Question

A New England Journal of Medicine study (November 1986) found that a substantial portion of acute hospital care is reported to be unnecessary. The physicians who conducted
the study reviewed the medical records of 1,132 patients hospitalized at six different locations across the country. Overall, 60% of admissions in the sample were judged to be appropriate, 23% were deemed inappropriate, whereas 17% could have been avoided by the use of ambulatory surgery. Let p1, p2, and p3 represent the true percentages of hospital admissions in the three aforementioned categories: appropriate, inappropriate, and avoidable by ambulatory surgery, respectively. Using the techniques from categorical data analysis in a “one-way table”, answer the following questions:
(a) Construct 95% confidence intervals for p1, p2, and p3.
(b) Let H0 be the null hypothesis stating that p1=0.65, p2=0.20, p3=0.15. Formulate an
appropriate alternative hypothesis Ha, and then test H0 using α =0.05.

Answer 1

A New England Journal of Medicine study (November 1986) found that a substantial portion of acute hospital care is reported to be unnecessary. The physicians who conducted
the study reviewed the medical records of 1,132 patients hospitalized at six different locations across the country. Overall, 60% of admissions in the sample were judged to be appropriate, 23% were deemed inappropriate, whereas 17% could have been avoided by the use of ambulatory surgery. Let p1, p2, and p3 represent the true percentages of hospital admissions in the three aforementioned categories: appropriate, inappropriate, and avoidable by ambulatory surgery, respectively. Using the techniques from categorical data analysis in a “one-way table”, answer the following questions:
(a) Construct 95% confidence intervals for p1, p2, and p3.

P1= 0.6, p2=0.23, p3=0.17

95% CI for p1 = (0.5715, 0.6285)

95% CI for p2 = (0.2055, 0.2545)

95% CI for p3= (0.1481, 0.1919)

calculations:

Confidence interval - proportion
	p1
	0.9500	confidence level
	0.6000	proportion
	1132.0000	n
	1.9600	z
	0.0285	half-width
	0.6285	upper confidence limit
	0.5715	lower confidence limit
Confidence interval - proportion
	p2
	0.9500	confidence level
	0.2300	proportion
	1132.0000	n
	1.9600	z
	0.0245	half-width
	0.2545	upper confidence limit
	0.2055	lower confidence limit
Confidence interval - proportion
	p3
	0.9500	confidence level
	0.1700	proportion
	1132.0000	n
	1.9600	z
	0.0219	half-width
	0.1919	upper confidence limit
	0.1481	lower confidence limit

(b) Let H0 be the null hypothesis stating that p1=0.65, p2=0.20, p3=0.15. Formulate an
appropriate alternative hypothesis Ha, and then test H0 using α =0.05.

Goodness of Fit Test
	observed	expected	O - E	(O - E)² / E
	0.60*1132 = 679.2000	0.65*1132 = 735.800	-56.600	4.354
	0.23*1132 = 260.3600	0.20*1132 = 226.400	33.960	5.094
	0.17*1132 = 192.4400	0.15*1132 = 169.800	22.640	3.019
Total	1132.0000	1132.000	0.000	12.467
	12.467	chi-square
	2.0000	df
	.0020	p-value

Critical value with 2 df at 0.05 level = 5.991

Calculated chi square = 12.467 > 5.991 the critical value.

Ho is rejected.