Question

A sample of 64 was taken from a population, with the mean of the sample being 119 and the standard deviation of the sample being 16. What is the 95% confidence interval for the mean of the population?
A. (115, 123)
B. (117, 121)
C. (113, 125)
D. (118, 120)

Answer 1

We have the sample size, sample mean and the sample standard deviation. Since the population standard deviation is not know, we will use t-distribution to find the confidence interval.

The critical t value for 95% confidence interval and 63 degrees of freedom is 1.998.

The 95% confidence for the population mean will be:

$(119-1.998 *\frac{16}{ \sqrt{64} } ,119+1.998 *\frac{16}{ \sqrt{64} }) \\ \\ =(115.004, 122.996) \\ \\=(115,123)$

Thus, the 95% confidence interval for the population mean will be (115,123)

So, option A is the correct answer