An article reports that in a sample of 50 microdriils drilli

An article reports that in a sample of 50 microdriils drilling a low?carbon alloy steel, the average Lifetime (expressed as the number of holes drilled before failure) was 12.68 with a standard deviation of 6.83. Based on this data, an engineer reported a confidence interval of (10.91, 14.45) but neglected to specify the level. What is the level of this confidence interval? Round z &2 to three decimal places and your final answer to the nearest percent. The level is __________ %

Solution

sample size,n = 50
sample mean,xbar = 12.68
standard deviation,s = 6.83

margin of error = (14.45-10.91)/2 = Za/2 * (s/sqrt(n))
Za/2 * (6.83/sqrt(50)) = 1.77
Za/2 = 1.77 / (6.83/sqrt(50))
Za/2 = 1.83
a/2 = 1 - 0.9664
a = 0.0672

level of confidence = (1-a)*100 %
= 93.28 %