Question

• 2-1 Your organization collects data on individual patients shown in Appendix Table 2. Identify whether each variable is measured nominally, ordinally, or as an interval/ratio variable.
• 2-2 What statistical measures would you use to summarize the variable for age? What about for gender? For convenience satisfaction? How would you present these graphically?
• 2-3 If you are interested in whether satisfaction scoring differed by the amount the individual paid as a co-pay, how would you state this inquiry as testable hypotheses (the null and alternative)? What statistical test would you run to test this hypothesis?
• 2-4 Perform the test defined in Table 2-3. State your conclusions.
• 2-5 Compare gender and having Rx coverage. State your hypotheses. Perform the appropriate statistical test and interpret.

Appendix Table 1 Chi-Square Values by Alpha Level and Degrees of Freedom

Df	0.5	0.1	0.05	0.02	0.01	0.001
1	0.455	2.706	3.841	5.412	6.635	10.827
2	1.386	4.605	5.991	7.824	9.210	13.815
3	2.366	6.251	7.815	9.837	11.345	16.268
4	3.357	7.779	9.488	11.668	13.277	18.465
5	4.351	9.236	11.070	13.388	15.086	20.510

Appendix Table 2 Patient Satisfaction, by Insurer, Same Day of Appointment, Prescription Coverage and Level of Co-pay

Gender	Age	Convenience Satisfaction	Insurer	Same day appointment?	Rx coverage	Co-pay ($)
M	22	5	Select	Y	Y	15
M	24	4	Select	Y	N	15
F	45	5	Select	Y	Y	10
F	38	5	Select	Y	Y	10
F	48	3	Select	Y	N	10
F	50	4	Select	Y	Y	20
M	67	4	Medicare	Y	Y	5
F	23	5	Tri-state	N	Y	10
F	19	5	Tri-state	N	Y	15
M	14	3	Tri-state	Y	Y	20
F	27	5	Reliant	Y	N	15
M	33	3	Reliant	N	N	0
F	39	4	Reliant	N	N	5
F	47	4	Tri-state	N	Y	5
M	42	5	Tri-state	N	N	5
M	31	5	Reliant	N	N	15
M	20	4	Reliant	N	Y	15
F	72	5	Medicare	Y	Y	15
F	44	4	Tri-state	N	N	20
M	45	5	Select	Y	N	20
F	60	3	Reliant	Y	Y	20
M	63	5	Tri-state	Y	Y	5
F	27	5	Tri-state	Y	Y	5
M	68	5	Medicare	N	Y	5
F	33	5	Reliant	N	Y	10
F	38	3	Reliant	N	Y	15
F	55	4	Medicare	N	Y	15
M	51	5	Select	Y	Y	15
F	48	4	Tri-state	Y	Y	20
M	49	5	Tri-state	Y	Y	0
M	55	5	Select	N	Y	5
F	61	4	Reliant	Y	N	0
F	23	5	Medicare	Y	N	0
M	69	5	Medicare	N	Y	20
F	41	4	Tri-state	Y	N	5
F	14	4	Medicare	Y	N	0

Answer 1

2-1

Gender-nominally

Age-ordinally

Convenience Satisfaction-ordinally

Insurer-nominally

Same day apt-nominally

Rx coverage-nominally

Co-pay ($)-interval/ratio

2-2

a) we will use a bar graph to summarise the variable of age.

group	age range	no of patients
1	0-10	0
2	11 to 20	4
3	21-30	6
4	31-40	6
5	41-50	10
6	51-60	4
7	61-70	5
8	71-80	1

b) for gender

for convenience satisfaction, again using a bar graph

2-3

to test whether satisfaction scoring differed by the amount the individual paid as a co-pay, the following hypothesis will be made

H₀: Satisfaction scoring doesn’t differ because of the amount the individual pays as a co-pay.

H₁: Satisfaction scoring does differ by the amount the individual pays as a co-pay.

we will use the chi-square test to test this difference in the satisfaction

2-4

Step 1.LAY OUT THE DATA IN A TABLE

OBSERVED FREQUENCY
CO PAY\ SATISFACTION LEVEL	3	4	5
0	1	2	2
5	0	4	5
10	1	0	4
15	1	3	6
20	2	3	2

Step 2.  Select the appropriate test statistic.

The formula for the test statistic is:

In the test statistic, O = observed frequency and E=expected frequency in each of the response categories

The condition for appropriate use of the above test statistic is that each expected frequency is at least 5. In Step 4 we will compute the expected frequencies and we will ensure that the condition is met.

Step3: Add up rows and columns:

CO PAY\ SATISFACTION LEVEL	3	4	5
0	1	2	2	5
5	0	4	5	9
10	1	0	4	5
15	1	3	6	10
20	2	3	2	7
	5	12	19	36

Which gives us:
Step4: Calculate "Expected Value" for each entry
Multiply each row total by each column total and divide by the overall total

EXPECTED FREQUENCY
CO PAY\ SATISFACTION LEVEL	3	4	5
0	0.6944	1.666	2.63
5	1.25	3	4.75
10	0.69	1.67	2.63
15	1.38	3.33	5.27
20	0.97	2.34	3.694

now Subtract expected from observed, square it, then divide by expected:

In other words, use formula (O−E)² /E where

• O = Observed (actual) value
• E = Expected value

Which gets us the respective chi-square values

CHI SQUARE VALUES
COPAY \ SATISFACTION LEVEL	3	4	5
0	0.134444	0.0666667	0.154678
5	1.25	0.333333	0.0131579
10	0.134444	1.66667	0.702047
15	0.108889	0.0333333	0.0988304
20	1.08651	0.190476	0.777151

WE ADD ALL THE Chi-Squared Values of the table

Chi-Square is = 6.75063

Step 5. Set up decision rule.

The decision rule depends on the level of significance and the degrees of freedom, defined as df = (r-1)(c-1), where r and c are the numbers of rows and columns in the two-way data table.

The row variable is the cut off pay which is 5, thus r=5. The column variable is satisfaction level and 3 responses are considered, thus c=3. For this test, df=(5-1)(3-1)=8.

With tests, there are no upper, lower, or two-tailed tests. If the null hypothesis is true, the observed and expected frequencies will be close in value and the statistic will be close to zero. If the null hypothesis is false, then the statistic will be large.

this value is found using  chi-square table

The rejection region for the χ² test of independence is always in the upper (right-hand) tail of the distribution. For df=8 and a 5% level of significance, the appropriate critical value is 15.51 and the decision rule is as follows: Fail to reject H₀ if chi ²  <15.51

Step 6. Conclusion

since chi square= 6.75< 15.51

Thus we fail to reject the null hypothesis at 5% level of significance and hence we conclude that Satisfaction scoring doesn’t differ because of the amount the individual pays as a co-pay.