In: Statistics and Probability
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?
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