Question

A 0.05 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender​ selection, the proportion of baby girls is less less than 0.5. Assume that sample data consists of 78 girls in 169 ​births, so the sample statistic of 6/13 results in a z score that is 1 standard deviation below 0. Complete parts​ (a) through​ (h) below.

Answer 1

As we are testing whether the proportion of baby girls is less less than 0.5, therefore the null and the alternate hypothesis here are given as:

The sample proportion here is computed as:

The z score is computed as:

Now as this is a one tailed test, the p-value here is computed form the standard normal tables as:

p = P(Z < -1) = 0.1587

As the p-value here is 0.1587 > 0.05 which is the level of significance, therefore the test is not significant and we cannot reject the null hypothesis here, therefore we don't have sufficient evidence to support the claim that when parents use a particular method of gender​ selection