Question

In: Statistics and Probability

A study of brain surgery patient who survived 6 months after the surgery found that 22...

A study of brain surgery patient who survived 6 months after the surgery found that 22 of 54 (22/54) men and 22 of 68 (22/68) women lived in an assisted living facility. Can we say that we are 95% confident that the underlaying population proportion are not the same between men and women lived in an assisted living facility? [Note: Write hypothesises for your test and proof if you can reject them].

Solutions

Expert Solution

For sample 1, we have that the sample size is N_1= 5

, the number of favorable cases is X_1 = 22,

so then the sample proportion is

For sample 2, we have that the sample size is N_2 = 68,

the number of favorable cases is X_2 = 22,

so then the sample proportion is

The value of the pooled proportion is computed as

Also, the given significance level is α=0.05.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: p1​=p2​

Ha: p1​̸​=p2​

This corresponds to a two-tailed test, for which a z-test for two population proportions needs to be conducted.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is

z_c = 1.96.

The rejection region for this two-tailed test is

R={z:z>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that

z=0.958≤zc​=1.96,

it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p = 0.3379,

and since p=0.3379≥0.05,

it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion Men​ is different than Woman​, at the 0.05 significance level.

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