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.