a sample of 50 drills had a mean lifetime of 12.68 holes drilled when drilling a low-carbon steel. assume the population standard deviation is 6.83. what sample size (i.e., how many) would you need to have so that a 95% confidence interval will have a margin of error of 1.0?

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06-12-2022
Mathematics

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a sample of 50 drills had a mean lifetime of 12.68 holes drilled when drilling a low-carbon steel. assume the population standard deviation is 6.83. what sample size (i.e., how many) would you need to have so that a 95% confidence interval will have a margin of error of 1.0?

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ShubhamSrivastava ShubhamSrivastava · Answer 1 · 2022-12-11T22:35:49-05:00

Using the z-distribution, it is found that the 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).

We are given the standard deviation for the population, hence, the z-distribution is used. The parameters for the interval is:

Sample mean of [tex]x = 12.7[/tex]

Population standard deviation of [tex]\sigma = 6.37[/tex]

Sample size of .n = 50

The margin of error is:

[tex]M = z \frac{\sigma }{\sqrt{n} }[/tex]

In which z is the critical value.

We have to find the critical value, which is z with a p-value of [tex]\frac{1+\alpha }{2}[/tex] , in which [tex]\alpha[/tex] is the confidence level.

In this problem, , thus, z with a p-value of , which means that it is z = 1.96.

Then:

[tex]M= 1.96\frac{6.37}{\sqrt{50} } = 1.77[/tex]

The confidence interval is the sample mean plus/minutes the margin of error, hence:

x- M = 12.17 - 1.77 =10.4

x+M = 12.17 + 1.77 = 13.94

The 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).

Learn more about confidence level at :

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