Question

A sample of 50 drills had a mean lifetime of 12.17 holes drilled when drilling a low-carbon steel. Assume the population standard deviation is 6.37.Construct a 95% confidence interval for the mean lifetime of this type of drill

Answer 1

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 $\overline{x} = 12.17$
• 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 sample mean plus/minutes the margin of error, 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