In: Statistics and Probability
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 90% confidence intervals for p1, p2, and p3.
(b) Let H0 be the null hypothesis stating that p1=0.60, p2=0.25, p3=0.15. Formulate an appropriate alternative hypothesis Ha, and then test H0 using α =0.10.
(a) Construct 90% confidence intervals for p1, p2, and p3.
n = 1132
Sample proportion is denoted by
This is a binomial proportion so we will use normal approximation since n > 30 so it is large.
The (1 - )% confidence interval for population proportion
=1 - 0.90 = 0.10
Therefore the critical value at
=
= 1.6449 .............using normal percentage tables with p = 0.05
Category | Sam Pro | C.V. | Lower L | Upper L |
Appropriate | 60% | 1.6449 | 0.5760 | 0.6240 |
Inappropriate | 23% | 1.6449 | 0.2094 | 0.2506 |
Avoidable | 17% | 1.6449 | 0.1516 | 0.1884 |
(b) Let H0 be the null hypothesis stating that p1=0.60, p2=0.25, p3=0.15. Formulate an appropriate alternative hypothesis Ha, and then test H0 using α =0.10.
Again we can make use of a table. Here the alternative hypothesis will be that the proportion is not equal to the given values
For the binomial test we have the following procedure
(The population proportion is )
(The population proportion is not )
Test Stat :
= 10% (level of significance
Therefore the critical value at
=
= 1.6449 .............using normal percentage tables with p = 0.05
Decision critieria: Reject the null hypothesis if |test Stat| > C.V.
Category | Sam Pro | Test Stat | Absolute TS | C.V. | Decision | Conclusion | |
Appropriate | 0.6 | 60% | 0.00 | 0.00 | 1.6449 | Not reject (T.S. < C.V.) | The population is significantly 0.6 |
Inappropriate | 0.25 | 23% | -1.60 | 1.60 | 1.6449 | Not reject (T.S. < C.V.) | The population is significantly 0.25 |
Avoidable | 0.15 | 17% | 1.79 | 1.79 | 1.6449 | Reject (T.S. > C.V.) | The population is not significantly 0.15 |