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KIM [24]
3 years ago
6

A sample of 50 drills had a mean lifetime of 12.17 holes drilled when drilling a low-carbon steel. Assume the population standar

d deviation is 6.37.Construct a 95% confidence interval for the mean lifetime of this type of drill
Mathematics
1 answer:
zmey [24]3 years ago
6 0

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 <u>standard deviation for the population</u>, hence, the z-distribution is used. The parameters for the interval is:

  • Sample mean of \overline{x} = 12.17
  • Population standard deviation of \sigma = 6.37
  • Sample size of n = 50.

The margin of error is:

M = z\frac{\sigma}{\sqrt{n}}

In which z is the critical value.

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

In this problem, \alpha = 0.95, thus, z with a p-value of \frac{1 + 0.95}{2} = 0.975, which means that it is z = 1.96.

Then:

M = 1.96\frac{6.37}{\sqrt{50}} = 1.77

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

\overline{x} - M = 12.17 - 1.77 = 10.4

\overline{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).

A similar problem is given at brainly.com/question/22596713

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Using the above formula to solve the first part, we have :

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Step-by-step explanation:

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Step 3: Distribute

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