Question

Assume that a sample is used to estimate a population mean μ . Find the 99.5% confidence interval for a sample of size 890 with a mean of 37.2 and a standard deviation of 7.6. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

Answer 1

Given:

n = 890, = 37.2, S = 7.6

Confidence level = C = 99.5% = 0.995

= 1 - C = 1 - 0.995 = 0.005

Degree of freedom = n-1= 890 - 1 = 889

Critical value:

                 ..................Using t table

99.5% Confidence interval: