Respuesta :
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]\overline{x} = 12.17[/tex]
- Population standard deviation of [tex]\sigma = 6.37[/tex]
- Sample size of [tex]n = 50[/tex].
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, [tex]\alpha = 0.95[/tex], thus, z with a p-value of [tex]\frac{1 + 0.95}{2} = 0.975[/tex], 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:
[tex]\overline{x} - M = 12.17 - 1.77 = 10.4[/tex]
[tex]\overline{x} + M = 12.17 + 1.77 = 13.94[/tex]
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 https://brainly.com/question/22596713
