Question

In a manufacturing process, a random sample of 9 manufactured bolts has a mean length of 3 inches with a variance of .09. What is the 90 percent confidence interval for the true mean length of the manufactured bolt?

Answer 1

Answer:

$3-1.86\frac{0.3}{\sqrt{9}}=2.81$    

$3+ 1.86\frac{0.3}{\sqrt{9}}=3.19$    

So on this case the 90% confidence interval would be given by (2.81;3.19)  

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".

$\bar X=3$ represent the sample mean

$\mu$ population mean (variable of interest)

$s= \sqrt{0.09}= 0.3$ represent the sample standard deviation

n=9 represent the sample size  

Solution to the problem

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

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

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=9-1=8$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,8)".And we see that $t_{\alpha/2}=1.86$

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

$3-1.86\frac{0.3}{\sqrt{9}}=2.81$    

$3+ 1.86\frac{0.3}{\sqrt{9}}=3.19$    

So on this case the 90% confidence interval would be given by (2.81;3.19)