Question

1. when 302 college students are randomly selected and surveyed, it is found that 122 own a car (a) find a 99% confidence interval for the true proportion of all college students who own a car. (b) if a study claims that a majority of college students own a car, do you think the claim is true? (0.331, 0.477), no

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 122$ (number of successes) and $n=302$ (sample size).
$\hat{p}=\frac{122}{302}\approx0.404$

Step2: Find $z$-score for 99% confidence interval

For a 99% confidence interval, the significance level $\alpha=1 - 0.99=0.01$, and $\frac{\alpha}{2}=0.005$. The $z$-score $z_{\frac{\alpha}{2}}$ is such that $P(Z>z_{\frac{\alpha}{2}})=0.005$. From the standard normal table, $z_{\frac{\alpha}{2}} = 2.576$

Step3: Calculate the margin of error $E$

The formula for the margin of error for a proportion is $E = z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.404$, $n = 302$, and $z_{\frac{\alpha}{2}}=2.576$
$\hat{p}(1-\hat{p})=0.404\times(1 - 0.404)=0.404\times0.596 = 0.240784$
$\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.240784}{302}}\approx\sqrt{0.0007973}\approx0.0282$
$E=2.576\times0.0282\approx0.0726$

Step4: Calculate the confidence interval

The confidence interval for the population proportion $p$ is $\hat{p}-E$0.404- 0.0726=0.3314$
$0.404 + 0.0726=0.4766$

Answer:

(a) The 99% confidence interval is $(0.331,0.477)$
(b) Since the upper - bound of the 99% confidence interval ($0.477$) is less than $0.5$, we do not have sufficient evidence to support the claim that a majority of college students own a car. So the claim is not true.