Question

in a previous year, 54% of women aged 15 and older lived alone. a sociologist tests whether this percentage is different today by conducting a random sample of 600 women aged 15 and older and finds that 330 are living alone. is there sufficient evidence at the \\( \alpha = 0.05 \\) level of significance to conclude the proportion has changed?
determine whether all of requirements for the hypothesis test are satisfied. complete the following sentence
because \\( n p _ { 0 } ( 1 - p _ { 0 } ) = 149.0 > 10 \\), the sample size is less than 5% of the population size, and the sample is a random sample, all of the requirements for testing the hypothesis are satisfied
(round to one decimal place as needed)
identify the null and alternative hypotheses for this test.
\\( h _ { 0 } : p = 0.54 \\) versus \\( h _ { 1 } : p \
eq 0.54 \\)
(type integers or decimals. do not round)
find the test statistic for this hypothesis test.
\\( z _ { 0 } = \square \\)
(round to two decimal places as needed)

Explanation:

Step1: Calculate the sample proportion $\hat{p}$

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 330$ and $n=600$. So, $\hat{p}=\frac{330}{600}=0.55$.

Step2: Calculate the test statistic $z_{0}$

The formula for the test statistic in a one - sample proportion test is $z_{0}=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}(1 - p_{0})}{n}}}$.
Here, $p_{0}=0.54$, $n = 600$, and $\hat{p}=0.55$.
First, calculate the denominator: $\sqrt{\frac{0.54\times(1 - 0.54)}{600}}=\sqrt{\frac{0.54\times0.46}{600}}=\sqrt{\frac{0.2484}{600}}\approx\sqrt{0.000414}\approx0.0203$.
Then, calculate the numerator: $\hat{p}-p_{0}=0.55 - 0.54=0.01$.
So, $z_{0}=\frac{0.01}{0.0203}\approx0.49$.

Answer:

$z_{0}\approx0.49$