Question

The Digest of Education Statistics reported that the mean salary for teach- ers in 2012 was $56,410. Suppose that the sample size was 87 and that the sample standard deviation was $15,000 (note that we only have the sample standard deviation now...)

(a) what is the 90% confidence interval for the population mean?

(b) for the same sample size, find the 99% confidence interval for the population mean.

Answer 1

Answer:

Given that:

The Digest of Education Statistics reported that the mean salary for teach- ers in 2012 was $56,410. Suppose that the sample size was 87 and that the sample standard deviation was $15,000

3. x̅ = 56410, s = 15000, n = 87

(a) what is the 90% confidence interval for the population mean?

90% Confidence interval :

At α = 0.1 and df = n-1 = 86, two tailed critical value, t-crit = T.INV.2T(0.1, 86) = 1.663

Lower Bound = x̅ - t-crit*s/√n = 56410 - 1.663 * 15000/√87 = 53735.99

Upper Bound = x̅ + t-crit*s/√n = 56410 + 1.663 * 15000/√87 = 59084.01

(b) for the same sample size, find the 99% confidence interval for the population mean.

99% Confidence interval :

At α = 0.01 and df = n-1 = 86, two tailed critical value, t-crit = T.INV.2T(0.01, 86) = 2.634

Lower Bound = x̅ - t-crit*s/√n = 56410 - 2.634 * 15000/√87 = 52173.74

Upper Bound = x̅ + t-crit*s/√n = 56410 + 2.634 * 15000/√87 = 60646.26