Question

Conduct the following test at the alpha = 0.10 level of significance by determining ​(a) the null and alternative​ hypotheses, ​(b) the test​ statistic, and​ (c) the​ P-value. Assume that the samples were obtained independently using simple random sampling.

Test whether p1 not equals p2. Sample data are x1 = 28​, n1 = 254​, x2 = 38​, and n2 = 301.

Answer 1

a) H0: p1 - p2 = 0

H1: p1 - p2 ≠ 0

b) z=-0.58

c) p-value = 0.562

Explanation:

We need to determine whether p1 is not equals p2, so the null and alternative hypothesis are:

H0: p1 - p2 = 0

H1: p1 - p2 ≠ 0

Where p1 and p2 are the proportions of the population. Additionally, the proportions of the sample p1' and p2' are calculated as:

`p1'=(x1)/(n1)=(28)/(254)=0.1102\\p2'=(x2)/(n2)=(38)/(301)=0.1262`

Then, the test statistic is calculated using the following equation:

`z=\frac{(p1'-p2')-(p1-p2)}{\sqrt{p'(1-p')((1)/(n1)+(1)/(n2))} }`

Where p' is calculated as:

`p'=(x1+x2)/(n1+n2)=(28+38)/(254+301)=0.1189`

So, replacing the values, we get that the test statistic is:

`z=\frac{(0.1102-0.1262)-(0)}{\sqrt{0.1189(1-0.1189)((1)/(254)+(1)/(301))}}=-0.58`

Finally, using the standard normal table, the p-value is equal to:

`p-value=2*P(z<-0.58)=2*0.281=0.562`

The p-value is greater that the value of alpha 0.1, so we can't reject the null hypothesis and there is evidence to said that p1 and p2 are equals.