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.)
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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