In a study of 200 students, 42 were found to have a GPA greater than 3.0. Find a

98% confidence interval estimate of the proportion, p, of all students that have a GPA greater than 3.0.

CI98 = p ± 2.33 √(pq/n)

p = 42/200 = .21
q = 1 - p = 1 - .21 = .79
n = 200

With your data:

CI98 = .21 + 2.33 √[(.21)(.79)/200]

I'll let you take it from here to finish.

To find a 98% confidence interval estimate of the proportion, p, of all students that have a GPA greater than 3.0, we can use the formula for confidence interval for proportions.

The formula for the confidence interval estimate of the proportion is:

CI = p̂ ± Z * √((p̂(1 - p̂))/n)

Where:
CI is the confidence interval,
p̂ is the sample proportion,
Z is the Z-score corresponding to the desired confidence level,
√ represents the square root,
n is the sample size.

In this case, the sample size (n) is 200 and the sample proportion (p̂) is 42/200 = 0.21.

To find the Z-score corresponding to a 98% confidence level, we need to find the critical value by using a Z-table or calculator. For a 98% confidence level, the Z-score is approximately 2.33.

Now, we can plug these values into the formula:

CI = 0.21 ± 2.33 * √((0.21(1 - 0.21))/200)

Calculating this expression:

CI = 0.21 ± 2.33 * √(0.1659/200)

CI = 0.21 ± 2.33 * 0.0204

CI = 0.21 ± 0.0475

Therefore, the 98% confidence interval estimate of the proportion of all students that have a GPA greater than 3.0 is approximately 0.1625 to 0.2575.