Question

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

a. 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

b. 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).

c. 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).

d. 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 _______________. (show calculation).

e. 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)?

a. 1/7

b. 3.5

c. 13.8

d. 24.5

f. 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.

g. 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.

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

a. The test is valid because the sample size is large.

b. The test is valid because the sample is random and the observed counts are large enough.

c. The test is valid because the sample is random and the expected counts are large enough.

d. The test is not valid because we do not know the true population proportions.

i. 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

j. Under which of the following conditions can a large P- value arise?

1. H₀ is indeed true.
2. H₀ is not actually true, but too close to the real population distribution for us to tell them apart statistically.
3. H₀ is definitely not true, but the sample size is too small or the variability is too great to reach significance.
4. All of the above

Answer 1

observed frequencey, O	expected proportion	expected frequency,E	(O-E)	(O-E)²	(O-E)²/E
42	0.143	35.00	7.00	49.00	1.400
31	0.143	35.00	-4.00	16.00	0.457
33	0.143	35.00	-2.00	4.00	0.114
29	0.143	35.00	-6.00	36.00	1.029
45	0.143	35.00	10.00	100.00	2.857
44	0.143	35.000	9.00	81.00	2.314
21	0.143	35.000	-14.00	196.00	5.600

a) H0: p1 = p2 = p3 = p4 = p5 = p6 = p7 = 1/7

b) b. 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 ____1/7*42=35_________

c)

Under the null hypothesis of a uniform distribution of prescriptions over the 7 days of the week, the chi-square contribution for Monday is _______1.40__________ (show calculation).

d. 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 _____6__________. (show calculation).

e. 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)? = 13.8

f)

P value =   0.032   [ excel function: =chisq.dist.rt(test-stat,df) ]
Decision: P value < α, Reject Ho      
      
g) There is significant evidence that prescriptions are not uniformly distributed over the 7 days of the week.

h)

.

c. The test is valid because the sample is random and the expected counts are large enough.

i)

When observed counts are far from expected counts, we have evidence against H0.
Large values of χ2 indicate evidence against H0.