Question

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.

Answer 1

To calculate the 90% confidence interval for the proportion of all students at the college who have met with a counselor to develop an educational plan, divide the number of students who have met with a counselor by the total number of students in the sample. Calculate the standard error and the margin of error using the given formulas. Finally, construct the confidence interval.

Step-by-step explanation:

To calculate the confidence interval for the proportion of all students at the college who have met with a counselor to develop an educational plan, we can use the formula:

Confidence interval = sample proportion ± margin of error

The sample proportion is calculated by dividing the number of students who have met with a counselor by the total number of students in the sample:

Sample proportion = Number of students who have met with a counselor / Total number of students in the sample

The margin of error can be calculated using the formula:

Margin of error = Z * SE

Z is the z-score corresponding to the desired level of confidence, which is 90% in this case. SE is the standard error, which can be calculated using the formula:

SE = sqrt((sample proportion * (1 - sample proportion)) / sample size)

Plugging in the given values:

Sample proportion = 17 / 25 = 0.68

SE = sqrt((0.68 * (1 - 0.68)) / 25) = 0.09 (rounded to 2 decimal places)

Using a standard normal distribution table, the z-score for a 90% confidence level is approximately 1.645.

Therefore, the margin of error = 1.645 * 0.09 = 0.15 (rounded to 2 decimal places)

The confidence interval can now be calculated as:

Confidence interval = 0.68 ± 0.15

So, the 90% confidence interval for the proportion of all students at the college who have met with a counselor to develop an educational plan is approximately 0.53 to 0.83.