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

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

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{(\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