Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x-bar, is found to be 108, and the sample standard deviation, s, is found to be 17. Construct a 95% confidence interval for μ if the sample size, n, is 81. Hint: Calculate E and round to 1 decimal place. Then construct the interval by putting parenthesis around the lower bound and upper bound separated by a comma).

Answer 1

Answer:

95% confidence interval is (104.2, 111.8)

Explanation:

Mean = 108, sd = 17, n = 81, degree of freedom = n-1 = 81 - 1 = 80. t-value corresponding to 80 degrees of freedom and 95% confidence level is 1.990

Confidence Interval = Mean + or - Error margin

Error margin (E) = (t×sd)/√n = (1.990×17)/√81 = 33.83/9 = 3.8

E = 3.8

Lower bound = mean - E = 108 - 3.8 = 104.2

Upper bound = mean + E = 108 + 3.8 = 111.8

95% confidence interval is (104.2, 111.8)