Question

A simple random sample of size nequals17 is drawn from a population that is normally distributed. The sample mean is found to be x overbar equals 61 and the sample standard deviation is found to be sequals11. Construct a 95% confidence interval about the population mean. The 95% confidence interval is ( nothing, nothing). (Round to two decimal places as needed.)

Answer #1

df = n -1 = 17 - 1 = 16

t critical value at 0.05 level with 16 df = 2.12

95% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

61 - 2.12 * 11 / sqrt(17) < < 61 + 2.12 * 11 / sqrt(17)

55.34 < < 56.66

95% CI is **( 55.34 , 66.66)**

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