In a previous year, $55 \%$ of females aged 15 and older lived alone. A sociologist tests whether this percentage is different today by conducting random sample of 650 females aged 15 and older and finds that 365 are living alone. Is there sulficient evidence at the $\alpha=0.1$ level of significance conclude the proportion has changed? Because $n p_{0}\left(1-p_{0}\right)=160.9>10$, 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. \[ \mathrm{H}_{0}: \mathrm{p}=.55 \text { versus } \mathrm{H}_{1}: \mathrm{p} \not=.55 \] (Type integers or decimas. Do not round.) Find the test statistic for this hypothesis test. \[ L_{0}=\square \] (Round to two decinal places as needed)

Step 1 ：The problem is asking whether the proportion of females aged 15 and older living alone has changed from a previous year, where it was 55%. A sociologist conducted a random sample of 650 females aged 15 and older and found that 365 are living alone. We are asked to test this at the 0.1 level of significance.

Step 2 ：The null hypothesis for this test is \(H_{0}: p = 0.55\), which states that the proportion of females aged 15 and older living alone is still 55%. The alternative hypothesis is \(H_{1}: p \neq 0.55\), which states that the proportion has changed.

Step 3 ：The test statistic for a proportion hypothesis test is calculated using the formula \(Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\), where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and \(n\) is the sample size.

Step 4 ：In this case, \(\hat{p} = \frac{365}{650} = 0.5615384615384615\), \(p_0 = 0.55\), and \(n = 650\).

Step 5 ：Substituting these values into the formula, we get \(Z = \frac{0.5615384615384615 - 0.55}{\sqrt{\frac{0.55(1-0.55)}{650}}} = 0.5913123959890805\).

Step 6 ：Final Answer: The test statistic for this hypothesis test is approximately \(\boxed{0.59}\).