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

`3-1.86(0.3)/(√(9))=2.81`

`3+ 1.86(0.3)/(√(9))=3.19`

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

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=3` represent the sample mean

`\mu` population mean (variable of interest)

`s= √(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)(s)/(√(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(0.3)/(√(9))=2.81`

`3+ 1.86(0.3)/(√(9))=3.19`

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