An article reports that in a sample of 50 microdrills drilling a low-carbon alloy steel, the...

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 ____%.

Solutions

Expert Solution

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%

orchestra answered 1 year ago

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