Assume that a sample is used to estimate a population mean. Find the 98% confidence interval for a sample of size 51 with a mean of 38.9 and a standard deviation of 6.5. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. 98% C.I. =

To find the 98% confidence interval for the population mean when given a sample mean, standard deviation, and sample size, we use the formula for the confidence interval:
C I = ( x ˉ − z n ​ s ​ , x ˉ + z n ​ s ​ )
Where:

x ˉ is the sample mean, which is 38.9.
s is the sample standard deviation, which is 6.5.
n is the sample size, which is 51.
z is the z-score corresponding to the desired confidence level.

Since we are looking for a 98% confidence interval, we need the z-score that corresponds to the middle 98% of the standard normal distribution. For a 98% confidence level, z ≈ 2.33 .
Let's plug these values into the formula:

Calculate the standard error (SE): SE = n ​ s ​ = 51 ​ 6.5 ​ ≈ 0.911

Calculate the margin of error (ME): ME = z × SE = 2.33 × 0.911 ≈ 2.123

Determine the confidence interval:

Lower limit = x ˉ − ME = 38.9 − 2.123 ≈ 36.777
Upper limit = x ˉ + ME = 38.9 + 2.123 ≈ 41.023



Thus, the 98% confidence interval for the population mean is ( 36.777 , 41.023 ) . This means we are 98% confident that the true population mean lies within this interval.

Answered by DanielJosephParker | 2025-07-21