Question

In: Statistics and Probability

In a survey of post-operative infection, 25 surgical patient records are randomly sampled at the Minto...

In a survey of post-operative infection, 25 surgical patient records are randomly sampled at the Minto Hospital. Of these, 21 of the patients were found to have not suffered from a post-operative infection. Nationally, 94% of surgeries do not result in the patient suffering any post-operative infections.

(a) Briefly describe the population, random variable, and parameter associated with this survey.

(b) Carry out a test to see if there is any evidence that the Minto Hospital deviates from the national rate of post-operative non-infection. If the null hypothesis is true and noInf is the number of records showing no post-operative infection in a sample of size 25, then

P r[noInf < 21] = 0.015

Pr[noInf = 21] = 0.045

Pr[noInf > 21] = 0.940

(c) Which of the two types of error (I or II) could you potentially be making with your conclusion from (b)? Explain what the error would be, in words in the context of the test.

(d) Explain the meaning of the p-value found in (b), in the context of 25,000 imag- inary civil servants all independently taking random samples of 25 surgical pa- tient records from the Minto Hospital.

(e) If you were told to do this test at level 0.05, what would the outcome be? Explain briefly, using the p-value and the level.

(f) Suppose that you are planning to repeat the survey and that you want to do the test with the same hypotheses at level 0.1. Also, suppose that you want to get at least 80% power in the test for finding deviances from 0.94 which could be as small as 0.03. What is the smallest sample size that you should use?

(g) If failing to discover that the post-operative infection rate was different from the national average can cost the hospital a great deal of money, is 0.05 the best level to be using for the test in (c)? Explain briefly why it is, or suggest another level and explain why that would be better.

Solutions

Expert Solution

We are given that out of 25 patients 21 do not suffer from post operative infection.

Thus, the sample proportion is 21/25 = 0.84

It is also given that the national population proportion is 0.94

So we need to test the following hypothesis:

H0: Minto hospital has the same rate as the national rate of post operative non-infection. p1 = p0

H1: Minto hospital deviates from the national rate of post operative non-infection. p1 is not equal to p2

a) The population in this case is the national population whose rate has been considered for analysis.

The random variable is the number of cases of post operative non-infection.

The parameter associated with the survey is the rate of post operative non-infection.

b) Under the null hypothesis the test statistic is:

z = (p0 - p1) / σ

where σ = sqrt[ p1 * ( 1 - p1 ) / n ]

Thus the value of sigma is 0.073321

Hence the value of the test statistic is:

z = 1.363862

The corresponding p-value of the test statistic is 0.086306

Depending on the level of significance we may accept or reject the null hypothesis. If we consider a level of significance of 5% then the null hypothesis may be accepted and we may conclude that the Minto hospital may have the same rate as the national rate of post-operative non-infection.

If one considers the null hypothesis to be true then in that case the required probability is given as:

Hence we may calculate

P(noinf<21) = 0.015

P(noinf = 21) = 0.045

P(noinf > 21) = 0.94

c) From the conclusion in (b) one is accepting the null hypothesis which indicates to a possibility of type II error that is the error of accepting the false null hypothesis.

In the context of the survey it is accepting that the rate of the post operative non-infection in Minto hospital is the same as the national rate, which in reality is far from the national average.

d) The p-value found in (b) is 0.086306. It is basically the probability of how much the observed results are true under the null hypothesis. In the context of 25000 imaginary civil servants taking independent samples of size 25 this means that out of these 25000

25000*0.086306 = 2157.65 ~ 2158 of the civil servants will find samples that will result in the post operative non-infection rate same as the national or the population rate.


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