The article “Anodic Fenton Treatment of Treflan MTF” describes a two-factor experiment designed to study the sorption of the herbicide trifluralin.

Question
Study design

The article “Anodic Fenton Treatment of Treflan MTF” describes a two-factor experiment designed to study the sorption of the herbicide trifluralin. The factors are the initial trifluralin concentration and the $$\displaystyle{F}{e}^{{{2}}}\ :\ {H}_{{{2}}}\ {O}_{{{2}}}$$ delivery ratio. There were three replications for each treatment. The results presented in the following table are consistent with the means and standard deviations reported in the article. $$\begin{array}{|c|c|}\hline \text{Initial Concentration (M)} & \text{Delivery Ratio} & \text{Sorption (%)} \\ \hline 15 & 1:0 & 10.90 \quad 8.47 \quad 12.43 \\ \hline 15 & 1:1 & 3.33 \quad 2.40 \quad 2.67 \\ \hline 15 & 1:5 & 0.79 \quad 0.76 \quad 0.84 \\ \hline 15 & 1:10 & 0.54 \quad 0.69 \quad 0.57 \\ \hline 40 & 1:0 & 6.84 \quad 7.68 \quad 6.79 \\ \hline 40 & 1:1 & 1.72 \quad 1.55 \quad 1.82 \\ \hline 40 & 1:5 & 0.68 \quad 0.83 \quad 0.89 \\ \hline 40 & 1:10 & 0.58 \quad 1.13 \quad 1.28 \\ \hline 100 & 1:0 & 6.61 \quad 6.66 \quad 7.43 \\ \hline 100 & 1:1 & 1.25 \quad 1.46 \quad 1.49 \\ \hline 100 & 1:5 & 1.17 \quad 1.27 \quad 1.16 \\ \hline 100 & 1:10 & 0.93 \quad 0.67 \quad 0.80\\ \hline \end{array}$$ a) Estimate all main effects and interactions. b) Construct an ANOVA table. You may give ranges for the P-values. c) Is the additive model plausible? Provide the value of the test statistic, its null distribution, and the P-value.

2021-01-16

Step 1 Given: $$\displaystyle{I}={3}$$
$$\displaystyle{J}={4}$$
$$\displaystyle{K}={3}$$ a) The mean is the sum of all values divided by the number of data values: $$\displaystyle\overline{{{X}}}_{{{1}.}}={3.6992}$$
$$\displaystyle\overline{{{X}}}_{{{2}.}}={2.6492}$$
$$\displaystyle\overline{{{X}}}_{{{3}.}}={2.575}$$
$$\displaystyle\overline{{{X}}}_{{{.1}}}={8.2011}$$
$$\displaystyle\overline{{{X}}}_{{{.2}}}={1.9656}$$
$$\displaystyle\overline{{{X}}}_{{{.3}}}={0.9322}$$
$$\displaystyle\overline{{{X}}}_{{{.4}}}={0.7989}$$
$$\displaystyle\overline{{{X}}}_{{{11}}}={10.6}$$
$$\displaystyle\overline{{{X}}}_{{{12}}}={2.8}$$
$$\displaystyle\overline{{{X}}}_{{{13}}}={0.7967}$$
$$\displaystyle\overline{{{X}}}_{{{14}}}={0.6}$$
$$\displaystyle\overline{{{X}}}_{{{21}}}={7.1033}$$
$$\displaystyle\overline{{{X}}}_{{{22}}}={1.6967}$$
$$\displaystyle\overline{{{X}}}_{{{23}}}={0.8}$$
$$\displaystyle\overline{{{X}}}_{{{24}}}={0.9967}$$
$$\displaystyle\overline{{{X}}}_{{{31}}}={6.9}$$
$$\displaystyle\overline{{{X}}}_{{{32}}}={1.4}$$
$$\displaystyle\overline{{{X}}}_{{{33}}}={1.2}$$
$$\displaystyle\overline{{{X}}}_{{{34}}}={0.8}$$
$$\displaystyle\overline{{{X}}}={2.9744}$$ Step 2 The main effects are the difference of the mean of the values in the category decreased by the overall mean. $$\displaystyle{\color{blue}{\text{Main effects concentration}}}$$
$$\displaystyle\overline{{{X}}}_{{{1}.}}\ -\ \overline{{{X}}}={3.6992}\ -\ {2.9744}={0.7248}$$
$$\displaystyle\overline{{{X}}}_{{{2}.}}\ -\ \overline{{{X}}}={2.6492}\ -\ {2.9744}=\ -{0.3252}$$
$$\displaystyle\overline{{{X}}}_{{{3}.}}\ -\ \overline{{{X}}}={2.575}\ -\ {2.9744}=\ -{0.3994}$$
$$\displaystyle{\color{blue}{\text{Main effects delivery ratio}}}$$
$$\displaystyle\overline{{{X}}}_{{{.1}}}\ -\ \overline{{{X}}}={8.2011}\ -\ {2.9744}={5.2267}$$
$$\displaystyle\overline{{{X}}}_{{{.2}}}\ -\ \overline{{{X}}}={1.9656}\ -\ {2.9744}=\ -{1.0088}$$
$$\displaystyle\overline{{{X}}}_{{{.3}}}\ -\ \overline{{{X}}}={0.9322}\ -\ {2.9744}=\ -{2.0422}$$
$$\displaystyle\overline{{{X}}}_{{{.4}}}\ -\ \overline{{{X}}}={0.7989}\ -\ {2.9744}=\ -{2.1755}$$ The main effects are difference of the mean of the values in the category decreased by the mean of the row and the mean of the column, and increased by the overall mean. $$\displaystyle\overline{{{X}}}_{{{11}}}\ -\ \overline{{{X}}}_{{{1}.}}\ -\ \overline{{{X}}}_{{{.1}}}\ +\ \overline{{{X}}}={10.6}\ -\ {3.6992}\ -\ {8.2011}\ +\ {2.9744}={1.6741}$$
$$\displaystyle\overline{{{X}}}_{{{12}}}\ -\ \overline{{{X}}}_{{{1}.}}\ -\ \overline{{{X}}}_{{{.2}}}\ +\ \overline{{{X}}}={2.8}\ -\ {3.6992}\ -\ {1.9656}\ +\ {2.9744}={0.1096}$$
$$\displaystyle\overline{{{X}}}_{{{13}}}\ -\ \overline{{{X}}}_{{{1}.}}\ -\ \overline{{{X}}}_{{{.3}}}\ +\ \overline{{{X}}}={0.7967}\ -\ {3.6992}\ -\ {0.9322}\ +\ {2.9744}=\ -{0.8603}$$
$$\displaystyle\overline{{{X}}}_{{{14}}}\ -\ \overline{{{X}}}_{{{1}.}}\ -\ \overline{{{X}}}_{{{.4}}}\ +\ \overline{{{X}}}={0.6}\ -\ {3.6992}\ -\ {0.7989}\ +\ {2.9744}=\ -{0.9237}$$
$$\displaystyle\overline{{{X}}}_{{{21}}}\ -\ \overline{{{X}}}_{{{2}.}}\ -\ \overline{{{X}}}_{{{.1}}}\ +\ \overline{{{X}}}={7.1033}\ -\ {2.6492}\ -\ {8.2011}\ +\ {2.9744}=\ -{0.7726}$$
$$\displaystyle\overline{{{X}}}_{{{22}}}\ -\ \overline{{{X}}}_{{{2}.}}\ -\ \overline{{{X}}}_{{{.2}}}\ +\ \overline{{{X}}}={1.6967}\ -\ {2.6492}\ -\ {1.9656}\ +\ {2.9744}={0.0563}$$
$$\displaystyle\overline{{{X}}}_{{{23}}}\ -\ \overline{{{X}}}_{{{2}.}}\ -\ \overline{{{X}}}_{{{.3}}}\ +\ \overline{{{X}}}={0.8}\ -\ {2.6492}\ -\ {0.9322}\ +\ {2.9744}={0.1930}$$
$$\displaystyle\overline{{{X}}}_{{{24}}}\ -\ \overline{{{X}}}_{{{2}.}}\ -\ \overline{{{X}}}_{{{.4}}}\ +\ \overline{{{X}}}={0.9967}\ -\ {2.6492}\ -\ {0.7989}\ +\ {2.9744}={0.5230}$$
$$\displaystyle\overline{{{X}}}_{{{31}}}\ -\ \overline{{{X}}}_{{{3}.}}\ -\ \overline{{{X}}}_{{{.1}}}\ +\ \overline{{{X}}}={6.9}\ -\ {2.575}\ -\ {8.2011}\ +\ {2.9744}=\ -{0.9017}$$
$$\displaystyle\overline{{{X}}}_{{{32}}}\ -\ \overline{{{X}}}_{{{3}.}}\ -\ \overline{{{X}}}_{{{.2}}}\ +\ \overline{{{X}}}={1.4}\ -\ {2.575}\ -\ {1.9656}\ +\ {2.9744}=\ -{0.1662}$$
$$\displaystyle\overline{{{X}}}_{{{33}}}\ -\ \overline{{{X}}}_{{{3}.}}\ -\ \overline{{{X}}}_{{{.3}}}\ +\ \overline{{{X}}}={1.2}\ -\ {2.575}\ -\ {0.9322}\ +\ {2.9744}={0.6672}$$
$$\displaystyle\overline{{{X}}}_{{{34}}}\ -\ \overline{{{X}}}_{{{3}.}}\ -\ \overline{{{X}}}_{{{.4}}}\ +\ \overline{{{X}}}={0.8}\ -\ {2.575}\ -\ {0.7989}\ +\ {2.9744}={0.4005}$$ Step 3 b) Let us first determine the sum of squares: $$\displaystyle{S}{S}{A}={J}{K}\ {\sum_{{{i}={1}}}^{{{I}}}}\ {\overline{{{X}}}_{{{i}.}}^{{{2}}}}\ -\ {I}{J}{K}\ \overline{{{X}}}^{{{2}}}$$
$$\displaystyle={4}{\left({3}\right)}{\left({3.6992}\right)}^{{{2}}}\ +\ {4}{\left({3}\right)}{\left({2.6492}\right)}^{{{2}}}\ +\ {4}{\left({3}\right)}{\left({2.575}\right)}^{{{2}}}\ -\ {3}{\left({4}\right)}{\left({3}\right)}{\left({2.9744}\right)}^{{{2}}}$$
$$\displaystyle\approx\ {9.49}$$
$$\displaystyle{S}{S}{B}={I}{K}\ {\sum_{{{j}={1}}}^{{{J}}}}\ {\overline{{{X}}}_{{.{j}}}^{{{2}}}}\ -\ {I}{J}{K}\ \overline{{{X}}}^{{{2}}}$$
$$\displaystyle={3}{\left({3}\right)}{\left({8.2011}\right)}^{{{2}}}\ +\ {3}{\left({3}\right)}{\left({1.9656}\right)}^{{{2}}}\ +\ {3}{\left({3}\right)}{\left({0.9322}\right)}^{{{2}}}$$
$$\displaystyle+\ {3}{\left({3}\right)}{\left({0.7989}\right)}^{{{2}}}\ -\ {3}{\left({4}\right)}{\left({3}\right)}{\left({2.9744}\right)}^{{{2}}}$$
$$\displaystyle\approx\ {335.16}$$
$$\displaystyle{S}{S}{A}{B}={K}\ {\sum_{{{i}={1}}}^{{{I}}}}\ {\sum_{{{j}={1}}}^{{{j}}}}\ {\overline{{{X}}}_{{{i}{j}}}^{{{2}}}}\ -\ {J}{K}\ {\sum_{{{i}={1}}}^{{{I}}}}\ {\overline{{{X}}}_{{{i}}}^{{{2}}}}$$
$$\displaystyle-\ {I}{K}\ {\sum_{{{j}={1}}}^{{{J}}}}\ {\overline{{{X}}}_{{.{j}}}^{{{2}}}}\ +\ {I}{J}{K}\ \overline{{{X}}}^{{{2}}}$$
$$\displaystyle={3}{\left({10.6}\right)}^{{{2}}}\ +\ {3}{\left({2.8}\right)}^{{{2}}}\ +\ {3}{\left({0.7967}\right)}^{{{2}}}\ +\ {3}{\left({0.6}\right)}^{{{2}}}$$
$$\displaystyle+\ {3}{\left({7.1033}\right)}^{{{2}}}\ +\ {3}{\left({1.6967}\right)}^{{{2}}}\ +\ {3}{\left({0.8}\right)}\ +\ {3}{\left({0.9967}\right)}^{{{2}}}$$
$$\displaystyle+\ {3}{\left({6.9}\right)}^{{{2}}}\ +\ {3}{\left({1.4}\right)}^{{{2}}}\ +\ {3}{\left({1.2}\right)}^{{{2}}}\ +\ {3}{\left({0.8}\right)}^{{{2}}}$$
$$\displaystyle-\ {4}{\left({3}\right)}{\left({3.6992}\right)}^{{{2}}}\ -\ {4}{\left({3}\right)}{\left({2.6492}\right)}^{{{2}}}\ -\ {4}{\left({3}\right)}{\left({2.575}\right)}^{{{2}}}$$
$$\displaystyle-\ {3}{\left({3}\right)}{\left({8.2011}\right)}^{{{2}}}\ -\ {3}{\left({3}\right)}{\left({1.9656}\right)}^{{{2}}}\ -\ {3}{\left({3}\right)}{\left({0.9322}\right)}^{{{2}}}\ -\ {3}{\left({3}\right)}{\left({0.7989}\right)}^{{{2}}}$$
$$\displaystyle+\ {3}{\left({4}\right)}{\left({3}\right)}{\left({2.9744}\right)}^{{{2}}}$$
$$\displaystyle\approx\ {20.3}$$
$$\displaystyle{S}{S}{E}=\ {\sum_{{{i}={1}}}^{{{I}}}}\ {\sum_{{{j}={1}}}^{{{J}}}}\ {\sum_{{{k}={1}}}^{{{K}}}}\ {\overline{{{X}}}_{{{i}{j}{k}}}^{{{2}}}}\ -\ {K}\ {\sum_{{{i}={1}}}^{{{I}}}}\ {\sum_{{{j}={1}}}^{{{J}}}}\ {\overline{{{X}}}_{{{i}{j}}}^{{{2}}}}$$
$$\displaystyle={10.90}^{{{2}}}\ +\ {8.47}^{{{2}}}\ +\ {12.43}^{{{2}}}\ +\ {3.33}^{{{2}}}\ +\ {2.40}^{{{2}}}\ +\ {2.67}^{{{2}}}$$
$$\displaystyle+\ {0.79}^{{{2}}}\ +\ {0.76}^{{{2}}}\ +\ {0.84}^{{{2}}}\ +\ {0.54}^{{{2}}}\ +\ {0.69}^{{{2}}}\ +\ {0.57}^{{{2}}}$$
$$\displaystyle+\ {6.84}^{{{2}}}\ +\ {7.68}^{{{2}}}\ {6.79}^{{{2}}}\ +\ {1.72}^{{{2}}}\ +\ {1.55}^{{{2}}}\ +\ {1.82}^{{{2}}}$$
$$\displaystyle+\ {0.68}^{{{2}}}\ +\ {0.83}^{{{2}}}\ +\ {0.89}^{{{2}}}\ +\ {0.58}^{{{2}}}\ +\ {1.13}^{{{2}}}\ +\ {1.28}^{{{2}}}$$
$$\displaystyle+\ {6.61}^{{{2}}}\ +\ {6.66}^{{{2}}}\ +\ {7.43}^{{{2}}}\ +\ {1.25}^{{{2}}}\ +\ {1.46}^{{{2}}}\ +\ {1.49}^{{{2}}}$$
$$\displaystyle+\ {1.17}^{{{2}}}\ +\ {1.27}^{{{2}}}\ +\ {1.16}^{{{2}}}\ +\ {0.93}^{{{2}}}\ +\ {0.67}^{{{2}}}\ +\ {0.80}^{{{2}}}$$
$$\displaystyle-\ {3}{\left({10.6}\right)}^{{{2}}}\ -\ {3}{\left({2.8}\right)}^{{{2}}}\ -\ {3}{\left({0.7967}\right)}^{{{2}}}\ -\ {3}{\left({0.6}\right)}^{{{2}}}$$
$$\displaystyle-\ {3}{\left({7.1033}\right)}^{{{2}}}\ -\ {3}{\left({1.6967}\right)}^{{{2}}}\ -\ {3}{\left({0.8}\right)}^{{{2}}}\ -\ {3}{\left({0.9967}\right)}^{{{2}}}$$
$$\displaystyle-\ {3}{\left({6.9}\right)}^{{{2}}}\ -\ {3}{\left({1.4}\right)}^{{{2}}}\ -\ {3}{\left({1.2}\right)}^{{{2}}}\ -\ {3}{\left({0.8}\right)}^{{{2}}}$$
$$\displaystyle={9.78}$$ Step 4 The degrees of freedom for the effect of a variable is the number of categories of the variable decreased by 1. $$\displaystyle{d}{f}_{{{A}}}={I}\ -\ {1}={3}\ -\ {1}={2}$$ The degrees of freedom for the effect of a variable is the number of categories of the variable decreased by 1. $$\displaystyle{d}{f}_{{{B}}}={J}\ -\ {1}={4}\ -\ {1}={3}$$ The degrees of freedom for the interactions is the product of degrees of freedom for the effect of each variable. $$\displaystyle{d}{f}_{{{A}{B}}}={d}{f}_{{{A}}}\ \cdot\ {d}{f}_{{{B}}}={2}\ \cdot\ {3}={6}$$ The degrees of freedom for error is the product of the number of categories for each variable, and the number of replicates per treatment decreased by 1. $$\displaystyle{d}{f}_{{{E}}}={I}{J}{\left({K}\ -\ {1}\right)}={3}\ \cdot\ {4}\ \cdot\ {\left({3}\ -\ {1}\right)}={12}\ \cdot\ {2}={24}$$ Step 5 $$\displaystyle{M}{S}{A}{i}{s}{S}{S}{A}\div{d}{b}{y}{d}{f}_{{{A}}}:$$
$$\displaystyle{M}{S}{A}=\ {\frac{{{S}{S}{A}}}{{{d}{f}_{{{A}}}}}}=\ {\frac{{{9.49}}}{{{2}}}}={4.74}$$ MSB is SSB divided by $$\displaystyle{d}{f}_{{{B}}}:$$
$$\displaystyle{M}{S}{B}=\ {\frac{{{S}{S}{B}}}{{{d}{f}_{{{B}}}}}}=\ {\frac{{{335.16}}}{{{3}}}}={111.72}$$ MSAB is SSAB divided by $$\displaystyle{d}{f}_{{{A}{B}}}:$$
$$\displaystyle{M}{S}{A}{B}=\ {\frac{{{S}{S}{A}{B}}}{{{d}{f}_{{{A}{B}}}}}}=\ {\frac{{{20.3}}}{{{6}}}}={3.38}$$ MSE is SSE divided by $$\displaystyle{d}{f}_{{{E}}}:$$
$$\displaystyle{M}{S}{E}=\ {\frac{{{S}{S}{E}}}{{{d}{f}_{{{E}}}}}}=\ {\frac{{{9.78}}}{{{24}}}}={0.41}$$ The value of the test statistic $$\displaystyle{F}_{{{A}}}$$ is then MSA divided by MSE: $$\displaystyle{F}_{{{A}}}=\ {\frac{{{M}{S}{A}}}{{{M}{S}{E}}}}=\ {\frac{{{4.74}}}{{{0.41}}}}\ \approx\ {11.64}$$ The value of the test statistic $$\displaystyle{F}_{{{B}}}$$ is then MSB divided by MSE: $$\displaystyle{F}_{{{B}}}=\ {\frac{{{M}{S}{B}}}{{{M}{S}{E}}}}=\ {\frac{{{111.72}}}{{{0.41}}}}\ \approx\ {274.16}$$ The value of the test statistic $$\displaystyle{F}_{{{A}{B}}}$$ is then MSAB divided by MSE: $$\displaystyle{F}_{{{A}{B}}}=\ {\frac{{{M}{S}{A}{B}}}{{{M}{S}{E}}}}=\ {\frac{{{3.38}}}{{{0.41}}}}\ \approx\ {8.3}$$ Step 6 The p-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value corresponding to the rows is the number (or interval) in the row title of the F-distribution table in the appendix containing the F-value in the row $$\displaystyle{d}{f}_{{{2}}}={d}{f}_{{{E}}}={24}\ \text{and in the column}\ {d}{f}_{{{1}}}={d}{f}_{{{A}}}={2}:$$
$$\displaystyle{P}\ {<}\ {0.001}$$ The P-value corresponding to the columns is the number (or interval) in the row title of the F-distribution table in the appendix containing the F-value in the row $$\displaystyle{d}{f}_{{{2}}}={d}{f}_{{{E}}}={24}\ \text{and in the column}\ {d}{f}_{{{1}}}={d}{f}_{{{B}}}={3}:$$
$$\displaystyle{P}\ {<}\ {0.001}$$ The P-value corresponding to the interaction is the number (or interval) in the row title of the F-distribution table in the appendix containing the F-value in the row $$\displaystyle{d}{f}_{{{2}}}={d}{f}_{{{E}}}={24}\ \text{and in the column}\ {d}{f}_{{{1}}}={d}{f}_{{{A}{B}}}={6}:$$
$$\displaystyle{P}\ {<}\ {0.001}$$ Combine the information in an ANOVA table: $$\begin{array}{c} \text{Sourse} & df & SS & MS & F & P \\ \hline \text{Rows} & 2 & 9.49 & 4.74 & 11.64 & P<0.001 \\ \text{Columns} & 3 & 335.16 & 111.72 & 274.16 & P<0.001 \\ \text{Interaction} & 6 & 20.3 & 3.38 & 8.3 & P<0.001 \\ \hline \text{Total} & 35 & 374.72 \\ \end{array}$$

Step 7 c) By part (b): $$\displaystyle{F}_{{{A}{B}}}={8.3}\ {\quad\text{and}\quad}\ {P}\ {<}\ {0.001}$$ If the P-value is less than the significance level, then reject the null hypothesis. $$\displaystyle{P}\ {<}\ {0.05}\ \Rightarrow\ {R}{e}{j}{e}{c}{t}\ {H}_{{{0}}}$$ There is sufficient evidence the claim that the additive model is plausible.

Relevant Questions

A researcher is conducting a study to examine the effects of cognitive behavior therapy for the treatment of social anxiety in a sample of 16 participants. He measures the social anxiety scores of participants before the tratment and then again after treatment and the resulting data is as follows:
n=16, M_0=5, s =4
a) What type of design is this study (single-sample, independent measures, repeated measures)
b)State the null and alternate hypotheses
c) Using an $$\displaystyle\alpha\ level\ of\ {.05}{\left(\alpha={.05}\right)}$$, identify the critical values of t for a 2-tailed test.

In a similar study using the same design (sample of $$n = 9$$ participants), the individuals who wore the shirt produced an average estimate of $$M = 6.4$$ with $$SS = 162$$.
The average number who said they noticed was (population mean) 3.1. Calculate a One Sample t-test using a two-tailed test with $$\displaystyle\alpha={.05}$$.

The most appropriate study design depends, among other things, on the distribution of
1.A) The risk factor in the population of interest
2.B) The participants
3.C) The outcome in the population of interest
4.D) $$A\ \&\ C$$

Response rates to Web surveys are typically low, partially due to users starting but not finishing the survey. The factors that influence response rates were investigated in Survey Methodology (Dec. 2013). In a designed study, Web users were directed to participate in one of several surveys with different formats. For example, one format utilized a welcome screen witb a white background, and another format utilized a welcome screen with a red background. The "break-off rates," i.e., the proportion of sampled users who break off the survey before completing all questions, for the two formats are provided in the table. $$\begin{array}{|c|c|}\hline \text{ } & \text{White Welcome Screen} & \text{Red Welcome Screen} \\ \hline \text{Number of Web users} & 190 & 183 \\ \hline \text{Number who break off survey} & 49 & 37 \\ \hline \text{Break-off rate} & 0.258 & 0.202 \\ \hline \end{array}$$ a) Verify the values of the break-off rates shown in the table. b) The researchers theorize that the true break-off rate for Web users of the red welcome screen will be lower than the corresponding break-off Tate for users of the white welcome screen. Give the null and alternative hypotbesis for testing this theory. c) Conduct the test, part b, at $$\displaystyle\alpha={0.10}.$$ What do you conclude?

In an experiment designed to study the effects of illumination level on task performance (“Performance of Complex Tasks Under Different Levels of Illumination,” J. Illuminating Eng., 1976: 235–242), subjects were required to insert a fine-tipped probe into the eyeholes of ten needles in rapid succession both for a low light level with a black background and a higher level with a white background. Each data value is the time (sec) required to complete the task.
$$\begin{array}{|c|c|} \hline Subject & (1) & (2) & (3) & (4) & (5) &(6) & (7) & (8) & (9) \\ \hline Black & 25.85 & 28.84 & 32.05 & 25.74 & 20.89 & 41.05 & 25.01 & 24.96 & 27.47 \\ \hline White & 18.28 & 20.84 & 22.96 & 19.68 & 19.509 & 24.98 & 16.61 & 16.07 & 24.59 \\ \hline \end{array}$$
Does the data indicate that the higher level of illumination yields a decrease of more than 5 sec in true average task completion time? Test the appropriate hypotheses using the P-value approach.

An experiment designed to study the relationship between hypertension and cigarette smoking yielded the following data.
$$\begin{array}{|c|c|} \hline Tension\ level & Non-smoker & Moderate\ smoker & Heavy\ smoker \\ \hline Hypertension & 20 & 38 & 28 \\ \hline No\ hypertension & 50 & 27 & 18 \\ \hline \end{array}$$
Test the hypothesis that whether or not an individual has hypertension is independent of how much that person smokes.

The article “Rocker Shoe Put to the Test: Can it Really Walk the Walk as a Way to Get in Shape?” (USA TODAY, October 12, 2009) describes claims made by Skechers about Shape-Ups, a shoe line introduced in 2009. These curved-sole sneakers are supposed to help you “get into shape without going to the gym” according to a Skechers advertisement. Briefly describe how you might design a study to investigate this claim. Include how you would select subjects and what variables you would measure. Is the study you designed an observational study or an experiment?

Prove that the reduced row echelon forms of the matrices$$\left(\begin{array}{c}1 & \ 1 & \ 4 & \ 8 & \ 0 & \ -1 & \ -1 & \\ 1 & \ 2 & \ 3 & \ 9 & \ 0 & \ -5 & \ -2 \\ 0 & \ -2 & \ 2 & \ -2 & \ 1 & \ 14 & \ 3 \\ 1 & \ 4 & \ 1 & \ 11 & \ 0 & \ -13 & \ -4 \end{array}\right)\ \left(\begin{array}{c}0 & \ -3 & \ 3 & \ 1 & \ 5 \\ 0 & \ 1 & \ -1 & \ 0 & \ 0 \\ 0 & \ 2 & \ -2 & \ 0 & \ -3 \end{array}\right)$$

are the two matrices.

The point of the reduced row echelon form is that the corresponding system of linear equations is in a particularly simple form, from which the solutions to the system $$AX=C\ in\ (4)$$ can be determined immediately.