What is the p-value of this test?

Question

In a random sample of 90 people, 8 of them are left-handed. Let pp be the true proportion of left-handed people in the population. Does the sample indicate that pp is higher than 0.05? Use a 0.012 level of significance.

Answer 1

The sample proportion is 8/90=0.089. The population proportion is 0.05.

The difference is 0.039. The null hypothesis is that the population proportion is 0.05. The alternative hypothesis is that the sample proportion is significantly different from 0.05. The significance level is 0.012 and we need to apply a 2-tail test, so 0.006 is in the left tail and 0.06 is in the right tail, making the non-reject region 1-0.012=0.988 between the left and right tails.

The critical Z value for this is 2.257. 

We need to divide the 0.039 difference by √(0.05(1-0.05)/90) where 90 is the sample size. 0.039/0.023, Z=1.70 approx. This is below the critical value so we don’t reject 0.05 as the population proportion, given the significance level of 0.012.