Respuesta :
Answer:
a) [tex]0.530 - 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.47615[/tex]
[tex]0.530 + 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.58385[/tex]
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:
[tex]\hat p = \frac{175}{330}=0.530[/tex]
Part a
The confidence level is 95% , and the significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical values would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The confidence interval for the true proportion is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
Replacing the info given we got:
[tex]0.530 - 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.47615[/tex]
[tex]0.530 + 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.58385[/tex]
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
