CHAPTER 6: SAMPLE PROBLEMS FOR HOMEWORK, CLASS OR EXAMS

These problems are designed to be done without access to a computer, but they may require a calculator.

1. For each scenario below, select the best method for comparing the experiment-wise error rate at 5%.

A. You are comparing the mean reaction time to four different colors of lights. Your only concern is to decide which colors have the shortest mean reaction time.

B. You are comparing mean hardness for epoxy resins cured for one of five different curing times (1 hour, 1.25, 1.5, 1.75 or 2 hours). You have planned in advance that you will test for a linear and for a quadratic trend, and that these are the only follow-up hypotheses that will be tested.

C. You are comparing mean hardness for epoxy resins cured for one of five different curing times (1 hours, 1.25, 1.5, 1.75 or 2 hours). You will do all pairwise comparisons, but also wish to make any other contrasts which occur to you after inspecting the data.

2. You are comparing mean hardness for epoxy resins cured under one of four different conditions:

#1: catalyst A for 1 hour#2: catalyst A for 2 hours

#3: catalyst B for 1 hour#4: catalyst B for 2 hours

A. Give the coefficients which correspond to the following statements.

1) The average of the means for catalyst A equals that for Catalyst B.

2) The difference between the means at 1 and 2 hours is the same for Catalyst A as it is for catalyst B.

B. Show that the two contrasts for part (A) are orthogonal.

C. Create a new contrast that is orthogonal to both those in part (A), and interpret it.

3. The ANOVA table below summarizes an experiment with 4 groups, each with 8 observations. Complete the table, and give the appropriate interpretation assuming  = 5%.

SourcedfSSMSF

Between__480______

Within______

Total__2160

4. You have an experiment in which 48 patients with mild diabetes are randomly assigned to one of 4 different protocols for controlling blood glucose levels. Each patient submits to extensive monitoring for a one week period. The dependent variable is the percentage of time that the patient’s blood glucose was high (TIMEHIGH). It is desirable that this value be low.

a. Is there significant evidence that at least one group has a different mean TIMEHIGH, assuming  = 5%? Cite the appropriate test statistic.

[Part b is stated in terms of Tukey’s HSD, as that is easier to carry out by hand than Duncan’s.]

b. Protocols 1 and 3 are similar except that protocol 3 includes an exercise plan. Do these two group means differ significantly, using Tukey’s HSD? Assume that all the pairwise comparisons are being conducted, not just this one, and that the experimentwise significance level is to be kept at 10%.

c. After examining the data, the researcher decides to test the null hypothesis that the mean for Protocol 4 equals the average of the means for the other 3 groups: .

What would the appropriate method be for deciding whether this test statistic is significant? Would it be significant at an experimentwise error rate of 5%?

Sum of

Source DF Squares Mean Square F Value

Model 528.000000

Error 554.776930

Corrected Total 47 1082.776930

Level of ---timehigh

protocol N Mean

1 12 27.5403924

2 12 27.5403924

3 12 23.5403924

4 12 19.5403924

5. (Requires table for Analysis of Means). The following table summarizes data for Sales per Week for random samples of supermarkets, in $100,000s. Plot the control lines using  = 10%, and the observed means. Identify any weeks for which the sales seemed anomalous.

Week nMean SalesStandard deviation

1 1018.65.9

2 1023.46.2

3 1021.86.0

4 1022.65.8

5 1026.26.1

6. The following table summarizes data for Sales per Week for random samples of supermarkets, in $100,000s. The weeks correspond to weeks of an advertising campaign. In advance, it was decided that the only hypothesis to be tested is that there is a linear trend in the data. The coefficients for a linear trend in five groups are (-2, -1, 0, 1, 2). Test the null hypothesis of no linear trend using  = 5%. Hint: SSW = 1620.9.

Week nMean SalesStandard deviation

1 1018.65.9

2 1023.46.2

3 1021.86.0

4 1022.65.8

5 1026.26.1

7. You are examining the effect of curing time on the mean hardness of epoxy resins. There are five levels of curing time (1, 1.25, 1.5, 1.75 and 2 hours). Four batches of resin are cured at each level of time, for a total of 20 observations. Prior to the experiment, it was hypothesized that a quadratic model would suffice to model the relationship between time and hardness. Use the ANOVA table below to

a. Perform a lack of fit test for the quadratic model.

b. Perform a lack of fit test for a linear model.

c. Would you advise a linear or a quadratic model?

SourceSS

Between90

Linear79

Quadratic 5

Within30

8. A hospital administrator randomly selects records for 50 emergency room patients every month to find the proportion who had to wait more than 1 hour for assistance. For six consecutive months, the number waiting more than an hour were: 8, 12, 2, 19, 10, 13. Use an ANOM procedure with a significance level of 5% to see whether any of the months seems to be different.

SOLUTIONS

1. a. Duncan’s MRT (or Tukey’s HSD)

b. Bonferroni’s for two hypotheses

c. Scheffe’s

2. A1: 1 1 -1 -1

A2: 1 -1 -1 1

b. 1(1) + 1(-1) + (-1)(-1) + (-1)1 = 0 so these contrasts are orthogonal

c. average of the means at two hours equals the average of the means at one hour

(1 -1 1 -1)

This is orthogonal to both A1 [1(1) + (-1)1 + (1)(-1) + (-1)(-1) = 0] and

A2 [1(1) + (-1)(-1) + 1(-1) + (-1)(1) = 0].

3.

SourcedfSSMSF

Between 34801602.67

Within28168060

Total312160

The critical value with 3 and 28 df is about 2.92. Hence, there is no significant evidence that any of the groups have a different mean value.

4. a) F=13.96 with 3 and 44 df. The critical value is about 2.81. There is significant evidence that at least one group has a different mean.

b) The critical value is . The difference in the sample means for these two groups is 4, so Protocols 1 and 3 do differ significantly.

c) Use Scheffe’s method, since the contrast was decided after examining the data.

The critical value is . The calculated value for the contrast is . There is significant evidence that the mean for Protocol 4 differs from the means for the other groups.

5. The control limits are

No weeks means plot outside the control limits.

6. with 45 degrees of freedom. (Note the sum of the squared coefficients is 10, and the sample size in each group is also 10.) At  = 5%, there is significant evidence of a linear trend.

7. a. The Lack of Fit test for the quadratic model would have SS = (90 – 79 – 5) = 6 and 2 df for the numerator, and SS = 30 with 15 df for the denominator. F = 1.5, which is not significant. Hence, there is no significant evidence that the quadratic model does not fit – that is, it is reasonable to believe a quadratic model is sufficient.

b. The Lack of Fit test for the linear model would have SS = (90 – 79) = 11 with 3 df for the numerator. F = 1.83. There is no significant evidence that a linear model does not fit – that is, it is reasonable to believe a linear model is sufficient.

c. Prefer the linear model as it is simpler.

8. The overall proportion is 0.2133, the control limits are

The 3rd week (2/50 = 0.04) and the 4th week (19/50=0.38) are outside the control limits.