Question

a poll of 1132 teens aged 13 to 17 showed that 55% of them have made new friends online. use a 0.05 significance level to test the claim that half of all teens have made new friends online. use the p - value method. use the normal distribution as an approximation to the binomial distribution
let p denote the population proportion of all teens aged 13 to 17 who have made new friends online. identify the null and alternative hypotheses
$h_0:p = 0.5$
$h_1:p
eq0.5$
(type integers or decimals. do not round.)
identify the test statistic
$z=square$
(round to two decimal places as needed.)

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=0.55$, sample size $n = 1132$, and hypothesized proportion $p_0=0.5$.

Step2: Calculate the test - statistic formula

The formula for the test - statistic in a proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$.
Substitute the values: $\hat{p}=0.55$, $p_0 = 0.5$, $n=1132$.
First, calculate the denominator: $\sqrt{\frac{0.5\times(1 - 0.5)}{1132}}=\sqrt{\frac{0.25}{1132}}\approx\sqrt{0.000221}\approx0.0149$.
Then, calculate the numerator: $0.55 - 0.5=0.05$.
Now, $z=\frac{0.05}{0.0149}\approx3.36$.

Answer:

$z = 3.36$