Question

In: Statistics and Probability

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

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

Solutions

Expert Solution

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


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