Question

a random sample of 25 was drawn from a normal distribution with a standard deviation of 5. the sample mean is 80. determine the 95% confidence interval estimate of the population mean.

Answer 1

For A, the confidence interval formula above cannot be applied because n is less than 30, so we use the t-score confidence interval value

C.I = Mean ± t score × Standard deviation/√n

Mean = 80

Standard deviation = 5

n = 25

Degree of freedom = 25 - 1 = 24

Hence, 95% confidence interval degree of freedom t score = 2.064

Hence, Confidence Interval =

80 ± 2.064 × 5/√25

80 ± 2.064 × 1

80 ± 2.064

Confidence Interval =

80 - 2.064

= 77.936

80 + 2.064

= 82.064

Confidence Interval: (77.936, 82.064)