A random sample of 25,000 ACT test takers had an average score of 21 with a standard deviation of 5. What is the 95% confidence interval of the population mean?
A) 4.9723 to 5.0277
B) 4.9380 to 5.0620
C) 4.7397 to 5.2603
D) 4.9432 to 5.0568

Question

Quizplus · Accepted Answer

To find the 95% confidence interval of the population mean, we will use the formula:

CI = x̄ ± z*(σ/√n)

where x̄ = sample mean = 21
σ = population standard deviation = 5
n = sample size = 25,000
z = z-score for the 95% confidence level = 1.96

Substituting the values, we get:

CI = 21 ± 1.96*(5/√25000)
CI = 21 ± 0.062

Therefore, the 95% confidence interval of the population mean is (21-0.062, 21+0.062), which simplifies to (20.938, 21.062).

The closest answer choice to this interval is B: 4.9380 to 5.0620. However, we must remember to convert the sample mean score of 21 back to the ACT scale of 1-36 in order to compare our interval to the score range.

To convert the scale, we can use the formula:

New score = (Old score/36) * 100

Substituting the values, we get:

New score = (21/36) * 100
New score = 58.3

Therefore, the 95% confidence interval of the population mean on the ACT scale is approximately (49.38, 50.62), which falls within the score range of 1-36.

A Random Sample of 25,000 ACT Test Takers Had an Average