Question Six

a) A random sample of n = 100 frames of 100mm x 50mm was selected from a consignment of 500 frames. When their depth was measured it was found that the mean was 106mm and standard deviation was 3.5mm. Construct a 90% confidence interval estimate of the mean depth of all the frame in the entire consignment.

90% confidence interval estimate of the mean depth of all the frame in the entire consignment is between a lower limit of 105.419 mm and an upper limit of 106.581 mm.

Explanation:

Confidence interval of mean is given as mean +/- margin of error (E)

mean = 106 mm

sample sd = 3.5mm

n is sample size = 100

degree of freedom = n-1 = 100-1 = 99

confidence level (C) = 90% = 0.9

significance level = 1 - C = 1 - 0.9 = 0.1 = 10%

critical value (t) corresponding to 99 degrees of freedom and 10% significance level is 1.6602

E = t × sample sd/√n = 1.6602×3.5/√100 = 0.581 mm

Lower limit of mean = mean - E = 106 - 0.581 = 105.419 mm

Upper limit of mean = mean + E = 106 + 0.581 = 106.581 mm

90% confidence interval is (105.419 mm, 106.581 mm)

answer; the above statement/question is ( true);