One step in the manufacture of a certain metal clamp involves the drilling of four holes. In a sample of 150 clamps, the average time needed to complete this step was 72 seconds and the standard deviation was 10 seconds. (a) Find an 87% confidence interval for the mean time needed to complete the step. (b) What is the confidence level of the interval (69, 75)

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=72[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]s=10[/tex] represent the sample standard deviation

n=150 represent the sample size

87% confidence interval

Part a

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

The degrees of freedom are given by:

[tex] df = 150-1=149[/tex]

Since the Confidence is 0.87 or 87%, the value of [tex]\alpha=0.13[/tex] and [tex]\alpha/2 =0.065[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.065,149)".And we see that [tex]z_{\alpha/2}=1.52[/tex]

Now we have everything in order to replace into formula (1):

[tex]72-1.52\frac{10}{\sqrt{150}}=70.76[/tex]

[tex]72+1.52\frac{10}{\sqrt{150}}=73.24[/tex]

So on this case the 87% confidence interval would be given by (70.76;73.24)

Part b

For this case the confidence interval is given by: (69,75)

We can estimate the margin of error like this:

[tex] ME= \frac{75-69}{2}= 3[/tex]

And the margin of error is given by:

[tex] ME=t_{\alpha/2}\frac{s}{\sqrt{n}} [/tex]

And replacing we can solve for the critical value [tex]t_{\alpha/2}[/tex]

And we got:

[tex] t_{\alpha/2}= \frac{ME*\sqrt{n}}{s}= \frac{3* \sqrt{150}}{10}= 3.674[/tex]

And we can find the significance level with this formula:

"=2*T.DIST(-3.674;149;TRUE)"

And the confidence level would be 1-0.000332= 0.999668

And then that correspond to 99.96%