In: Statistics and Probability
An article reports that in a sample of 50 microdrills 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 (11.35, 14.01) but neglected to specify the
level. What is the level of this confidence interval?
Give your final answer to the nearest percent.
The level is ____%.
Solution :
Given that,
Population standard deviation = = 6.83
n = 50
Lower confidence interval = 11.35
Upper confidence interval = 14.01
= (Lower confidence interval + Upper confidence interval ) / 2
= (11.35 + 14.01) / 2
= 25.36 / 2 = 12.68
= 12.68
Margin of error = E = Upper confidence interval - = 14.01 - 12.68 = 1.33
Margin of error = 1.33
sample size = n = (Z/2* / E) 2
Z 2/2 = n * E 2 / 2
= 50 * 1.332 / 6.832
Z 2/2 = 1.895971
Z /2 = 1.38
= 0.17
confidence level = 1 - = 1 - 0.17 = 0.83 = 83%
The level is 83%