Question

In: Statistics and Probability

The manager of a pharmacy wants to know if prescriptions are filled uniformly over the 7...

  1. The manager of a pharmacy wants to know if prescriptions are filled uniformly over the 7 days of the week. The manager takes a simple random sample of 245 prescription receipts and finds that they are distributed as follows:

Day

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

Prescriptions

42

31

33

29

45

44

21

Which of the following is the appropriate null hypothesis for this test?

  1. H0: p1 = p2 = p3 = p4 = p5 = p6 = p7 = 1/7
  2. H0: p1 = p2 = p3 = p4 = p5 = 5/7 and p6 = p7 = 2/7
  3. H0: p1 = 0.17, p2 = 0.13, p3 = 0.13, p4 = 0.12, p5 = 0.18, p6 = 0.18, p7 = 0.09
  4. None of the above
  1. Under the null hypothesis of a uniform distribution of prescriptions over the 7 days of the week, the expected count of prescriptions for Monday is _____________ (show calculation).
  1. Under the null hypothesis of a uniform distribution of prescriptions over the 7 days of the week, the chi-square contribution for Monday is _________________ (show calculation).
  1. Under the null hypothesis of a uniform distribution of prescriptions over the 7 days of the week, the degrees of freedom for the chi-square test is _______________.
  1. What is the chi-square statistic for testing this null hypothesis of a uniform distribution of prescriptions over the 7 days of the week (show calculation)?

  1. 1/7
  2. 3.5
  3. 13.8
  4. 24.5
  1. What is the P-value for testing this null hypothesis of a uniform distribution of prescriptions over the 7 days of the week? Specify the distribution used and all relevant parameters.

  1. Using a significance level of 0.05, what is the appropriate conclusion for this test?
  2. All 7 days of the week have different prescription rates.
  3. There is significant evidence that prescriptions are not uniformly distributed over the 7 days of the week.
  4. Weekdays and weekends have significantly different prescription rates.
  5. The data are consistent with prescriptions being uniformly distributed over the 7 days of the week.
  1. What can we state about the chi-square test in this situation?
  1. The test is valid because the sample size is large.
  2. The test is valid because the sample is random and the observed counts are large enough.
  3. The test is valid because the sample is random and the expected counts are large enough.
  4. The test is not valid because we do not know the true population proportions.
  1. Which of the following statements about a chi-square hypothesis test is true?
  1. When observed counts are far from expected counts, we have evidence against H0.
  2. Large values of χ2 indicate evidence against H0.
  3. Expected counts are hypothetical, and do not have to be whole numbers.
  4. All of the above
  1. True/False: The P-value of the chi-square test is the area to the left of the calculated C2 statistic under this chi-square distribution
  1. True
  2. False

Solutions

Expert Solution

Expected Probability if prescriptions are filled uniformly over the 7 days of the week = 1/7

Thus, the appropriate null hypothesis for this test is as follows:

H0: p1 = p2 = p3 = p4 = p5 = p6 = p7 = 1/7 (Option A)

Expected count of prescriptions for Monday = Total sample size* Expected Probability = 245*1/7

Expected count of prescriptions for Monday = 35

Chi-square contribution for Monday = (Observed - Expected)^2/Expected = (42-35)^2/35

Chi-square contribution for Monday = 1.4

Degrees of freedom for the chi-square test = No. of Days in a week - 1 = 7 - 1

Degrees of freedom for the chi-square test = 6

Chi Square test statistic is equat to sum of contribution of each day in the week

Chi Sq.statistic = (42-35)^2/35+(31-35)^2/35 +(33-35)^2/35+(29-35)^2/35+(45-35)^2/35+(44-35)^2/35+(21-35)^2/35

Chi Sq.statistic = 13.8 (Option C)

P-value corresponding to df = 6 and Chi Sq.statistic = 13.8 is obtianed using p-value calculator. Screenshot below:

P-value = 0.0323

Since p-value less than 0.05, we reject null hypothesis

Thus, at 0.05 significance level, there is significant evidence that prescriptions are not uniformly distributed over the 7 days of the week.

What can we state about the chi-square test in this situation?

Correct Answer: The test is valid because the sample is random and the expected counts are large enough. (Option C)

Which of the following statements about a chi-square hypothesis test is true?

Correct Answer: All of the above (Option D)

The P-value of the chi-square test is the area to the left of the calculated C2 statistic under this chi-square distribution FALSE


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