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 the 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 } = 0.49 \\)

(round to two decimal places as needed.)

determine the p - value for this hypothesis test.

\\( p \text { - value } = \square \\)

(round to three decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}=\frac{330}{600} = 0.55$.

Step2: Recall the formula for the P - value in a two - tailed z - test

For a two - tailed z - test, the P - value is $P = 2\times(1 - \Phi(|z_0|))$, where $\Phi$ is the cumulative distribution function of the standard normal distribution and $z_0$ is the test statistic.
Given $z_0=0.49$.
We know that $\Phi(0.49)$ can be found using a standard normal table or a calculator with a normal distribution function. Using a standard normal table or calculator, $\Phi(0.49)=0.6879$.

Step3: Calculate the P - value

Substitute into the formula:
$P = 2\times(1 - 0.6879)=2\times0.3121 = 0.6242\approx0.624$

Answer:

$0.624$