Answer:
The confidence interval is given for this case [tex] (0.262,0.438)[/tex]
[tex]0.262 \leq p \leq 0.438[/tex]
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:
[tex] \hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
The confidence level for this case is 0.95 and the significance level is [tex]\alpha=0.05[/tex] and the critical value for this case is [tex]z_{\alpha/2}=1.96[/tex]
The confidence interval is given for this case [tex] (0.262,0.438)[/tex]
[tex]0.262 \leq p \leq 0.438[/tex]
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
