according to the u.s. census bureau, in the first quarter of 2018, the homeownership rate in the u.s. was…

according to the u.s. census bureau, in the first quarter of 2018, the homeownership rate in the u.s. was 64.2%. researchers believe that this rate is higher for residents of a certain town and decide to to a study. a random sample of 100 households in the town was studied and the homeownership rate in the sample was found to be 73%. answer options assuming that the conditions for conducting the appropriate significance test are satisfied, do the data support the researchers belief that the homeownership rate in this town is greater than the rate reported by the u.s. census bureau? elimination tool select one answer a yes. since the sample proportion is more than 0.05 greater than the proportion reported by the u.s. census bureau, the results are statistically significant at the α = 0.05 level. b yes. since the probability that 73 or more households in the sample will own their own home is 0.024 and since 0.024 < 0.05, the results are statistically significant at the α = 0.05 level. c yes. since the p - value of 0.033 is less than 0.05, the results are statistically significant at the α = 0.05 level.
Answer
Explanation:
Step1: Set up hypotheses
Let $p$ be the true home - ownership rate in the town. The null hypothesis $H_0:p = 0.642$ and the alternative hypothesis $H_1:p>0.642$.
Step2: Calculate the test - statistic
The sample proportion $\hat{p}=0.73$, sample size $n = 100$. The standard deviation of the sampling distribution of the sample proportion under the null hypothesis is $\sigma_{\hat{p}}=\sqrt{\frac{p_0(1 - p_0)}{n}}=\sqrt{\frac{0.642\times(1 - 0.642)}{100}}\approx\sqrt{\frac{0.642\times0.358}{100}}\approx\sqrt{\frac{0.230836}{100}}\approx0.048$. The z - test statistic is $z=\frac{\hat{p}-p_0}{\sigma_{\hat{p}}}=\frac{0.73 - 0.642}{0.048}=\frac{0.088}{0.048}\approx1.83$.
Step3: Calculate the p - value
For a one - tailed test with $z\approx1.83$, the p - value is $P(Z>1.83)=1 - P(Z\leq1.83)$. Looking up in the standard normal table, $P(Z\leq1.83)=0.9664$, so the p - value is $1 - 0.9664 = 0.0336\approx0.033$.
Step4: Make a decision
Since the significance level $\alpha = 0.05$ and the p - value $0.033<0.05$, we reject the null hypothesis.
Answer:
C. Yes. Since the p - value of 0.033 is less than 0.05, the results are statistically significant at the $\alpha = 0.05$ level.