Question

We want to test the hypothesis that more than 30% of U.S. households have internet access (with a significance level of 5%). We collect a sample of 150 households and find that 57 have access.

Answer 1

Let the proportion of US households having internet access be P.

Then, Sample proportion (p) = 57/150

= 0.38

Null hypothesis (Ho) : P = 0.30

Alternative hypothesis (H1) : P 0.30

Test statistic is given by-

where, po is the value of population proportion under the null hypothesis = 0.30;

n is sample size = 150

Therefore,

z = 2.162

Now, critical value of z at 5% level of significance for the one tailed test is 1.645 as obtained from the z table.

Since, z calculated > critical value of z, we may reject the null hypothesis. So, proportion of US households having internet access is greater than 30%