Community college students survey students at their college and ask, "Have you met with a counselor to develop an educational plan?" Of the 25 randomly selected students, 17 have met with a counselor to develop an educational plan. What is the 90% confidence interval for the proportion of all students at the college that have met with a counselor to develop an educational plan? (Answers may vary slightly do to rounding. Round SE to 2 decimal places before calcluating the margin of error.) Group of answer choices 0.53 to 0.83 0.50 to 0.86 0.45 to 0.91 We should not calculate the 90% confidence interval because normality conditions are not met.

We have to develop a 90% confidence interval for the proportion of all students at the college that have met with a counselor to develop an educational plan.

We have a sample of size 25, where the sample proportion is:

[tex]p=X/n=17/25=0.68[/tex]

The critical value of z for a 90% CI is z=1.645.

The margin of error is then:

[tex]E=z\cdot\sqrt{\dfrac{p(1-p)}{n}}=1.645*\sqrt{\dfrac{0.68*0.32}{25}}=1.645*\sqrt{0.008704}=1.645*0.093\\\\\\E=0.15[/tex]

Then, the lower and upper bound of the confidence interval are:

[tex]LL=\hat p-E=0.68-0.15=0.53\\\\UL=\hat p+E=0.68+0.15=0.83[/tex]

The 90% CI for p is (0.53, 0.83)