Question

In a sample of 330 randomly selected PCC students it was found that 175 of them have a car. Assume this sample comes from a population that is normally distributed. a. Calculate the EBP then construct a 95% confidence interval estimate of the percentage of PCC students have a car (round the values of the interval to three decimal places). b. Based on your confidence interval, can we conclude that most students at PCC own a car?

Answer 1

Answer:

a) $0.530 - 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.47615$

$0.530 + 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.58385$

And if we convert this interval into % we got (47.615%, 58.385%)

b) For this case since the lower value for the confidence interval is a value lower than 0.5 (50%) we don't have enough evidence to conclude at 10% of significance than  most students at PCC own a car

Step-by-step explanation:

We can begin calculating the best estimator for the true proportion of students at PCC who own a car:

$\hat p = \frac{175}{330}=0.530$

Part a

The confidence level is 95% , and the significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical values would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the true proportion is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing the info given we got:

$0.530 - 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.47615$

$0.530 + 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.58385$

And if we convert this interval into % we got (47.615%, 58.385%)

Part b

For this case since the lower value for the confidence interval is a value lower than 0.5 (50%) we don't have enough evidence to conclude at 10% of significance than  most students at PCC own a car