Question

In a random sample of 60 shoppers from a large suburban mall, 36 indicated that they had been to a movie in the past month. In an independent random sample of 50 shoppers chosen from the downtown shopping area, 31 indicated that they had been to a movie in the past month. What significance test should be used to determine whether these data provide sufficient evidence to reject the hypothesis that the proportion of suburban mall shoppers who had been to a movie in the same month is the same as the proportion downtown

Answer 1

Answer:

There is no enough statistical evidence to reject the hypothesis that the proportion of suburban mall shoppers who had been to a movie in the same month is the same as the proportion downtown.

Step-by-step explanation:

We have to perform an hypothesis test on the difference of proportions to see if there is enough evidence to reject the null hypothesis.

The null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a: \pi_1-\pi_2\neq0$

The significance level is assumed to be α=0.05

The suburban sample has  a proportion of

$p_1=36/60=0.6$

The downtown sample has a proportion of

$p_2=31/50=0.62$

The average of proportion is:

$p=\frac{n_1p_1+n_2p_2}{n_1+n_2}=\frac{60*0.60+50*0.62}{60+50}=\frac{67}{110}=0.61$

The standard deviation of the difference of proportions is:

$s_{p1-p2}=\sqrt{\frac{p(1-p)}{n_1} +\frac{p(1-p)}{n_2} } =\sqrt{\frac{0.61*0.39}{60} +\frac{0.61*0.39}{50} }\\\\s_{p1-p2}=\sqrt{0.003965+0.004758}=0.0934$

Then, we can calculate the value of the statistic z

$z=\frac{p_1-p_2}{s}=\frac{0.60-0.62}{0.0934}=\frac{-0.02}{0.0934} = 0.214$

The P-value for this z is:

$P(|z|>0.214)=0.83054$

The P-value is bigger than the significance level, so there is no enough evidence to reject the null hypothesis.