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Chapter 10: Q. 10.98 (page 430)

L. Petersen et al. evaluated the effects of integrated treatment for patients with a first episode of psychotic illness in the paper “A Randomised Multicentre Trial of Integrated Versus Standard Treatment for Patients With a First Episode of Psychotic Illness” (British Medical Journal, Vol. 331, (7517):602). Part of the study included a questionnaire that was designed to measure client satisfaction for both the integrated treatment and a standard treatment. The data on the WeissStats CD are based on the results of the client questionnaire. Use the technology of your choice to do the following.

(a) Obtain normal probability plots, boxplots, and the standard deviations for the two samples.

(b) Based on your results from part (a), which would you be inclined to use to compare the population means: a pooled or a nonpooled t-procedure? Explain your answer.

(c) Do the data provide sufficient evidence to conclude that, on average, clients preferred the integrated treatment? Perform the required hypothesis test at the 1% significance level by using both the pooled t-test and the nonpooled t-test. Compare your results.

(d) Find a 98% confidence interval for the difference between mean client satisfaction scores for the two treatments. Obtain the required confidence interval by using both the pooled t-interval procedure and the nonpooled t-interval procedure. Compare your results

Short Answer

Expert verified

(a) The normal probability by using MINITAB output is:

The boxplot is gven by:

(b) From the probability plots and boxplots, we can conclude that there is no outlier for standard treatment and one outlier for integrated treatment.

Therefore, the graph reveals, the nonpoold t-test is moderately assumption of normality, therefore assumption 3 is wrong.

Hence it is not satisfied, we cannot do the parts (c)-(d).

Step by step solution

01

Part (a) Step 1. GIven Information. 

The data on the WeissStats CD are based on the results of the client questionnaire.

02

Part (a) Step 2. Summary statement.

The summary statistics for two samples of the integrated treatment and standdard treatment is:

03

Part (a) Step 3. Normal probability plot.

Using MINITAB output, the normal probability is:

04

Part (a) Step 4. BOX plot for the treatment.

The boxplot by using MINITAB is:

05

Part (b) Step 1. A  pooled or a nonpooled t-procedure 

Fromm the summary statistics for the two samples, integrated and standard treatment are considerably different. So the pooled t-test is inappropriate.

To check three conditions for non-pooled t-test:

The data obtained from a randomized comparative experiment. Therefore, assumptions 1 and 2are satisfied.

From the probability plots and boxplots, we can conclude that there is no outlier for standard treatment and one outlier for integrated treatment.

Therefore, the graph reveals, the nonpoold t-test is moderately assumption of normality, therefore assumption 3 is wrong.

Hence it is not satisfied, we cannot do the parts (c)-(d).

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Most popular questions from this chapter

The sample standard deviations are 23.6 and 59.2.

In each of exercise 10.13-10.18, we have presented a confidence interval for the difference,μ1-μ2, between two population means. interpret each confidence intervaL

90% CI from5to10

Suppose that you want to perform a hypothesis test to compare the means of two populations, using independent simple random samples. Assume that the two distributions of the variable under consideration have the same shape, but are not normal, and both sample sizes are large. Answer the following questions and explain your answers.

a. Is it permissible to use the pooled t-test to perform the hypothesis test?

b. Is it permissible to use the Mann-Whitney test to perform the hypothesis test?

c. Which procedure is preferable, the pooled t-test or the Mann-Whitney test?

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