Question

the manager at a sports radio station will be covering several football games over the weekend. she knows that most of her listeners are at least 22 years old and wants to know what age group she should gear her advertisements to for the games. she takes a random sample of 46 listeners from age 22 - 39 and 57 listeners who are 40 and older and asks them if they are likely to tune in to football games. the \yes\ responses are recorded below.

	age 22 - 39 (group 1)	age 40+ (group 2)
sample size n	46	57

at the 0.05 level of significance, we are attempting to investigate if there is a significant difference in the proportion of listeners based on age.
enter the p - value - round to 4 decimal places.
p - value =

Explanation:

Step1: Calculate sample proportions

Let $n_1 = 46$, $x_1=32$, $n_2 = 57$, $x_2 = 47$.
The sample proportion for group 1 is $\hat{p}_1=\frac{x_1}{n_1}=\frac{32}{46}\approx0.6957$.
The sample proportion for group 2 is $\hat{p}_2=\frac{x_2}{n_2}=\frac{47}{57}\approx0.8246$.
The pooled - proportion $\hat{p}=\frac{x_1 + x_2}{n_1 + n_2}=\frac{32+47}{46 + 57}=\frac{79}{103}\approx0.7670$.

Step2: Calculate the test - statistic

The test - statistic for testing the difference between two proportions is $z=\frac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}$.
Substitute the values:
\[

$$\begin{align*} z&=\frac{0.6957 - 0.8246}{\sqrt{0.7670\times(1 - 0.7670)\times(\frac{1}{46}+\frac{1}{57})}}\\ &=\frac{- 0.1289}{\sqrt{0.7670\times0.2330\times(\frac{57 + 46}{46\times57})}}\\ &=\frac{-0.1289}{\sqrt{0.1787\times\frac{103}{2622}}}\\ &=\frac{-0.1289}{\sqrt{0.1787\times0.0393}}\\ &=\frac{-0.1289}{\sqrt{0.007023}}\\ &=\frac{-0.1289}{0.0838}\\ &\approx - 1.54 \end{align*}$$

\]

Step3: Calculate the p - value

Since this is a two - tailed test, the p - value is $2\times P(Z\lt|z|)$.
$P(Z\lt1.54) = 0.9382$ (from standard normal table).
The p - value $=2\times(1 - 0.9382)=2\times0.0618 = 0.1236$.

Answer:

$0.1236$