\tau_i \ne 0 </math>

Similarly for factor B and the interaction effect:

<math> H_0\!:~\beta_1 = \beta_2 = ... = \beta_b = 0 </math><br>
<math> H_1\!:~\beta_i \ne 0 </math><br>

<math> H_0\!:~(\tau \beta)_{ij} = 0 </math> (for all <math>i</math>, <math>j</math>)<br>
<math> H_1\!:~(\tau \beta)_{ij} \ne 0 </math> (for at least one combination of <math>i</math> and <math>j</math>)<br>

To test these hypotheses, we are interested in the following sums of squares. Each of the following equations estimates the effect on our data due to the subscripted factor (<math>A </math> → factor A, <math>B</math> → factor B, <math>AB</math> → interaction, and <math>E</math> → random error).

<math>SS_A = bn \Sigma^a_{i=1} (\bar{y}_i.. - \bar{y}...)^2 </math><br>
<math>SS_B = an \Sigma^b_{j=1} (\bar{y}._j. - \bar{y}...)^2 </math><br>
<math>SS_{AB} = n \Sigma^a_{i=1} \Sigma^b_{j=1} (\bar{y}_{ij}. - \bar{y}_{i}.. - \bar{y}._j. + \bar{y}...)^2 </math><br>
<math> SS_E = \Sigma^a_{i=1} \Sigma^b_{j=1} \Sigma^n_{k=1} (y_{ijk} - \bar{y}_{ij}.)^2 </math><br>

Before we can compare these quantities to each other, we must divide by the degrees of freedom to normalize. We calculate the normalized ''mean squares'' with:

<math> MS_A = \frac{SS_A}{a-1} </math><br>
<math> MS_B = \frac{SS_B}{b-1} </math><br>
<math> MS_{AB} = \frac{SS_{AB}}{(a-1)(b-1)} </math><br>
<math> MS_E = \frac{SS_E}{ab(n-1)} </math><br>

We test whether our factor and interaction effects are significant by dividing their mean squares by the mean square of our random error to get <math>F_0</math>. This value must be large for us to conclude that the effect we're testing is significant (the effect must be large relative to the random error). We pick a value from the F distribution based on <math>\alpha</math> and our degrees of freedom – this determines what "large" is. Do this in Excel with <tt>FINV(α,numerator's DOF,denominator's DOF)</tt> and in R with <tt>qf(1-alpha, numerator's DOF, denominator's DOF)</tt> (DOF = degrees of freedom).

<table cellspacing=0 cellpadding=5 style="border-top: 1px solid black; border-bottom: 1px solid black">
<tr align="center">
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Source of Variation </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Sum of Squares </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Degrees of Freedom </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Mean Square </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> <math>F_0</math> </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> <math>F_{\alpha}</math> </th>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Factor A (Temperature) </th>
<td> 0.301 </td>
<td> 2 </td>
<td> 0.151 </td>
<td> 8.469 </td>
<td> 4.256 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Factor B (Pressure) </th>
<td> 0.768 </td>
<td> 2 </td>
<td> 0.384 </td>
<td> 21.594 </td>
<td> 4.256 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Interaction </th>
<td> 0.069 </td>
<td> 4 </td>
<td> 0.017 </td>
<td> 0.969 </td>
<td> 3.633 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Error </th>
<td> 0.160 </td>
<td> 9 </td>
<td> 0.018 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee" style="border-top: 1px solid black;"> Total </th>
<td style="border-top: 1px solid black;"> 1.298 </td>
<td style="border-top: 1px solid black;"> 17 </td>
<td style="border-top: 1px solid black;"></td>
<td style="border-top: 1px solid black;"></td>
<td style="border-top: 1px solid black;"></td>
</tr>
</table>

We conclude from these tests that the effects from factors A and B (temperature and pressure) are significant, and there is no significant interaction effect. However, these tests are only accurate if the model is correct.

===Residual Plots and Model Adequacy Testing===
[[Image:5-1norm_prob.png|thumb|360px|left|'''Figure 3:''' These plots indicate that our data is sufficiently gaussian and we can believe the results of our hypothesis testing.]]

To test whether our data is sufficiently gaussian that our hypothesis testing is valid, we create a normal probability plot. As before, we calculate the residuals (by subtracting the expected values from the datapoints, <math>y_{ijk} - \bar{y}_{ij}.</math>) and sorting them in an array. Then we plot the sorted residuals against z-values, where the z-values are calculated in Excel by doing <tt>NORMSINV(percent)</tt> or in R with <tt>qnorm(percent)</tt>. <tt>Percent</tt> is equal to index_of_array/(DOF + 1). If the plot does not roughly fall along a straight line, the data is not from a Gaussian distribution.

There are several other useful plots we can make with residuals. Plotting <math>\bar{y}_{ij}.</math> against residuals produces a plot that should have data randomly scattered throughout its entire area, if not the data may not be gaussian. Plotting the residuals against either of the factors should indicate that the data is more or less equally random in each level of the factors. Note that because <math>n = 2</math>, there is some symmetry present in each of these plots.

Given that our data fits the model reasonably well, and that each of the factors are significant, the optimum conditions for producing high yield are when temperature is 170 °C and pressure is 200 psig.

MediaWiki:Mainpage

2010-08-06T01:41:55Z

Statadmin:

Statistics_Tutorial:_Design_and_Analysis_of_Experiments

MediaWiki:Mainpage

2010-08-06T01:39:35Z

Statadmin: Created page with "Statistics Tutorial: Design and Analysis of Experiments"

Statistics Tutorial: Design and Analysis of Experiments

Main Page

2010-08-06T01:28:56Z

Statadmin: moved Main Page to Statistics Tutorial: Design and Analysis of Experiments: Replace "Main Page" with meaningful title.

#REDIRECT [[Statistics Tutorial: Design and Analysis of Experiments]]

Statistics Tutorial: Design and Analysis of Experiments

2010-08-06T01:28:56Z

Statadmin: moved Main Page to Statistics Tutorial: Design and Analysis of Experiments: Replace "Main Page" with meaningful title.

This is a tutorial for Dr. Edward Brash's statistics class, at Christopher Newport University. Detailed solutions are presented for several problems from the first few chapters of ''Design and Analysis of Experiments'', 6th ed. by Douglas Montgomery. Solutions were prepared using the <tt>R</tt> statistical programming language, which is much more powerful than <tt>MS Excel</tt>, but has a significantly steeper learning curve.

==Solutions==
[[Problem 2-1, Comparing a single mean to a specified value|Problem 2-1, Comparing a single mean to a specified value]]

[[Problem 2-2, Comparing a single mean to a specified value (second example)|Problem 2-2, Comparing a single mean to a specified value (second example)]]

[[Problem 2-4, Determining required sample size|Problem 2-4, Determining required sample size]]

[[Problems 3-1 through 3-3, Analysis of variance|Problems 3-1 through 3-3, Analysis of variance]]

[[Problem 4-1, Randomized blocks|Problem 4-1, Randomized blocks]]

[[Problem 5-1, Factorial designs|Problem 5-1, Factorial designs]]

Statistics Tutorial: Design and Analysis of Experiments

2010-07-31T19:15:34Z

Statadmin:

File:5-1norm prob.png

2010-07-31T19:03:06Z

Statadmin: These plots indicate that our data is sufficiently gaussian and we can believe the results of our hypothesis testing.

These plots indicate that our data is sufficiently gaussian and we can believe the results of our hypothesis testing.

File:5-1dataplot.png

2010-07-31T19:02:16Z

Statadmin: This plot illustrates the various effects on the data in problem 5-1. The hats (^) indicate that these are approximate effects calculated from our sample data.

This plot illustrates the various effects on the data in problem 5-1. The hats (^) indicate that these are approximate effects calculated from our sample data.

File:5-1yield.png

2010-07-31T19:01:46Z

Statadmin: This plot of yield vs. pressure, at the different levels of temperature, helps to visualize the different effects in our problem.

This plot of yield vs. pressure, at the different levels of temperature, helps to visualize the different effects in our problem.

Problem 5-1, Factorial designs

2010-07-31T19:01:07Z

Statadmin: Created page with '==Problem Statement== ''The yield of a chemical process is being studied. The two most important variables are thought to be the pressure and the temperature. Three levels of eac…'

==Problem Statement==
''The yield of a chemical process is being studied. The two most important variables are thought to be the pressure and the temperature. Three levels of each factor are selected, and a factorial experiment with two replicates is performed. The yield data follow:''
<table cellpadding=2 cellspacing=0 style="border-top: 1px solid black; border-bottom: 1px solid black"><tr><th rowspan=2 valign=bottom style="border-bottom: 1px solid black;">Temperature (°C)</th><th colspan=3 style="border-bottom: 1px solid black;">Pressure (psig)</th></tr>
<tr><td style="border-bottom: 1px solid black;">200</td><td style="border-bottom: 1px solid black;">215</td><td style="border-bottom: 1px solid black;">230</td></tr>
<tr><td rowspan=2 valign=top>150</td><td>90.4</td><td>90.7</td><td>90.2</td></tr>
<tr><td>90.2</td><td>90.6</td><td>90.4</td></tr>
<tr><td rowspan=2 valign=top>160</td><td>90.1</td><td>90.5</td><td>89.9</td></tr>
<tr><td>90.3</td><td>90.6</td><td>90.1</td></tr>
<tr><td rowspan=2 valign=top>170</td><td>90.5</td><td>90.8</td><td>90.4</td></tr>
<tr><td>90.7</td><td>90.9</td><td>90.1</td></tr></table>

<ol style="list-style-type:lower-latin">
<li>''Analyze the data and draw conclusions. Use α=0.05.''</li>
<li>''Prepare appropriate residual plots and comment on the model's adequacy.''</li>
<li>''Under what conditions would you operate this process?''</li>
</ol>

==Solution==
<div style="float:left; vertical-align: top; padding-right: 20px; padding-bottom: 20px;">
[[Image:5-1yield.png|thumb|180px|none|'''Figure 1:''' This plot of yield vs. pressure, at the different levels of temperature, helps to visualize the different effects in our problem.]]
<br>
[[Image:5-1dataplot.png|thumb|180px|none|'''Figure 2:''' This plot illustrates the various effects on the data in problem 5-1. The hats (^) indicate that these are approximate effects calculated from our sample data.]]</div>

We are given a set of data that describes the yield of a chemical process, but the yield is affected by two factors: temperature and pressure. At each combination of <math>a</math> levels of factor A (temperature) and <math>b</math> levels of factor B (pressure), <math>n</math> data points have been collected (see chart in problem statement). Not only do these two factors influence the yield data directly, but they might also interact to affect the yield in an unexpected manner.

The average of the <math>n</math> data points at a specific temperature and pressure in our problem is <math>\bar{y}_{ij}.</math>. An interaction effect would cause the difference between two <math>\bar{y}_{ij}.</math> values in a column of our chart to vary as pressure is changed. For example, say we set the pressure to 200 psig and observe the average yield increase from 90 to 95 as temperature is increased from 150 °C to 160 °C. But when we set the pressure to 215 psig the yield drops from 90 to 85 as temperature is again increased from 150 °C to 160 °C. This would be an interaction effect, since the effect on the yield cannot be separated into independent temperature and pressure effects. Of course, what appears to be an interaction effect may just be large random error, so we'll have to check for that.

The plots shown in Figure 1 are very useful for visualizing the factor and interaction effects. We can see an apparent temperature effect that indicates 170 °C produces the highest yield. 215 psig also appears to be the best pressure if we want high yield, as the green line is higher than the others. Interaction effects in this type of plot are indicated by non-parallel lines, so there is not much apparent interaction here. Before we draw any conclusions, we must test that this plot provides a true image of the chemical process, and is not just a product of high random error.

Each data point in our problem can be thought of as the sum of several effects:

<center><math>y_{ijk} = \mu + \tau_i + \beta_j + (\tau \beta)_{ij} + \epsilon_{ijk}</math></center>

Where:<br>
<math> i = 1,2,...,a </math><br>
<math> j = 1,2,...,b </math><br>
<math> k = 1,2,...,n </math><br>
and:<br>
<math> y_{ijk} = </math> data point from the <math>i</math>th level of factor A and the <math>j</math>th level of factor B<br>
<math> \mu = </math> the grand mean<br>
<math> \tau_i = </math> the effect of the <math>i</math>th level of factor A<br>
<math> \beta_j = </math> the effect of the <math>j</math>th level of factor B<br>
<math> (\tau \beta)_{ij} = </math> the effect of interaction between factors A and B<br>
<math> \epsilon_{ijk} = </math> a random error affecting the data point from the <math>i</math>th level of factor A and the <math>j</math>th level of factor B<br>

Figure 2 illustrates how these effects sum up to each data point. The blue line indicates the grand mean <math>\bar{y}.. \approx \mu </math>, the red X indicates the sample mean of the <math>i</math>th level of temperature <math>\bar{y}_i.</math>, the green # indicates the #th pressure level effect indicator, the brown triangle indicates the interaction effect indicator, and the black # indicates the data point from the #th level of pressure. The projection of each line onto the x-axis tells you the quantitatively what its estimated effect is, and each point can be thought of as where the data would be if effects closer to the real data point were not present.

=== Hypothesis Testing ===
We would like to know if any of our estimated effects are significant. Therefore, we have three tests to do, one for each type of effect. For factor A, our null hypothesis will be that all the treatment means are equal (they all come from the same distribution). Our alternative hypothesis is that they are not all equal:

<math> H_0\!:~\tau_1 = \tau_2 = ... = \tau_a = 0 </math><br>
<math> H_1\!:~\tau_i \ne 0 </math>

Similarly for factor B and the interaction effect:

<math> H_0\!:~\beta_1 = \beta_2 = ... = \beta_b = 0 </math><br>
<math> H_1\!:~\beta_i \ne 0 </math><br>

<math> H_0\!:~(\tau \beta)_{ij} = 0 </math> (for all <math>i</math>, <math>j</math>)<br>
<math> H_1\!:~(\tau \beta)_{ij} \ne 0 </math> (for at least one combination of <math>i</math> and <math>j</math>)<br>

To test these hypotheses, we are interested in the following sums of squares. Each of the following equations estimates the effect on our data due to the subscripted factor (<math>A </math> → factor A, <math>B</math> → factor B, <math>AB</math> → interaction, and <math>E</math> → random error).

<math>SS_A = bn \Sigma^a_{i=1} (\bar{y}_i.. - \bar{y}...)^2 </math><br>
<math>SS_B = an \Sigma^b_{j=1} (\bar{y}._j. - \bar{y}...)^2 </math><br>
<math>SS_{AB} = n \Sigma^a_{i=1} \Sigma^b_{j=1} (\bar{y}_{ij}. - \bar{y}_{i}.. - \bar{y}._j. + \bar{y}...)^2 </math><br>
<math> SS_E = \Sigma^a_{i=1} \Sigma^b_{j=1} \Sigma^n_{k=1} (y_{ijk} - \bar{y}_{ij}.)^2 </math><br>

Before we can compare these quantities to each other, we must divide by the degrees of freedom to normalize. We calculate the normalized ''mean squares'' with:

<math> MS_A = \frac{SS_A}{a-1} </math><br>
<math> MS_B = \frac{SS_B}{b-1} </math><br>
<math> MS_{AB} = \frac{SS_{AB}}{(a-1)(b-1)} </math><br>
<math> MS_E = \frac{SS_E}{ab(n-1)} </math><br>

We test whether our factor and interaction effects are significant by dividing their mean squares by the mean square of our random error to get <math>F_0</math>. This value must be large for us to conclude that the effect we're testing is significant (the effect must be large relative to the random error). We pick a value from the F distribution based on <math>\alpha</math> and our degrees of freedom – this determines what "large" is. Do this in Excel with <tt>FINV(α,numerator's DOF,denominator's DOF)</tt> and in R with <tt>qf(1-alpha, numerator's DOF, denominator's DOF)</tt> (DOF = degrees of freedom).

<table cellspacing=0 cellpadding=5 style="border-top: 1px solid black; border-bottom: 1px solid black">
<tr align="center">
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Source of Variation </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Sum of Squares </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Degrees of Freedom </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Mean Square </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> <math>F_0</math> </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> <math>F_{\alpha}</math> </th>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Factor A (Temperature) </th>
<td> 0.301 </td>
<td> 2 </td>
<td> 0.151 </td>
<td> 8.469 </td>
<td> 4.256 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Factor B (Pressure) </th>
<td> 0.768 </td>
<td> 2 </td>
<td> 0.384 </td>
<td> 21.594 </td>
<td> 4.256 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Interaction </th>
<td> 0.069 </td>
<td> 4 </td>
<td> 0.017 </td>
<td> 0.969 </td>
<td> 3.633 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Error </th>
<td> 0.160 </td>
<td> 9 </td>
<td> 0.018 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee" style="border-top: 1px solid black;"> Total </th>
<td style="border-top: 1px solid black;"> 1.298 </td>
<td style="border-top: 1px solid black;"> 17 </td>
<td style="border-top: 1px solid black;"></td>
<td style="border-top: 1px solid black;"></td>
<td style="border-top: 1px solid black;"></td>
</tr>
</table>

We conclude from these tests that the effects from factors A and B (temperature and pressure) are significant, and there is no significant interaction effect. However, these tests are only accurate if the model is correct.

===Residual Plots and Model Adequacy Testing===
[[Image:5-1norm_prob.png|thumb|360px|left|'''Figure 3:''' These plots indicate that our data is sufficiently gaussian and we can believe the results of our hypothesis testing.]]

To test whether our data is sufficiently gaussian that our hypothesis testing is valid, we create a normal probability plot. As before, we calculate the residuals (by subtracting the expected values from the datapoints, <math>y_{ijk} - \bar{y}_{ij}.</math>) and sorting them in an array. Then we plot the sorted residuals against z-values, where the z-values are calculated in Excel by doing <tt>NORMSINV(percent)</tt> or in R with <tt>qnorm(percent)</tt>. <tt>Percent</tt> is equal to index_of_array/(DOF + 1). If the plot does not roughly fall along a straight line, the data is not from a Gaussian distribution.

There are several other useful plots we can make with residuals. Plotting <math>\bar{y}_{ij}.</math> against residuals produces a plot that should have data randomly scattered throughout its entire area, if not the data may not be gaussian. Plotting the residuals against either of the factors should indicate that the data is more or less equally random in each level of the factors. Note that because <math>n = 2</math>, there is some symmetry present in each of these plots.

Given that our data fits the model reasonably well, and that each of the factors are significant, the optimum conditions for producing high yield are when temperature is 170 °C and pressure is 200 psig.

File:4-1dataplot.png

2010-07-31T18:59:19Z

Statadmin:

Problem 4-1, Randomized blocks

2010-07-31T18:58:16Z

Statadmin: Created page with '==Problem Statement== ''A chemist wishes to test the effect of four chemical agents on the strength of a particular type of cloth. Because there might be variability from one bo…'

==Problem Statement==
''A chemist wishes to test the effect of four chemical agents on the strength of a particular type of cloth. Because there might be variability from one bolt to another, the chemist decides to use a randomized block design, with the bolts of cloth considered as blocks. She selects five bolts and applies all four chemicals in random order to each bolt. The resulting tensile strengths follow. Analyze data from this experiment (use <math>\alpha = 0.05</math>) and draw appropriate conclusions.

<table border=1 cellspacing=0 cellpadding=4>
<tr><th bgcolor="#eeeeee"></th><th colspan=5 bgcolor="#eeeeee">Bolt</th></tr>
<tr><th bgcolor="#eeeeee">Chemical</th><th bgcolor="#eeeeee">1</th><th bgcolor="#eeeeee">2</th><th bgcolor="#eeeeee">3</th><th bgcolor="#eeeeee">4</th><th bgcolor="#eeeeee">5</th></tr>
<tr><th bgcolor="#eeeeee">1</th><td>73</td><td>68</td><td>74</td><td>71</td><td>67</td></tr>
<tr><th bgcolor="#eeeeee">2</th><td>73</td><td>67</td><td>75</td><td>72</td><td>70</td></tr>
<tr><th bgcolor="#eeeeee">3</th><td>75</td><td>68</td><td>78</td><td>73</td><td>68</td></tr>
<tr><th bgcolor="#eeeeee">4</th><td>73</td><td>71</td><td>75</td><td>75</td><td>69</td></tr>
</table>

==Solution==
[[Image:4-1dataplot.png|thumb|360px|left|'''Figure 1:''' Illustration of data.]]

In this problem we need to compare <math>a = 4</math> treatments (chemical agents), but the data is being influenced by <math>b = 5</math> blocks (different bolts of fabric). Each datapoint can be thought of as the sum of several factors:<br>
<math> y_{ij} = \mu + \tau_i + \beta_j + \epsilon_{ij} </math><br>
Where:<br>
<math> i = 1,2,...,a </math><br>
<math> j = 1,2,...,b </math><br>
and:<br>
<math> y_{ij} = </math> datapoint from the <math>i</math>th treatment and the <math>j</math>th block<br>
<math> \mu = </math> the grand mean<br>
<math> \tau_i = </math> the effect of the <math>i</math>th treatment<br>
<math> \beta_j = </math> the effect of the <math>j</math>th block<br>
<math> \epsilon_{ij} = </math> a random error affecting the datapoint from the <math>i</math>th treatment and the <math>j</math>th block<br>

In the left plot of Figure 1, this equation is illustrated. The blue line marks the grand sample mean <math>\bar{y}.. \approx \mu</math>, and the red X marks the treatment mean <math>\bar{y}_i.</math> (mean of all data treated by the same chemical). The green number marks where the datapoint would be if the treatment effect and block effect were present, but there was no random error. Finally, the black number marks the location of the datapoint – the sum of the grand mean and each effect. The value of the numbers indicates which block that point was affected by.

In order to examine only the effect of the chemical agent, each bolt was divided into four pieces, each of which was treated by a different chemical agent. This serves to average out the effect of the different bolts. This is displayed in the left plot in Figure 1. The green lines deviate from each treatment mean identically, illustrating that the block effect is constant across chemical treatments.

The plot on the right in Figure 1 illustrates the data similarly, but the squares indicate the block means, and the treatment effects are shown deviating from them. This is merely an alternate visualization of the sum of the grand mean and each effect.

===Hypothesis Testing===

As usual, our null hypothesis will be that all the treatment means are equal (and come from the same distribution). Our alternative hypothesis is that they are not all equal:

<math>H_0\!:~\tau_1 = \tau_2 = ... = \tau_a = 0 </math><br>
<math>H_1\!:~\tau_i \ne 0 </math>

In order to test these hypotheses, we are interested in the sums of squares of all the error effects.

<math> SS_{Total} = \sum_{i=1}^a \sum_{j=1}^b y^2_{ij} - \frac{y^2\!\!..}{N} = 191.75</math><br>
<math> SS_{Treatments} = \frac{1}{b}\sum_{i=1}^a y^2_i\!. - \frac{y^2\!\!..}{N} = 12.95</math><br>
<math> SS_{Blocks} = \frac{1}{a} \sum_{j=1}^b y^2\!\!._j - \frac{y^2\!\!..}{N} = 157.00</math><br>
<math> SS_{Error} = SS_{Total} - SS_{Treatments} - SS_{Blocks} = 21.80</math>

Each of these is a measure of its respective type of error. Graphically, in the left plot of Figure 1 <math>SS_{Total}</math> is the sum of the squares of the distance along the x-axis from the grand mean to each of the black numbers. <math>SS_{Treatments}</math> is similar, but for the x-components of the red lines. For <math>SS_{Blocks}</math> we are summing the squares of the x-component of the green lines. And for <math>SS_{Error}</math> it is the x-component of the black lines that are squared and summed (it is easier to calculate this by arithmetic as in the equation above).

By dividing the statistics above by their respective degrees of freedom, we get a normalized measure of the error effects, or a ''mean sum of squares''.

<math>MS_{Treatments} = \frac{SS_{Treatments}}{a-1} = 4.32</math><br>
<math>MS_{Blocks} = \frac{SS_{Blocks}}{b-1} = 39.25</math><br>
<math>MS_{Error} = \frac{SS_{Error}}{(a-1)(b-1)} = 1.81</math>

In order to test our hypothesis, we must compare the treatment error to random error. We will do this with the F statistic, <math>F_0 = \frac{MS_{Treatments}}{MS_{Error}} = 2.38</math>. We compare this quantity to a <math>F_{\alpha,~a-1,~(a-1)(b-1)} = 3.49</math>, which comes from the F-distribution (this can be calculated in Excel with <tt>FINV(α,a−1,N−a)</tt>; in R use <tt>qf(1-alpha, a-1, (a-1)*(b-1))</tt>. If <math>F_{0} > F_{\alpha,~a-1,~(a-1)(b-1)}</math>, we reject the null hypothesis. As this is not the case, we cannot reject our null hypothesis.

<table cellspacing=0 cellpadding=5 style="border-top: 1px solid black; border-bottom: 1px solid black">
<tr align="center">
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Source of Variation </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Sum of Squares </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Degrees of Freedom </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Mean Square </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> <math>F_0</math> </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> <math>F_{\alpha}</math> </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> P-Value </th>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Treatments (Chemical agent) </th>
<td> 12.95 </td>
<td> 3 </td>
<td> 4.317 </td>
<td> 2.376 </td>
<td> 3.490 </td>
<td> 0.121 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Blocks (Bolts) </th>
<td> 157.00 </td>
<td> 4 </td>
<td> 39.250 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Error </th>
<td> 21.80 </td>
<td> 12 </td>
<td> 1.817 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee" style="border-top: 1px solid black;"> Total </th>
<td style="border-top: 1px solid black;"> 191.75 </td>
<td style="border-top: 1px solid black;"> 19 </td>
<td style="border-top: 1px solid black;"></td>
<td style="border-top: 1px solid black;"></td>
<td style="border-top: 1px solid black;"></td>
<td style="border-top: 1px solid black;"></td>
</tr>
</table>

Statistics Tutorial: Design and Analysis of Experiments

2010-07-31T18:57:51Z

Statadmin:

This is a tutorial for Dr. Brash's statistics class, at Christopher Newport University. Here, we will present detailed solutions to several problems from the first few chapters of ''Design and Analysis of Experiments'', 6th ed. by Douglas Montgomery. Solutions are available in Microsoft Excel format and [http://www.r-project.org R] format. Although beginners typically find Microsoft Excel easier to use, it proves to be very limiting for more advanced statistical analysis. R is a much more flexible and powerful software package for statistical analysis which is freely available, but has a somewhat steeper learning curve than Excel.

==Solutions==
[[Problem 2-1, Comparing a single mean to a specified value|Problem 2-1, Comparing a single mean to a specified value]]

[[Problem 2-2, Comparing a single mean to a specified value (second example)|Problem 2-2, Comparing a single mean to a specified value (second example)]]

[[Problem 2-4, Determining required sample size|Problem 2-4, Determining required sample size]]

[[Problems 3-1 through 3-3, Analysis of variance|Problems 3-1 through 3-3, Analysis of variance]]

[[Problem 4-1, Randomized blocks|Problem 4-1, Randomized blocks]]

[[Problem 5-1, Factorial designs|Problem 5-1, Factorial designs]]

File:3-1fake4.png

2010-07-31T18:39:51Z

Statadmin: Confidence interval on the differences in means.

Confidence interval on the differences in means.

File:3-1fake3.png

2010-07-31T18:39:14Z

Statadmin: Confidence interval on the mean tensile strength for each mixing technique.

Confidence interval on the mean tensile strength for each mixing technique.

File:3-1e.png

2010-07-31T18:23:51Z

Statadmin: Residuals vs. predicted tensile strength.

Residuals vs. predicted tensile strength.

File:Npp.png

2010-07-31T18:20:51Z

Statadmin: Normal probability plot.

Normal probability plot.

File:3-1b.png

2010-07-31T18:19:58Z

Statadmin: Comparison of data to a T distribution.

Comparison of data to a T distribution.

File:3-1error comparison.png

2010-07-31T18:18:05Z

Statadmin: Comparison of hypothetical data sets for which most of the error is between the treatment means (left) and within the treatment means (right).

Comparison of hypothetical data sets for which most of the error is between the treatment means (left) and within the treatment means (right).

File:3-1dataplot.png

2010-07-31T18:17:19Z

Statadmin: Illustration of data.

Illustration of data.

Problems 3-1 through 3-3, Analysis of variance

2010-07-31T18:11:59Z

Statadmin: Created page with 'We will provide solutions for problems 3-1, 3-2 and 3-3 in this tutorial. ==Problem statement== ===Problem 3-1=== ''The tensile strength of portland cement is being studied. Fou…'

We will provide solutions for problems 3-1, 3-2 and 3-3 in this tutorial.

==Problem statement==
===Problem 3-1===
''The tensile strength of portland cement is being studied. Four different mixing techniques can be used economically. A completely randomized experiment was conducted and the following data collected.''

<table border=1 cellspacing=0 cellpadding=4>
<tr><th bgcolor="#eeeeee">Mixing<br>Technique</th><th colspan=4 bgcolor="#eeeeee">Tensile Strength (lb/in<sup>2</sup>)</th></tr>
<tr><td bgcolor="#eeeeee">1</td><td>3129</td><td>3000</td><td>2865</td><td>2890</td></tr>
<tr><td bgcolor="#eeeeee">2</td><td>3200</td><td>3300</td><td>2975</td><td>3150</td></tr>
<tr><td bgcolor="#eeeeee">3</td><td>2800</td><td>2900</td><td>2985</td><td>3050</td></tr>
<tr><td bgcolor="#eeeeee">4</td><td>2600</td><td>2700</td><td>2600</td><td>2765</td></tr>
</table>

<ol style="list-style-type:lower-latin">
<li>''Test the hypothesis that mixing techniques affect the strength of the cement. Use α = 0.05.''</li>
<li>''Construct a graphical display as described in section 3-5.3 to compare the mean tensile strengths for the four mixing techniques. What are your conclusions?''</li>
<li>''Use the Fisher LSD method with α = 0.05 to make comparisons between pairs of means.''</li>
<li>''Construct a normal probability plot of the residuals. What conclusion would you draw about the validity of the normality assumption?''</li>
<li>''Plot the residuals vs. the predicted tensile strength. Comment on the plot.''</li>
<li>''Prepare a scatter plot of the results to aid the interpretation of the results of this experiment.''</li>
</ol>

===Problem 3-2===
<ol style="list-style-type:lower-latin">
<li>''Rework part (b) of problem 3-1 using Tukey's test with α = 0.05. Do you get the same conclusions from Tukey's test that you did from the graphical procedure and/or the Fisher LSD method?''</li>
<li>''Explain the difference between the Tukey and Fisher procedures.''</li>
</ol>

===Problem 3-3===
''Reconsider the experiment in problem 3-1. Find a 95% confidence interval on the mean tensile strength of the portland cement produced by each of the four mixing techniques. Also find a 95% confidence interval on the difference in means for techniques 1 and 3. Does this aid you in interpreting the results of the experiment?''

==Solution==
[[Image:3-1dataplot.png|thumb|left|'''Figure 1:''' Illustration of data.]]

In these problems, we are given a data set with <math>a=4</math> subsets, each containing <math>n=4</math> values for a total of <math>N=a n=16</math> data points. Each subset contains measurements of tensile strength of cement samples that were produced with a different mixing technique, or '''treatment'''. We define the mean over all data points as the '''grand mean''', and the mean of each point within a given treatment as the '''treatment mean'''. To compare these data subsets, it is useful to think of each data point as the sum of the grand mean <math>\mu</math>, the ith treatment mean <math>\tau_i</math>, and a random error ϵ<sub>ij</sub> specific to the jth data point in the ith treatment (refer to figure 1).

<center><math>y_{ij}=\mu+\tau_i+\epsilon_{ij}\begin{cases}i=1,2,\ldots,a\\j=1,2,\ldots,n\end{cases}</math></center>

Since we are given a finite set of data, we must approximate these means by calculating sample means. The grand sample mean is given by:

<center><math>\bar{y}..=\frac{1}{N}\sum_{i=1}^a \sum_{j=1}^n y_{ij} = 2932</math></center>

The sample treatment means are given by:

<center><math>\bar{y}_i.=\frac{1}{n}\sum_{j=1}^n y_{ij}</math> where <math>i=1, 2, \ldots, a</math></center>

The dot indicates that you are summing over the variable it replaces.

===Section 3-1 (A): Hypothesis testing===
[[Image:3-1error_comparison.png|thumb|360px|left|'''Figure 2:''' Comparison of hypothetical data sets for which most of the error is between the treatment means (left) and within the treatment means (right).]]
We would like to know if one of our data subsets is significantly different from the others, as this may indicate that one of our manufacturing techniques is superior (or inferior) to the others. To compare our data subsets we are interested in whether most of the error is within the treatments (ϵ) or between the treatments (<math>\tau</math>). If most of the error is between the treatment means, then we can claim there are significant differences between them. If there is too much error within the treatment means we cannot claim that they are significantly different (see figure x). Mathematically, we can approximate the error between the treatment means as

<center><math>\mathrm{MS_{Treatments}}=\frac{n \sum_{i=1}^a (\bar{y}_i.-\bar{y}..)^2}{a-1}=163,247</math></center>

To approximate the error within the treatment means, it is easiest to subtract the error between the means from the total error:

<center><math>\mathrm{MS_{Error}}=\frac{\sum_{i=1}^a \sum_{j=1}^n (y_{ij}-\bar{y}..)^2-n \sum_{i=1}^a (\bar{y}_i.-\bar{y}..)^2}{N-a}</math><math>=\frac{\sum_{i=1}^a \sum_{j=1}^n y_{ij}^2-\frac{1}{n}\sum_{i=1}^{a} y_i.^2}{N-a}=12,826</math></center>

We are interested in the ratio

<center><math>F_0=\frac{\mathrm{MS_{Treatments}}}{\mathrm{MS_{Error}}}=12.7</math></center>

To determine whether or not there are significant differences between our treatments, we will compare F<sub>0</sub> to <math>F_{\alpha,\,a-1,\,N-a}=3.5</math> from the F distribution. In Excel this value can be found using the function <tt>FINV(α,a−1,N−a)</tt>; in R it can be found using <tt>qf(1−α,a−1,N−a)</tt>. If <math>F_0>F_{\alpha,\,a-1,\,N-a}</math>, which is the case here, then the error between treatment means is large enough compared to the error within treatment means to conclude that there is a significant difference between at least one treatment and the others.

<table cellspacing=0 cellpadding=5 style="border-top: 1px solid black; border-bottom: 1px solid black">
<tr align="center">
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Source of Variation </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Sum of Squares </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Degrees of Freedom </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> Mean Square </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> <math>F_0</math> </th>
<th bgcolor="#eeeeee" style="border-bottom: 1px solid black;"> <math>F_{\alpha}</math> </th>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Mixing Technique </th>
<td> 489740 </td>
<td> 3 </td>
<td> 163247 </td>
<td> 12.728 </td>
<td> 3.490 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee"> Error </th>
<td> 153908 </td>
<td> 12 </td>
<td> 12826 </td>
</tr>
<tr align="center">
<th bgcolor="#eeeeee" style="border-top: 1px solid black;"> Total </th>
<td style="border-top: 1px solid black;"> 643648 </td>
<td style="border-top: 1px solid black;"> 15 </td>
<td style="border-top: 1px solid black;"></td>
<td style="border-top: 1px solid black;"></td>
<td style="border-top: 1px solid black;"></td>
</tr>
</table>

===Section 3-1 (B): Graphical display to compare mean tensile strengths===
[[Image:3-1b.png|thumb|left|Comparison of data to a T distribution.]]
A relatively simple way to visualize the treatment means, and see whether or not they are statistically equal (qualitatively), is to simply plot the four averages on the same graph as a T distribution. We need to know what mean and standard deviation to use for our T distribution. For the mean, we will simply use the grand mean, and we will approximate the standard deviation with <math>\sqrt{\mathrm{MS_{Error}}/n}=\sqrt{12826/4}=56.6</math>. This approximation for the standard deviation relies on <math>\mathrm{MS_{Error}}</math>, which does not take into account the differences between the treatment means. It assumes that the treatment means are all equal – if they are not statistically equivalent, it will be obvious when we plot the treatment means on the same plot as the distribution.

Looking at the plot on the left, we see that for our data it is unlikely that all of the treatment means come from the plotted distribution. The two treatment means under the tails of the T distribution appear to be significantly different from those under the center.

===Section 3-1 (C): Fisher LSD comparisons===
Fisher LSD comparisons allow each pair of treatment means to be compared. This is done using a t-test as we did in problem 2.11 B (solution: [http://www.jlab.org/~pcarter/stats/2-11.xlsx Excel], [http://www.jlab.org/~pcarter/stats/2-11.R R]), but replacing <math>S_p</math> with <math>\sqrt{\mathrm{MS_E}}</math>:

<center><math>t_0=\frac{\bar{y}_i.-\bar{y}_j.}{\sqrt{\frac{2\mathrm{MS_E}}{n}}}</math></center>

Solving for <math>\bar{y}_{i.} - \bar{y}_{j.}</math> yields:

<center><math>\bar{y}_i.-\bar{y}_j.=t_0\sqrt{\frac{2\mathrm{MS_E}}{n}}</math></center>

We will compare this to a theoretical value called the least significant difference:

<center><math>\mathrm{LSD}=t_{\alpha/2,~N-a} \sqrt{\frac{2\mathrm{MS_E}}{n}}=174.5</math></center>

If <math>|\bar{y}_i-\bar{y}_j| > \mathrm{LSD}</math> then the treatment means <math>i</math> and <math>j</math> are significantly different.

In Excel, you can calculate LSD using <tt>TINV(α,N-a)*sqrt(2*MSe/n)</tt>, where <tt>MSe</tt> is <math>\mathrm{MS_{Error}}</math>. In R, the equivalent command is <tt>qt(1-α/2,N-a)*sqrt(2*MSe/n)</tt>.

The following table shows the differences between each pair of treatment means. Differences highlighted in <span style="background-color: #ddddff">blue</span> are large enough for that pair to be considered significantly different.

<table border=1 cellspacing=0 cellpadding=4>
<tr><th colspan=4 align=center><math>|\bar{y}_i-\bar{y}_j|</math></th></tr>
<tr><th bgcolor="#eeeeee"></th><th bgcolor="#eeeeee">2</th><th bgcolor="#eeeeee">3</th><th bgcolor="#eeeeee">4</th></tr>
<tr><th bgcolor="#eeeeee">1</th><td align='right' bgcolor="#ddddff">185.25</td><td align='right'>37.25</td><td align='right' bgcolor="#ddddff">304.75</td></tr>
<tr><th bgcolor="#eeeeee">2</th><td align='right'></td><td align='right' bgcolor="#ddddff">222.50</td><td align='right' bgcolor="#ddddff">490.00</td></tr>
<tr><th bgcolor="#eeeeee">3</th><td align='right'></td><td align='right'></td><td align='right' bgcolor="#ddddff">267.50</td></tr>
</table>

To apply the Fisher LSD method to our data, we will compare LSD to each of the numbers in the chart to the left. For example, we see that 185.25 > 174.5, so there is a statistically significant difference between treatment 1 and 2.

===Section 3-1 (D): Normal probability plot===
[[Image:npp.png|thumb|left|Normal probability plot.]]

We have been assuming that our data is distributed normally (on a Gaussian), and that it is therefore valid to do t-tests. To be sure, we should check our normality assumption by creating a normal probability plot. This is done by plotting the residuals against values from a z-distribution. Residuals are calculated by subtracting the corresponding treatment mean from each data point, and must be sorted before using them to make the plot. The values we seek from a z-distribution are obtained by doing <tt>NORMSINV(percent)</tt> in Excel, or <tt>qnorm(percent)</tt> in R. In these commands, <tt>percent</tt> is a number from 1/(dof+1) to dof/(dof+1) where <tt>dof</tt> is the degrees of freedom. These commands return z-distribution values that represent ideal residual values. If the resulting plot is roughly linear, then the normality assumption is valid.
<br clear="all" />

===Section 3-1 (E): Plot of residuals vs. predicted tensile strength===
[[Image:3-1e.png|thumb|left|Residuals vs. predicted tensile strength.]]
As an estimate of the tensile strength for each treatment, we use the treatment mean. The plot of residuals vs. their treatment means gives an indication of the relative sizes of errors between (x-axis) and within (y-axis) treatments.

<br clear="all" />

===Section 3-1 (F): Plot of all data===
[[Image:3-1dataplot.png|Plot of all data.]]

===Section 3-2 (A): Tukey test===
Tukey's test is similar to Fisher LSD comparisons in that they allow pairs of treatment means to be compared. However, instead of using the t-statistic, Tukey's test uses the Studentized range statistic q:

<center><math>q=\frac{\bar{y}_{max}-\bar{y}_{min}}{\sqrt{\mathrm{MS_{Error}/n}}}</math></center>

Solving for <math>\bar{y}_{max}-\bar{y}_{min}</math> yields:

<center><math>\bar{y}_{max}-\bar{y}_{min} = q\sqrt{\mathrm{MS_{Error}}/n}</math></center>

We will compare this with the theoretical value:

<center><math>T_\alpha = q_\alpha(a,~f) \sqrt{\mathrm{MS_{Error}}/n}=4.2 \sqrt{12826/4}=237.75</math></center>

This can be calculated in R using <tt>qtukey(1-alpha,a,N-a)*sqrt(MSe/n)</tt> where <tt>MSe</tt> is <math>\mathrm{MS_{Error}}</math>. If <math>|\bar{y}_i - \bar{y}_j| > T_\alpha</math>, there is a significant difference between the two treatments.

We now compare this statistic to the differences between the treatment means. Differences highlighted in <span style="background-color: #ddddff">blue</span> are large enough for that pair to be considered significantly different.

<table border=1 cellspacing=0 cellpadding=4>
<tr><th colspan=4 align=center><math>|\bar{y}_i-\bar{y}_j|</math></th></tr>
<tr><th bgcolor="#eeeeee"></th><th bgcolor="#eeeeee">2</th><th bgcolor="#eeeeee">3</th><th bgcolor="#eeeeee">4</th></tr>
<tr><th bgcolor="#eeeeee">1</th><td align='right'>185.25</td><td align='right'>37.25</td><td align='right' bgcolor="#ddddff">304.75</td></tr>
<tr><th bgcolor="#eeeeee">2</th><td align='right'></td><td align='right'>222.50</td><td align='right' bgcolor="#ddddff">490.00</td></tr>
<tr><th bgcolor="#eeeeee">3</th><td align='right'></td><td align='right'></td><td align='right' bgcolor="#ddddff">267.50</td></tr>
</table>

===Section 3-2 (B): Difference between Tukey and Fisher procedures===
The Fisher procedure uses the T-statistic to compare pairs of treatment means, while the Tukey test uses the Studentized range statistic. One consequence of this is that the Fisher procedure controls the error rate <math>\alpha</math> for each individual pairwise comparison, whereas the Tukey test controls the overall error rate.

===Section 3-3: Confidence intervals===
<div style="float:left; vertical-align: top; padding-right: 20px; padding-bottom: 20px;">
[[Image:3-1fake3.png|thumb|none|'''Figure 7:''' Confidence interval on the mean tensile strength for each mixing technique.]]
<br>
[[Image:3-1fake4.png|thumb|none|'''Figure 8:''' Confidence interval on the differences in means.]]</div>
We want to find a 95% confidence interval on the mean tensile strength for each mixing technique. The upper bound of each confidence interval is the treatment mean plus the least significant difference, <math>\bar{y}_i.+\mathrm{LSD}</math>. The lower bound is <math>\bar{y}_i.-\mathrm{LSD}</math>. LSD was calculated for <math>\alpha</math>=0.95, so this gives us a 95% confidence interval (see figure 7).

<center><math>\mathrm{LSD}=t_{\alpha/2,~N-a} \sqrt{\frac{2\mathrm{MS_E}}{n}}=174.5</math></center>

<table border=1 cellspacing=0 cellpadding=4>
<tr><th bgcolor="#eeeeee"></th><th bgcolor="#eeeeee">lower bound</th><th bgcolor="#eeeeee">treatment mean</th><th bgcolor="#eeeeee">upper bound</th></tr>
<tr><th bgcolor="#eeeeee">Treatment 1</th><td align='right'>2848</td><td align='right'>2971</td><td align='right'>3094</td></tr>
<tr><th bgcolor="#eeeeee">Treatment 2</th><td align='right'>3033</td><td align='right'>3156</td><td align='right'>3280</td></tr>
<tr><th bgcolor="#eeeeee">Treatment 3</th><td align='right'>2810</td><td align='right'>2933</td><td align='right'>3057</td></tr>
<tr><th bgcolor="#eeeeee">Treatment 4</th><td align='right'>2543</td><td align='right'>2666</td><td align='right'>2790</td></tr>
</table>
To find the confidence interval on the differences in means, we simply subtract to get the difference between our treatment means, and then use the formula above to calculate the confidence interval (see figure 8).

<table border=1 cellspacing=0 cellpadding=4>
<tr><th bgcolor="#eeeeee"></th><th bgcolor="#eeeeee">lower bound</th><th bgcolor="#eeeeee"><math>\bar{y}_{i.} - \bar{y}_{j.}</math></th><th bgcolor="#eeeeee">upper bound</th></tr>
<tr><th bgcolor="#eeeeee">Treatment 1 - 2</th><td align='right'>-359</td><td align='right'>-185</td><td align='right'>-10</td></tr>
<tr><th bgcolor="#eeeeee">Treatment 1 - 3</th><td align='right'>-137</td><td align='right'>37</td><td align='right'>211</td></tr>
<tr><th bgcolor="#eeeeee">Treatment 1 - 4</th><td align='right'>130</td><td align='right'>304</td><td align='right'>479</td></tr>
<tr><th bgcolor="#eeeeee">Treatment 2 - 3</th><td align='right'>48</td><td align='right'>222</td><td align='right'>396</td></tr>
<tr><th bgcolor="#eeeeee">Treatment 2 - 4</th><td align='right'>315</td><td align='right'>490</td><td align='right'>664</td></tr>
<tr><th bgcolor="#eeeeee">Treatment 3 - 4</th><td align='right'>93</td><td align='right'>267</td><td align='right'>441</td></tr>
</table>

Statistics Tutorial: Design and Analysis of Experiments

2010-07-31T18:10:51Z

Statadmin:

This is a tutorial for Dr. Brash's statistics class, at Christopher Newport University. Here, we will present detailed solutions to several problems from the first few chapters of ''Design and Analysis of Experiments'', 6th ed. by Douglas Montgomery. Solutions are available in Microsoft Excel format and [http://www.r-project.org R] format. Although beginners typically find Microsoft Excel easier to use, it proves to be very limiting for more advanced statistical analysis. R is a much more flexible and powerful software package for statistical analysis which is freely available, but has a somewhat steeper learning curve than Excel.

==Solutions==
[[Problem 2-1, Comparing a single mean to a specified value|Problem 2-1, Comparing a single mean to a specified value]]

[[Problem 2-2, Comparing a single mean to a specified value (second example)|Problem 2-2, Comparing a single mean to a specified value (second example)]]

[[Problem 2-4, Determining required sample size|Problem 2-4, Determining required sample size]]

[[Problems 3-1 through 3-3, Analysis of variance|Problems 3-1 through 3-3, Analysis of variance]]

[[User:Pcarter/4-1|Problem 4-1]]

[[User:Pcarter/5-1|Problem 5-1]]

File:2-4fig1.png

2010-07-31T16:52:45Z

Statadmin: Confidence interval of the sample mean.

Confidence interval of the sample mean.

Problem 2-4, Determining required sample size

2010-07-31T16:52:17Z

Statadmin: Created page with '===Problem Statement=== ''A normally distributed random variable has an unknown mean <math>\mu</math> and a known variance <math>\sigma^2=9</math>. Find the sample size required…'

===Problem Statement===

''A normally distributed random variable has an unknown mean <math>\mu</math> and a known variance <math>\sigma^2=9</math>. Find the sample size required to contruct a 95 percent confidence interval on the mean that has a total length of 1.0.''

===Solution===
[[Image:2-4fig1.png|thumb|'''Figure 1:''' Confidence interval of the sample mean.]]

The problem is asking us how much data we need in order to say with 95% certainty that the mean <math>\mu</math> is within a range of length one. We know from previous problems how to calculate the boundaries of a confidence interval for the sample mean, and we are told the total length of the confidence interval should be one. So we begin by subtracting the equation for the left edge of the confidence interval from the equation for the right edge, and setting this equal to one:

<math>(\mu + \frac{z_{\alpha/2}\sigma}{\sqrt{N}} ) - (\mu - \frac{z_{\alpha/2}\sigma}{\sqrt{N}}) = 1</math>

Simplifying and solving for N, we get:

<math>N = (2 z_{\alpha/2} \sigma)^2</math>

We can plug in for <math>\sigma</math> with the value given in the problem statement, and we can solve for <math>z_{alpha/2}</math> with <tt>NORMSINV(1-0.05)</tt> in Excel, or <tt>qnorm(1-0.05/2)</tt> in R (we assumed an <math>\alpha</math> of 0.05).

<math>N = (2 \cdot 1.96 \cdot 3)^2 = 138.29</math>

We can't take a .29th of a datapoint, so we round up to <math>N = 139</math>.

Statistics Tutorial: Design and Analysis of Experiments

2010-07-31T16:51:50Z

Statadmin:

This is a tutorial for Dr. Brash's statistics class, at Christopher Newport University. Here, we will present detailed solutions to several problems from the first few chapters of ''Design and Analysis of Experiments'', 6th ed. by Douglas Montgomery. Solutions are available in Microsoft Excel format and [http://www.r-project.org R] format. Although beginners typically find Microsoft Excel easier to use, it proves to be very limiting for more advanced statistical analysis. R is a much more flexible and powerful software package for statistical analysis which is freely available, but has a somewhat steeper learning curve than Excel.

==Solutions==
[[Problem 2-1, Comparing a single mean to a specified value|Problem 2-1, Comparing a single mean to a specified value]]

[[Problem 2-2, Comparing a single mean to a specified value (second example)|Problem 2-2, Comparing a single mean to a specified value (second example)]]

[[Problem 2-4, Determining required sample size|Problem 2-4, Determining required sample size]]

[[User:Pcarter/3-1 through 3-3|3-1 through 3-3]]

[[User:Pcarter/4-1|Problem 4-1]]

[[User:Pcarter/5-1|Problem 5-1]]

File:2-2fig5.png

2010-07-31T16:46:30Z

Statadmin: The confidence interval about the theoretical mean.

The confidence interval about the theoretical mean.

File:2-2fig4.png

2010-07-31T16:45:22Z

Statadmin: The confidence interval about the sample mean.

The confidence interval about the sample mean.

File:2-2fig2.png

2010-07-31T16:44:08Z

Statadmin: Our plot after normalizing.

Our plot after normalizing.

File:2-2fig1.png

2010-07-31T16:43:33Z

Statadmin: Our data compared to a theoretical Gaussian distribution.

Our data compared to a theoretical Gaussian distribution.

Problem 2-2, Comparing a single mean to a specified value (second example)

2010-07-31T16:43:07Z

Statadmin: Created page with '==Problem Statement== ''The viscosity of a liquid detergent is supposed to average 800 centistokes at 25 °C. A random sample of 16 batches of detergent is collected, and the ave…'

==Problem Statement==
''The viscosity of a liquid detergent is supposed to average 800 centistokes at 25 °C. A random sample of 16 batches of detergent is collected, and the average viscosity is 812. Suppose we know that the standard deviation of viscosity is σ = 25 centistokes.''

<ol style="list-style-type:lower-latin">
<li>''State the hypotheses that should be tested.''</li>
<li>''Test these hypotheses using α = 0.05. What are your conclusions?''</li>
<li>''What is the P-value for the test?''</li>
<li>''Find a 95 percent confidence interval on the mean.''</li>
</ol>

==Solution==
[[Image:2-2fig1.png|thumb|left|'''Figure 1:''' Our data compared to a theoretical Gaussian distribution.]]

===Section A: Choosing hypotheses===

In this problem we are told we would like our liquid detergent to have a mean of <math>\mu_0 = 800</math> and a standard deviation of <math>\sigma = 25</math>: this is our theoretical distribution. We are also told that a sample of <math>N = 16</math> batches of detergent have an average viscosity of <math>\overline{y} = 812</math>, which estimates the mean of our true distribution. We would like to know if the means of our true and theoretical distributions are likely to be the same. There are two hypotheses to consider here. Our null hypothesis is that the means of the two distributions are equal, and our alternative hypothesis is that they are not equal.

<center>H<sub>0</sub>: μ = μ<sub>0</sub><br />
H<sub>1</sub>: μ ≠ μ<sub>0</sub></center>

The alternate hypothesis is called ''two-tailed'' because it is true if <math>\mu < \mu_0</math> and if <math>\mu > \mu_0</math>.

<br clear='all'>
[[Image:2-2fig2.png|thumb|left|'''Figure 2:''' Our plot after normalizing.]]

===Section B: Z-values===

To compare the mean of the true distribution to that of the theoretical distribution, we test the null hypothesis with a z-test. The z-value is calculated as in problem 2-1:

<center><math>z=\frac{\bar{y}-\mu_0}{\sigma/\sqrt{n}}=1.92</math><br/></center>

Since we have a two-tailed alternative hypothesis, we must define a rejection region at both extremes of our theoretical distribution. Our value for <math>\alpha</math> determines the total size of the rejection region, so we simply declare that 2.5% (since <math>\alpha</math> = .05, or 5%) of the area on the left of our theoretical distribution is a rejection region, and 2.5% of the area on the right is also a rejection region (see Figure 2). We calculate <math>z_{\alpha/2}</math> to determine the x-value that corresponds to the rightmost edge of the rejection region on the left, and <math>z_{1-\alpha/2}</math> to find the leftmost edge of the rejection region on the right (see Figure 2 again).

<center><math>z_{\alpha/2}=-1.96</math>, <math>z_{1-\alpha/2} = 1.96</math></center>

If z is between <math>{z_\alpha}</math> and <math>z_{1-\alpha/2}</math>, it is not in the rejection region and we claim the null hypothesis to be true (with a confidence of 95%). Otherwise we claim the alternative hypothesis to be true (with the same confidence).

Note that <math>|z_{\alpha/2}| = z_{1-\alpha/2}</math>. This is always true. You can simply compare your z-value to <math>|z_{\alpha/2}|</math> to perform a z-test. If <math>z < |z_{\alpha/2}|</math>, claim your null hypothesis to be true, otherwise claim that your alternative hypothesis is true.

===Section C: P-values===
We now calculate a P-value the same way we did in problem 2-1. Graphically, we extend the rejection region inwards from both tails until we run into our z-value. The P-value is the area of the shaded area, calculated in Excel with <tt>=2*NORMSDIST(-ABS(z))</tt> or in R with <tt>2*pnorm(-abs(z))</tt>. These functions integrate a normal distribution from negative infinity to the number we give it (in this case the negative absolute value of z), so we give it our negative z-value and multiply by two in order to get the total area of the shaded regions on the graph.

<center>P-value <math>= 0.0549</math></center>

Our z-value is very close to the rejection region, so a plot illustrating the extended rejection regions will look nearly identical to Figure 2.

<div style="float:left; vertical-align: top">[[Image:2-2fig4.png|thumb|left|'''Figure 4:''' The confidence interval about the sample mean.]]<br>
[[Image:2-2fig5.png|thumb|left|'''Figure 5:''' The confidence interval about the theoretical mean.]]</div>

===Section D: Confidence intervals===
To calculate the limits of the confidence interval for the sample mean, we use the following formula:
<center><math>\bar{y}</math> confidence interval limits = <math>\mu_0\pm z_{\alpha/2} \sigma/\sqrt{n}</math></center>

This tells us the range in which a sample mean could lie in order for us to accept our null hypothesis:
<center><math>787.75<\bar{y}<812.25</math></center>

We can calculate the confidence interval for <math>\mu_0</math> in a similar way, which tells us the range in which the mean of our theoretical distribution (given a sample mean of 812) could lie in order for us to accept our null hypothesis:
<center><math>\mu_0</math> confidence interval limits = <math>\bar{y}\pm z_{\alpha/2} \sigma/\sqrt{n}</math><br />
<math>799.75<\mu_0<824.25</math></center>

Statistics Tutorial: Design and Analysis of Experiments

2010-07-31T16:42:21Z

Statadmin:

This is a tutorial for Dr. Brash's statistics class, at Christopher Newport University. Here, we will present detailed solutions to several problems from the first few chapters of ''Design and Analysis of Experiments'', 6th ed. by Douglas Montgomery. Solutions are available in Microsoft Excel format and [http://www.r-project.org R] format. Although beginners typically find Microsoft Excel easier to use, it proves to be very limiting for more advanced statistical analysis. R is a much more flexible and powerful software package for statistical analysis which is freely available, but has a somewhat steeper learning curve than Excel.

==Solutions==
[[Problem 2-1, Comparing a single mean to a specified value|Problem 2-1, Comparing a single mean to a specified value]]

[[Problem 2-2, Comparing a single mean to a specified value (second example)|Problem 2-2, Comparing a single mean to a specified value (second example)]]

[[User:Pcarter/2-4|Problem 2-4]]

[[User:Pcarter/3-1 through 3-3|3-1 through 3-3]]

[[User:Pcarter/4-1|Problem 4-1]]

[[User:Pcarter/5-1|Problem 5-1]]

Statistics Tutorial: Design and Analysis of Experiments

2010-07-31T16:40:59Z

Statadmin:

This is a tutorial for Dr. Brash's statistics class, at Christopher Newport University. Here, we will present detailed solutions to several problems from the first few chapters of ''Design and Analysis of Experiments'', 6th ed. by Douglas Montgomery. Solutions are available in Microsoft Excel format and [http://www.r-project.org R] format. Although beginners typically find Microsoft Excel easier to use, it proves to be very limiting for more advanced statistical analysis. R is a much more flexible and powerful software package for statistical analysis which is freely available, but has a somewhat steeper learning curve than Excel.

==Solutions==
[[Problem 2-1, Comparing a single mean to a specified value|Problem 2-1]]

[[Problem 2-2], Comparing a single mean to a specified value (second example)|Problem 2-2]]

[[User:Pcarter/2-4|Problem 2-4]]

[[User:Pcarter/3-1 through 3-3|3-1 through 3-3]]

[[User:Pcarter/4-1|Problem 4-1]]

[[User:Pcarter/5-1|Problem 5-1]]

File:Gaussian5.png

2010-07-31T16:09:16Z

Statadmin: The confidence interval about the theoretical mean.

The confidence interval about the theoretical mean.

File:Gaussian4.png

2010-07-31T16:08:40Z

Statadmin: The confidence interval about the sample mean.

The confidence interval about the sample mean.

File:Gaussian3.png

2010-07-31T16:07:40Z

Statadmin: Illustrating the P-value.

Illustrating the P-value.

File:Gaussian2.png

2010-07-31T16:06:19Z

Statadmin: Our plot after normalizing.

Our plot after normalizing.

File:Gaussian1.png

2010-07-31T16:04:59Z

Statadmin: Our data compared to a theoretical Gaussian distribution.

Our data compared to a theoretical Gaussian distribution.

Problem 2-1, Comparing a single mean to a specified value

2010-07-31T15:57:26Z

Statadmin: Created page with '==Problem statement== ''The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that the standard deviation of breaking strength is σ …'

==Problem statement==
''The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that the standard deviation of breaking strength is σ = 3 psi. A random sample of four speciments is tested, and the results are y<sub>1</sub> = 145, y<sub>2</sub> = 153, y<sub>3</sub> = 150, and y<sub>4</sub> = 147.''

<ol style="list-style-type:lower-latin">
<li>''State the hypotheses that you think should be tested in this experiment.''</li>
<li>''Test these hypothesis using α = 0.05. What are your conclusions?''</li>
<li>''Find the ''P''-value for the test in part (b).''</li>
<li>''Construct a 95 percent confidence interval on the mean breaking strength.''</li>
</ol>

==Solution==
[[Image:Gaussian1.png|thumb|left|'''Figure 1:''' Our data compared to a theoretical Gaussian distribution.]]

===Section A: Choosing hypotheses===
In this problem we are given a set of four data points. These data points all come from a distribution of breaking strengths which has an unknown mean μ. We will call this the ''true distribution''. Previous experience indicates that breaking strengths follow a Gaussian ''theoretical distribution'' with a standard deviation of 3 psi, so we assume this for our distribution also. Our task is to determine whether or not the true mean, which is impossible to know exactly, is greater than or equal to 150 psi. We plot this data and the distribution in figure 1.

Since the sample mean is an approximation of the true mean, we define the ''standard error of the mean'' (SEM) to be <math>\sigma/\sqrt{n}=1.5</math>, where n=4 is the number of data points.

<table border=1 cellspacing=0 cellpadding=4>
<tr><th bgcolor="#eeeeee"></th><th bgcolor="#eeeeee">Distribution type</th><th bgcolor="#eeeeee">Mean</th><th bgcolor="#eeeeee">Standard deviation</th></tr>
<tr><th bgcolor="#eeeeee">True distribution</th><td>Normal</td><td><math>\mu\approx\overline{y}=148.25</math></td><td><math>\sigma=3</math></td></tr>
<tr><th bgcolor="#eeeeee">Theoretical distribution</th><td>Normal</td><td><math>\mu_0=150</math></td><td><math>\sigma_0=3</math></td></tr>
</table>

We first state two hypotheses. The null hypothesis is that our data does come from the theoretical distribution: the true mean μ = μ<sub>0</sub>. Our alternative hypothesis states that the data comes from a distribution centered around a different mean.

There are three choices for the alternative hypothesis: μ < 150, μ > 150, and μ ≠ 150. We adopt the convention that the alternative hypothesis will be true if the data does not meet the requirements. In this case, the breaking strength of the fiber is required to be at least 150 psi, so we choose μ < 150 as our alternative hypothesis.

Formally, we state our hypotheses as:<br/>
<center>H<sub>0</sub>: μ = 150<br/>
H<sub>1</sub>: μ < 150</center>

<br clear='all'>
[[Image:Gaussian2.png|thumb|left|'''Figure 2:''' Our plot after normalizing.]]

===Section B: Z-values===
For convenience, we start by standardizing our theoretical distribution to have a mean of zero and a standard deviation of one. To do this, we first center the distribution around zero by subtracting the theoretical mean (150) from each point in the distribution. We then divide each point by the standard deviation (3). The sample mean can be standardized in the same manner. We plot the normalized distribution and sample mean in figure 2.

We now assume that the null hypothesis is true and ask whether or not this assumption makes sense. Given that this assumption is true, the sample mean is most likely to be close to zero. To test this, we define a range over which we consider our sample mean to be unacceptable, the ''rejection region''. If the sample mean is in the rejection region, it is too far from zero and we reject the null hypothesis.

We will define the lower limit to be z<sub>α</sub>, where α=0.05. Graphically, given a standard Gaussian distribution, the area under the curve left of z<sub>0.05</sub> is equal to 5% of the total area. You can either look up z<sub>α</sub> in a table or calculate it using a software package. Using Excel, the appropriate function is <tt>=NORMSINV(alpha)</tt>. The corresponding function in R is <tt>qnorm(alpha)</tt>. Using one of these methods, we find that z<sub>0.05</sub>=−1.64.

We then find the z-value of our data and compare the z-value to z<sub>α</sub>. The formula for the z-value is as follows:

<center><math>z=(\overline{y}-\mu_0)(\frac{1}{\sigma})(\sqrt{n})</math></center>

Because we have already standardized our data, μ<sub>0</sub>=0 and σ=1, so this formula simplifies to <math>\overline{y} \sqrt{n}=-0.417\cdot2=-0.833</math>. Note that the formula above normalizes the data, if it has not already been normalized. The z-value can be interpreted as the distance between the sample mean and μ<sub>0</sub>, scaled by a factor which makes the z-value more extreme with large sample sizes. If we take many samples, our z-value is more likely to fall in the rejection region, because we are more certain of the accuracy of our sample mean.

The rejection region for our z-value is from negative infinity to z<sub>α</sub>. We see that our z-value is greater than z<sub>α</sub>. Therefore, we cannot reject the null hypothesis.

[[Image:Gaussian3.png|thumb|left|'''Figure 3:''' Illustrating the P-value.]]

===Section C: P-values===
Another way to judge how likely it is that our null hypothesis is true is to calculate the P-value. If we were to redo the experiment, taking four new data points, the P-value gives us the probability of our new sample mean being at least as extreme as our original sample mean. Graphically, if we extend the critical region until it reaches our z-value, the P-value is equal to the area of the shaded region (see figure 3).

To calculate the P-value in Excel, use <tt>=NORMSDIST(-ABS(z))</tt>. In R, use <tt>pnorm(-abs(z))</tt>. (We use the negative absolute value because <tt>NORMSDIST</tt> and <tt>pnorm</tt> integrate from negative infinity to the z-value. If the z-value is positive, we instead want to integrate from the z-value to positive infinity, which is mathematically equivalent to integrating from negative infinity to the negative of the z-value.)

For this problem, we find that the P-value is 0.202. Note that a P-value of 0.5 indicates that the sample mean is equal to the mean of the theoretical distribution. You can see this graphically by noting that the z-value will be zero in this case, and integrating the theoretical distribution to zero covers half of the area. (Recall that the total area under a standard Gaussian curve is one.) The further the P-value is from 0.5, the greater the distance between the two means.

<div style="float:left; vertical-align: top; padding-right: 20px; padding-bottom: 20px;">[[Image:Gaussian4.png|thumb|none|'''Figure 4:''' The confidence interval about the sample mean.]]<br>
[[Image:Gaussian5.png|thumb|none|'''Figure 5:''' The confidence interval about the theoretical mean.]]</div>

===Section D: Confidence intervals===
We now return to our original data set and theoretical distribution with the mean of 150 psi; that is, we will no longer use our normalized space.

We will now calculate the range of sample means that would lead us to conclude that the breaking strength of our fiber is at least 150 psi, given an α of 0.05. This range is known as the confidence interval about the sample mean.

To calculate this interval, we ask what sample mean would give us a z-value equal to z<sub>α</sub>. We can determine this by substituting z<sub>α</sub> for z into the formula for z, and solving for <math>\overline{y}</math>:

<center><math>z_\alpha=(\overline{y}-\mu_0)(\frac{1}{\sigma})(\sqrt{n}) \Rightarrow \overline{y} = \mu_0+\frac{z_\alpha \sigma}{\sqrt{n}}=147.53</math></center>

This is the lower limit of our confidence interval. Because any sample mean greater than 150 is acceptable, the upper limit of the confidence interval is infinity. We plot this interval in figure 4. Formally, our confidence interval about the sample mean is

<center><math>147.53 < \overline{y} < \infty</math></center>

We next calculate a confidence interval about the mean of the theoretical distribution, μ<sub>0</sub>. This will give us the range of minimum breaking strengths we could have specified and still found our data acceptable. We can calculate this in much the same way as the previous confidence interval: substitute z<sub>α</sub> for z in the formula for z, but this time solve for μ<sub>0</sub>:

<center><math>\mu_0=\overline{y}-\frac{z_\alpha \sigma}{\sqrt{n}}=151.22</math></center>

This is the upper limit of our confidence interval. The lower limit is zero, because we simply require the theoretical mean to be less than this number. Formally, our confidence interval about the theoretical mean is

<center><math>0 \le \mu_0 < 151.22</math></center>

Statistics Tutorial: Design and Analysis of Experiments

2010-07-31T15:56:38Z

Statadmin:

This is a tutorial for Dr. Brash's statistics class, at Christopher Newport University. Here, we will present detailed solutions to several problems from the first few chapters of ''Design and Analysis of Experiments'', 6th ed. by Douglas Montgomery. Solutions are available in Microsoft Excel format and [http://www.r-project.org R] format. Although beginners typically find Microsoft Excel easier to use, it proves to be very limiting for more advanced statistical analysis. R is a much more flexible and powerful software package for statistical analysis which is freely available, but has a somewhat steeper learning curve than Excel.

==Solutions==
[[Problem 2-1, Comparing a single mean to a specified value|Problem 2-1]]

[[User:Pcarter/2-2|Problem 2-2]]

[[User:Pcarter/2-4|Problem 2-4]]

[[User:Pcarter/3-1 through 3-3|3-1 through 3-3]]

[[User:Pcarter/4-1|Problem 4-1]]

[[User:Pcarter/5-1|Problem 5-1]]

Statistics Tutorial: Design and Analysis of Experiments

2010-07-31T15:37:25Z

Statadmin: