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

Mohammed Ghanem · Accepted Answer

Final Answer : B, Explanation :  We use the formula:CI = x̄ ± z α/2 * (σ /√n)where x̄ is the sample mean, z α/2 is the z-score for the desired confidence level (α/2 = 0.025 for 95% confidence), σ is the population standard deviation, and n is the sample size.Plugging in the values given:CI = 21 ± 1.96 * (5 / √25000)CI = 21 ± 0.062CI = (20.938, 21.062)So the 95% confidence interval for the population mean is (4.9380 to 5.0620). The correct choice is B.

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

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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 5.0277B) 4.9380 to 5.0620C) 4.7397 to 5.2603D) 4.9432 to 5.0568

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Learning Objectives

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Related questions

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