Question

You are evaluating a new rapid pregnancy test for a local clinic. Urine samples from 40 females who visited the clinic are collected.

By your current method, 38 out of 40 samples are positive.

By the new method, 34 out of 40 samples are positive.

Is there a significant difference in performance between the two methods?

- State the null hypothesis

What is the basis for your answer?

Answer 1

For current method 1, we have that the sample size is N1=40, the number of favorable cases is X1=38, so then the sample proportion is p^1 =X1/N1=38/40=0.95
For sample 2, we have that the sample size is N2=40, the number of favorable cases is X2=34, so then the sample proportion is p^2= X2/N2= 34/40=0.85

The value of the pooled proportion is computed as pˉ = X1+X2/N1+N2= 38+34/40+40= 0.9

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 zc=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| = 1.491 ∣z∣=1.491≤zc=1.96, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.136, and since p=0.136≥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 p1 is different than p2, at the 0.05 significance level.

Basis of my answer is comparing Z cal value with Z critical value at 0.05 (assumed) level of significance.