Question

In a survey of 700 community college students, 481 indicated that they have read a book for personal enjoyment during the school year (based on data from the Community College Survey of Student Engagement).

Construct a 90% confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year. Explain your steps, and write how you would interpret this confidence interval.

Answer 1

To construct a 90% confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year, we can use the formula CI = p +- z * sqrt((p * (1-p))/n), where p is the sample proportion, z is the z-score corresponding to the desired confidence level, and n is the sample size. Plugging in the given values, we find that the 90% confidence interval is approximately (0.656, 0.718).

Step-by-step explanation:

To construct a 90% confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year, we can use the formula:

CI = p +- z * sqrt((p * (1-p))/n)

where:

• CI is the confidence interval
• p is the sample proportion
• z is the z-score corresponding to the desired confidence level
• n is the sample size

Using the given information:

• p = 481/700 = 0.687
• z = 1.645 (corresponding to a 90% confidence level)
• n = 700

Plugging in these values, we get:

CI = 0.687 +- 1.645 * sqrt((0.687 * (1-0.687))/700)

Simplifying the equation, the 90% confidence interval is approximately (0.656, 0.718). This means that we can be 90% confident that the true proportion of community college students who have read a book for personal enjoyment during the school year falls within this range.