Question

several months ago while shopping, i was interviewed to see whether or not id be interested in signing up for a subscription to a yoga app. i fall into the category of people who have a membership at a local gym, and guessed that, like me, many people in that category would not be interested in the app. my friend christine falls in the category of people who do not have a membership at a local gym, and i was thinking that she might like a subscription to the app. after being interviewed, i looked at the interviewers results. of the 91 people in my market category who had been interviewed, 14 said they would buy a subscription, and of the 111 people in christines market category, 36 said they would buy a subscription. assuming that these data came from independent, random samples, can we conclude, at the 0.05 level of significance, that the proportion p1 of all mall shoppers in my market category who would buy a subscription is less than the proportion p2 of all mall shoppers in christines market category who would a subscription? perform a one - tailed test. then complete the parts below. carry your intermediate computations to three or more decimal places and round your answers as specified in the parts below. (if necessary, consult a list of formulas.) (a) state the null hypothesis h0 and the alternative hypothesis h1. h0:□ h1:□ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) □ (d) find the p - value. (round to three or more decimal places.) □ (e) can we conclude that the proportion of mall shoppers in my market category who would buy a subscription is less than the proportion in christines market category who would? yes no

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the proportions are equal, i.e., $H_0:p_1 - p_2=0$. The alternative hypothesis $H_1$ is that the proportion in the first - category is less than the proportion in the second category, so $H_1:p_1 - p_2<0$.

Step2: Determine test - statistic type

We are comparing two proportions from independent samples, so we use a two - proportion z - test.

Step3: Calculate sample proportions

Let $n_1 = 91$, $x_1=14$, $n_2 = 111$, $x_2 = 36$. The sample proportions are $\hat{p}_1=\frac{x_1}{n_1}=\frac{14}{91}\approx0.154$, $\hat{p}_2=\frac{x_2}{n_2}=\frac{36}{111}\approx0.324$. The pooled proportion $\hat{p}=\frac{x_1 + x_2}{n_1 + n_2}=\frac{14 + 36}{91+111}=\frac{50}{202}\approx0.248$.

Step4: Calculate the test - statistic

The formula for the two - proportion z - test statistic is $z=\frac{(\hat{p}_1-\hat{p}_2)-0}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}$. Substituting the values:
\[

$$\begin{align*} z&=\frac{(0.154 - 0.324)-0}{\sqrt{0.248\times(1 - 0.248)\times(\frac{1}{91}+\frac{1}{111})}}\\ &=\frac{- 0.17}{\sqrt{0.248\times0.752\times(\frac{111 + 91}{91\times111})}}\\ &=\frac{-0.17}{\sqrt{0.186496\times\frac{202}{10101}}}\\ &=\frac{-0.17}{\sqrt{0.186496\times0.02}}\\ &=\frac{-0.17}{\sqrt{0.00372992}}\\ &=\frac{-0.17}{0.0611}\\ &\approx - 2.782 \end{align*}$$

\]

Step5: Calculate p - value

Since this is a one - tailed test with $z\approx - 2.782$, the p - value is $P(Z < - 2.782)$. Looking up in the standard normal table, the p - value is approximately $0.003$.

Step6: Make a decision

Since the p - value ($0.003$) is less than the significance level $\alpha = 0.05$, we reject the null hypothesis.

Answer:

(a) $H_0:p_1 - p_2 = 0$, $H_1:p_1 - p_2<0$
(b) Two - proportion z - test
(c) $-2.782$
(d) $0.003$
(e) Yes