A90% confidence interval for a population mean is (65, 77). the population distribution is approximately normal and the population standard deviation is unknown. this confidence interval is based on a simple random sample of 25 observations. calculate the sample mean, the margin of error, and the sample standard deviation.

Part A:

The sample mean is at the midpoint of the lower and the upper values of the confidence interval.

Thus, the sample mean is given by:

(77 + 65) / 2 = 71

Therefore, the sample mean is 71.

Part 2:

The margin of error is the distance between the upper and the lower confidence interval values from the sample mean.

Thus, margin of error is given by 77 - 71 = 71 - 65 = 6.

Therefore, margin of error is 6.

Part 3:

The margin of error is given by



where is the value of the t-distribution given the degree of freedom; s is the sample standard deviation and n is the sample size.

Since the sample size is 25, the degree of freedom is 25 - 1 = 24.


From the t-distrbution table, we have:



Thus, margin of error is given by:

Okay i already have it though