Question

You supervise two outpatient facilities and want to compare the patient satisfaction at each. You decide that a score of 8.5, or better, out of 10 will be considered “good satisfaction”, and a score of less than 8.5, out of 10, will be considered “poor satisfaction”. You conduct a survey of patients at each facility over the same three month interval and gather the following data:

	# with average score 8.5 or higher	# with average score 8.4 or lower	Total
Facility #1	437	154	591
Facility #2	353	164	517
Total	790	318	1,108

You perform a chi square test to determine whether the differences in scores are independent of the facility from which they came.

You will have to calculate the expected values for each of the cells (lecture 14, slides 4, 5). It would be nice if Excel could calculate the expected values from the actual data … but it doesn’t L.

In Excel, create two 2x2 tables; one with the actual values and one with the expected values. Use the CHISQ.TEST function to calculate the chi square value for the data.

When selecting the “Actual range” and the “Expected range” for your data entry, select the 2x2 values (not including the totals) for *both* of the facilities for the actual range and select the 2x2 values (not including the totals) for *both* of the facilities for the expected range.

E.g., for the actual range, you’d select the bolded cells:

	# with average score 8.5 or higher	# with average score 8.4 or lower	Total
Facility #1	437	154	591
Facility #2	353	164	517
Total	790	318	1,108

Similarly, for the expected range (selected from the table of expected values that you created).

The result that Excel displays when you run the chi square test is the p-value. You don’t have to do anything else. ]

1. Using Excel, determine whether the results are independent of the facility from which they came. Copy and paste your Excel data and result. (Just copy and paste the portion of the Excel page that has your work on it; do not copy and paste the entire Excel page … they’re huge!) (10)
2. What is your conclusion about patient satisfaction at each of the facilities? (5)

Answer 1

Chi square test using excel:

	# with average score 8.4 or lower	Total					# with average score 8.4 or lower	Total
# with average score 8.5 or higher						# with average score 8.5 or higher
Facility #1	437	154	591		Facility #1	421.3809	169.6191	591
Facility #2	353	164	517		Facility #2	368.6191	148.3809	517
Total	790	318	1,108		Total	790	318	1108
			p value=	0.037597

Since p value is less than alpha=0.05. Hence we reject null hypothesis at 5% level of significance and conclude that patients satisfaction depends upon facilities provided.