In: Statistics and Probability
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 |
What is the chi-square statistic for testing this null hypothesis of a uniform distribution of prescriptions over the 7 days of the week? ********************PLEASE SHOW WORK!!**********************
a. 1/7
b. 3.5
c. 13.8
d. 24.5
Solution:
Here, we have to find the expected count of prescriptions for Monday.
We are given that the number of prescriptions distributed uniformly among the seven days of the week.
Sum of all frequencies = 42+31+33+29+45+44+21 = 245
Total number of days = 7
Expected count of prescriptions for all days = 245/7 = 35
Test statistic formula is given as below:
Chi square = ∑[(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
Calculation tables for test statistic are given as below:
Day |
O |
E |
(O - E)^2/E |
Monday |
42 |
35 |
1.4000 |
Tuesday |
31 |
35 |
0.4571 |
Wednesday |
33 |
35 |
0.1143 |
Thursday |
29 |
35 |
1.0286 |
Friday |
45 |
35 |
2.8571 |
Saturday |
44 |
35 |
2.3143 |
Sunday |
21 |
35 |
5.6000 |
Total |
245 |
245 |
13.7714 |
Test Statistic = Chi square = ∑[(O – E)^2/E] = 13.7714
χ2 statistic = 13.8
Answer: c. 13.8