Question

In a sample of 120 randomly selected people in City A, 5% prefer eating salad after the main course. In a sample of 120 randomly selected people in City B, 8% prefer eating salad after the main course. Check that the conditions are met for a hypothesis test to compare the population proportion of people that prefer eating salad after the main course in City A to the proportion in City B.

Answer 1

Answer:

There is no enough evidence that the proportions are different.

Step-by-step explanation:

We have to perform a hypothesis test on the difference of proportions.

In this case, the sample size is equal.

The null and alternative hypothesis are

$H_0: \pi_1=\pi_2\\\\ H_1: \pi_1\neq\pi_2$

The significance level is assumed to be 0.05.

The weighted average p, as the sample sizes are the same, is the average of both proportions:

$p=\frac{p_1+p_2}{2} =\frac{0.05+0.08}{2}=0.065$

The standard deviation is

$s=\sqrt{\frac{2p(1-p)}{n} } =\sqrt{\frac{2*0.065(1-0.065)}{120}} =0.032$

The z-value for this sample is:

$z=\frac{p_1-p_2}{s} =\frac{0.05-0.08}{0.032} =-0.9375$

The P-value for z=-0.9375 is P=0.3485.

The P-value (0.35) is greater than the significance level (0.05), so the null hypothesis is failed to reject.

There is no enough evidence that the proportions are different.