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In each of Exercises 3.51-3.52,

(a) use the technology of your choice to determine the mean and median of each of the two data sets.

(b) compare the two data sets by using your results from part (a).

Treating Psychotic Illness. 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 site are based on the results of the client questionnaire.

Short Answer

Expert verified

Part (a) The mean and median for Integrated treatment are 24.916 and 25, respectively, whereas the mean and median for Standard treatment are 23 and 23.

Part (b) The Integrated treatment groups exceed the STANDARD treatment, and both distributions are roughly symmetric.

Step by step solution

01

Part (a) Step 1. Given information.

The given statement is:

The information is for a questionnaire that was created to assess client satisfaction with both the integrated and standard treatments.

02

Part (a) Step 2. Using MINITAB to calculate the mean and median.

MINITAB is used to calculate the mean and median values.

The following is the procedure for calculating the mean, median, and mode.

  • Select Stat >Basic Statistics> Descriptive Statistics from the drop-down menu.
  • In the Variables section, type INTEGRATED, and STANDARD.
  • Select Mean and Median from the Statistic menu.
  • Click OK
03

Part (a) Step 3. MINITAB output.

The output of MINITAB is displayed below:

The figure shows that the median is 25 and the mean is 24.916 for Integrated Treatment.

According to the figure, the Standard Treatment mean is 23 and the median is 23.

04

Part (b) Step 1. Compare the results of part (a).

Integrated treatment's mean and median values are higher than Standard treatment's mean and median values.

The mean and median values for the integrated treatment are nearly comparable. As a result, the distribution is roughly symmetric.

The mean and median values for the standard treatment are nearly comparable. As a result, the distribution is roughly symmetric.

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Most popular questions from this chapter

Consider the following sample of exam scores, arranged in increasing order.

The sample mean and sample standard deviation of these exam scores are 85and 16.1, respectively.
a. Compare the percentage of the observations that actually lie within two standard deviations to either side of the mean with that given by Chebyshev's rule withk=2
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Prolonging Life. Vitamin C(ascorbate) boosts the human immune system and is effective in preventing a variety of illnesses. In a study by E. Cameron and L. Pauling titled "Supplemental Ascorbate in the Supportive Treatment of Cancer; Reevaluation of Prolongation of Survival Times in Terminal Human Cancer" (Proceedings of the National Acadeny of Seience, Vol. 75, No. 9, pp. 4538-4542), patients in advanced stages of cancer were given a vitamin Csupplement. Patients were grouped according to the organ affected by cancer: stomach, bronchus, colon, ovary, or breast. The study yielded the survival times, in days, given on the WeissStats site.

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