Question

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)

Answer 1

Answer:

a) $72-1.52\frac{10}{\sqrt{150}}=70.76$  

$72+1.52\frac{10}{\sqrt{150}}=73.24$  

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

b) $ME= \frac{75-69}{2}= 3$

And the margin of error is given by:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$

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

And we got:

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

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%

Step-by-step explanation:

Previous concepts  

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".

$\bar X=72$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$s=10$ 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:  

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

The degrees of freedom are given by:

$df = 150-1=149$

Since the Confidence is 0.87 or 87%, the value of $\alpha=0.13$ and $\alpha/2 =0.065$, 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 $z_{\alpha/2}=1.52$  

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

$72-1.52\frac{10}{\sqrt{150}}=70.76$  

$72+1.52\frac{10}{\sqrt{150}}=73.24$  

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:

$ME= \frac{75-69}{2}= 3$

And the margin of error is given by:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$

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

And we got:

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

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%