Question

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

2. 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).

3. 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).

4. 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 _______________.

5. 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

            245/7 = 35

6. 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.

7. Using a significance level of 0.05, what is the appropriate conclusion for this test?

1. All 7 days of the week have different prescription rates.
2. There is significant evidence that prescriptions are not uniformly distributed over the 7 days of the week.
3. Weekdays and weekends have significantly different prescription rates.
4. The data are consistent with prescriptions being uniformly distributed over the 7 days of the week.

8. 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.

Answer 1

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

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 245/7 = 35

since sum of all observed frequencies 42+31+33+29+45+44+21 = 245

Under the null hypothesis of a uniform distribution of prescriptions over the 7 days of the week, the chi-square contribution for Monday is (42-35)^2 / 35 = 1.40

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 7-1 = 6

From the given data

	Observed	Expected
Day	Freq (Oi)	Freq Ei	(Oi-Ei)^2 /Ei
Monday	42	35.00	1.4000
Tuesday	31	35.00	0.4571
Wednesday	33	35.00	0.1143
Thursday	29	35.00	1.0286
Friday	45	35.00	2.8571
Saturday	44	35.00	2.3143
Sunday	21	35.00	5.6000
Total:	245	245	13.7714

the chi-square statistic for testing this null hypothesis of a uniform distribution of prescriptions over the 7 days of the week is 13.7714

P-Value: 0.0323

since P-value < alpha 0.05 so we reject H0

thus we conclude that Option (B) There is significant evidence that prescriptions are not uniformly distributed over the 7 days of the week.

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