Question

.) Is the type of health care provider related to the patient’s satisfaction with care? Data are from a study of 10 patients who were seen for health care at an HMO. Each patient had three different urgent care visits. On one visit, the patient was seen by a physician (MD), at another visit by a nurse practitioner (NP), and at the third visit by a physician’s assistant (PA). After each visit, the patient completed a patient satisfaction survey. Satisfaction was measured by a 10-item scale, which returns scores from 0 to 10, with 0 indicating the least satisfied and 10 indication the most satisfied. The data:

Patient	NP	MD	PA
1	9	7	6
2	9.5	6.5	8
3	5	7	4
4	8.5	8.5	6
5	9.5	5	7
6	7.5	8	6
7	8	6.5	6.5
8	7	6.5	4
9	8.5	7	6.5
10	6	7	3

          Complete the appropriate test by hand. If appropriate,   

         complete a Post Hoc test. (To be done by hand)

Answer 1

One way ANOVA analysis,

From the data values,

.	NP	MD	PA
.	9	7	6
.	9.5	6.5	8
.	5	7	4
.	8.5	8.5	6
.	9.5	5	7
.	7.5	8	6
.	8	6.5	6.5
.	7	6.5	4
.	8.5	7	6.5
.	6	7	3
.	.	.	.
Sum, Xij	78.5	69	57
Average,	7.85	6.9	5.7
\sum_i Xij^2	636.25	484	346.5
	20.025	7.9	21.6
n	10	10	10

Where 'i' represent rows and 'j' represents the column

The degree of freedom values are,

The sum of square values are,

Where,

The means squares are computed using the formula,

The F-statistic value is obtained using the formula,

The F-critical value

The F-critical value is obtained from the F-distribution table for significance level = 0.05, numerator degree of freedom = 2 and denominator degree of freedom = 27

Conclusion,

Since,

The null hypothesis is rejected, hence we can conclude that there is a significant difference among the 4 groups

Post Hoc Test

Now, the Tukey multiple comparisons procedure is used to test all the pairwise comparisons and identify which pair is significantly different.

The Tukey method uses the formula,

Since ni = nj, the HSD value will be the same for each comparison

The q value is obtained using the q distribution table for significance level = 0.05, number of groups, k = 4, degree of freedom = N - k = 40 - 4 = 36.

The HSD value is,

Decision Rule: If,

Now,

Comparison	Difference	HSD
NP-MD	0.95	<	1.502	Not significant
NP-PA	2.15	>	1.502	Significant
MD-PA	1.2	<	1.502	Not significant

Only the pairwise comparison of means is significant between NP and PA.