Assignment

Note:

• Unless otherwise stated, use 𝛼 = 0.05 .
• Show your work to receive credit
• Please show m anual calculation
  1. A materials engineer is interested in comparing the hardness of six paints. The experiment involved dividing a metal sheet into 24 sections and randomly assigning four of these sections to each paint. All sections were painted (attempting to keep the coat thickness as equal as possible) and, after 180 days of weathering, tested using an automatic scratch hardness tester. The load causing rupture of the coating, measured in newtons, was converted into a load measured in grams.
  a ( 6 pts) Complete the following ANOVA table ( 7 dashed cells) based on this information.
  
  b (6 pts) State the model conditions that must be approximately true in order for the standard analysis to provide trustworthy results. For each condition, also state at least one diagnostic you can use to evaluate it.
  c (5 pts) State the null and alternative hypotheses associated with the ANOVA table \(F\) statistic in the table above. Also, using \(\alpha=0.05\) perform the test and state your conclusion?
  d (5 pts) Turns out the experimenter considered only three paints each with and without an additive. Given the treatment order
  Paint1 Paint1+Additive Paint2 Paint2+Additive Paint3 Paint3+Additive
  Write down the coefficients for each of the following linear combinations of means:
  1. The difference between Paint1 and Paint2
  2. The difference between Paint1+Additive and Paint2+Additive
  3. The average hardness of the six paints
  4. The increase in hardness due to the additive in Paint3
  5. The average increase in hardness due to the additive

e (3 pts) Which of these linear combinations are contrasts? Explain your answer.

f ( 5 pts) Specify the standard error for each of these five linear combinations?

2. Indicate whether the statement is True or False by circling the appropriate letter.

a (5 pts) A two-sample \(t\) test \(\left(H_{0}: \mu_{1}=\mu_{2}\right)\) is performed and the \(P\) -value is $0.068$.

T F : If \(\alpha=.05\), we conclude the Null hypothesis is true

T F : If \(\alpha=.05\), a Type II error is possible

T F : If \(\alpha=.01\), a Type I error is possible

T F : There is a \(6.8 \%\) chance that the Null hypothesis is true

T F : If \(H_{A}\) is one-sided, the \(P\) -value for the two-sided alternative is 0.034

b (5 pts) The cholesterol level in eggs can be controlled by adding certain chemicals to the chicken feed. Suppose a researcher decides to investigate eight different combinations of chemicals using analysis of variance ( \(n=5\) eggs for each combination) and if the \(F\) test is significant at \(\alpha=.05\), compare the means using Tukey's method. Based on this information,

T F : The treatment (or model) degrees of freedom are 7

T F : The error degrees of freedom are 33

T F: If \(F_{0}=2.42\), the experimenter should reject \(H_{0}\)

T F : For the overall \(F\) test, the null hypothesis is \(H_{0}: \sigma_{\tau}^{2}=0\)

T F : The Tukey adjustment is used to protect against Type II errors

c (5 pts) A company wishes to improve their standard office chairs to reduce back pain among the staff. They want to compare three types of chair and need help determining the sample size \(n\) for each group. The necessary sample size \(n\)

T F : Will be smaller using \(\alpha=0.05\) than \(\alpha=0.01\)

T F : Will be smaller for the means \((50,25,75)\) than \((50\,\,20\,\,80)\)

T F: Will be larger for variance \(\sigma=15\) than \(\sigma=12\)

T F : Will be larger for \(90 \%\) power than for \(80 \%\) power

T F : Will be smaller for the means \((50,25,75)\) than \((55\,\,30\,\,80)\)

d (5 pts) You are considering the Tukey and Bonferonni adjustments for comparing all pairs of means when \(a=5\).

T F : The Tukey procedure will result in fewer Type I errors

T F : The Tukey procedure will result in more Type I errors

T F : The Bonferonni procedure uses the \(\alpha / 5\) significance level for each comparison

T F : The Tukey procedure controls the familywise error rate

T F : The Tukey procedure is more powerful

3. Rare Blokes Dairy is concerned that certain cow herds are not producing milk adequately. To assess this, nine herds were randomly selected and from each herd, the milk production of three randomly selected cows was recorded.

a (3 pts) Explain why this is a single factor random effects experiment.

b ( 7 pts) Given that \(\mathrm{SS}_{\mathrm{E}}=27\) and \(\mathrm{SS}_{\mathrm{Herd}}=48\), what proportion of the total variability in milk production is due to the herd (i.e., intraclass correlation coefficient)?

c ( 5 pts) On average, a cow produces $6.5$ gallons of milk per day. The average of the study is \(\bar{y} . .=5.8\). Is there enough evidence \((\alpha=0.05)\) to conclude the average production from a Rare Blokes Dairy cow is less than $6.5$ gallons?

4. Short answer questions.

a (5 pts) Based on the two diagnostic plots below, describe any possible violations of model conditions and appropriate remedies.

b (5 pts) Explain why a procedure that controls the false discovery rate (FDR) will result in at least as many rejections of the null hypothesis as a procedure that controls the family-wise experiment rate (FWER).

d (5 pts) Describe the differences between Tukey's multiple comparison adjustment and the Newman-Keuls multiple comparison adjustment, making sure to describe which has more power.

e (5 pts) Explain when and why nonconstant variance can be a problem in the analysis of a one-way fixed effects model (i.e., in both the overall \(F\) test and mean comparisons)

f (5 pts) Suppose you were told that \(N=30\) experimental units were divided equally among \(a=5\) treatments. Describe how to construct the one-way fixed effects ANOVA model from this information along with the treatment sample means \(\bar{y}_{i \text {. }}\) and treatment sample variances \(s_{i}^{2}\).

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