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Chapter 12: Q2CQQ (page 566)

Cola Weights For the four samples described in Exercise 1, the sample of regular Coke has a mean weight of 0.81682 lb, the sample of Diet Coke has a mean weight of 0.78479 lb, the sample of regular Pepsi has a mean weight of 0.82410 lb, and the sample of Diet Pepsi has a mean weight of 0.78386 lb. If we use analysis of variance and reach a conclusion to reject equality of the four sample means, can we then conclude that any of the specific samples have means that are significantly different from the others?

Short Answer

Expert verified

No, it cannot be inferred from the analysis of variance test that which of the specific samples have a statistically significant difference in their sample means.

Step by step solution

01

Given information

Refer to the results in Exercise 1 for establishing the difference between the four sample means. The different sample mean weights are provided for four different samples of cola content. The sample of regular Coke has a mean weight of 0.81682 lb.

The sample of diet Coke has a mean weight of 0.78479 lb.

The sample of regular Pepsi has a mean weight of 0.82410 lb.

The sample of diet Pepsi has a mean weight of 0.78386 lb.

02

Difference in the specific sample means

The analysis of variance tests if there is a difference between the mean of different groups.

Under the analysis of variance test, if there is a significant difference in the sample means due to a factor, it cannot be used to infer which of the samples have a statistically significant difference in their means.

An additional test, such as ‘Tukey’s pairwise comparison test’, is conducted to determine which of the specific samples have a significant difference in their means.

Then,by observing the sample means, it can be guessed that there appears to be a difference in the mean weights of the regular samples and diet samples, but this cannot be determined statistically using the ANOVA table.

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Most popular questions from this chapter

Estimating Length Using the same results displayed in Exercise 8, does it appear that the length estimates are affected by the subject’s major?

Suppose that a one-way ANOVA is being performed to compare the means of three populations and that the sample sizes are 10,12 , and 15. Determine the degrees of freedom for the F-statistic.

Male Pulse Rates and Age Using the pulse rates of males from Data Set 1 “Body Data” in Appendix B after they are partitioned into the three age brackets of 18–25, 26–40, and 41–80, we get the following SPSS display. Using a 0.05 significance level, test the claim that males from the three age brackets have the same mean pulse rate. What do you conclude?Male Pulse Rates and Age Using the pulse rates of males from Data Set 1 “Body Data” in Appendix B after they are partitioned into the three age brackets of 18–25, 26–40, and 41–80, we get the following SPSS display. Using a 0.05 significance level, test the claim that males from the three age brackets have the same mean pulse rate. What do you conclude?

Does It Pay to Plead Guilty? The accompanying table summarizes randomly selected sample data for San Francisco defendants in burglary cases (based on data from “Does It Pay to Plead Guilty? Differential Sentencing and the Functioning of the Criminal Courts,” by Brereton and Casper, Law and Society Review, Vol. 16, No. 1). All of the subjects had prior prison sentences. Use a 0.05 significance level to test the claim that the sentence (sent to prison or not sent to prison) is independent of the plea. If you were an attorney defending a guilty defendant, would these results suggest that you should encourage a guilty plea?


Guilty Plea

Not Guilty Plea

Sent to Prison

392

58

Not Sent to Prison

564

14

Bonferroni Test Shown below are weights (kg) of poplar trees obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the table below. The data are from a study conducted by researchers at Pennsylvania State University and were provided by Minitab, Inc. Also shown are partial results from using the Bonferroni test with the sample data.

No Treatment

Fertilizer

Irrigation

Fertilizer and Irrigation

1.21

0.94

0.07

0.85

0.57

0.87

0.66

1.78

0.56

0.46

0.10

1.47

0.13

0.58

0.82

2.25

1.30

1.03

0.94

1.64

  1. Use a 0.05 significance level to test the claim that the different treatments result in the same mean weight.
  1. What do the displayed Bonferroni SPSS results tell us?
  1. Use the Bonferroni test procedure with a 0.05 significance level to test for a significant difference between the mean amount of the irrigation treatment group and the group treated with both fertilizer and irrigation. Identify the test statistic and either the P-value or critical values. What do the results indicate?
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