Question

We survey a random sample of American River College students and ask if they drink coffee on a regular basis. The 90% confidence interval for the proportion of all American River College students who drink coffee on a regular basis is (0.262, 0.438). What will be true about the 95% confidence interval for these data

Answer 1

Answer:

The confidence interval is given for this case $(0.262,0.438)$

$0.262 \leq p \leq 0.438$

And we can conclude that the true proportion of all American River College students who drink coffee on a regular basis is between 0.262 and 0.438 with a confidence of 0.95

Step-by-step explanation:

For this case we want to determine the confidence interval for the proportion of all American River College students who drink coffee on a regular basis and is given by thsi general formula:

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

The confidence level for this case is 0.95 and the significance level is $\alpha=0.05$ and the critical value for this case is $z_{\alpha/2}=1.96$

The confidence interval is given for this case $(0.262,0.438)$

$0.262 \leq p \leq 0.438$

And we can conclude that the true proportion of all American River College students who drink coffee on a regular basis is between 0.262 and 0.438 with a confidence of 0.95