Question

A precision control engineer is interested in the mean length of tubing being cut automatically by machine. It is known that the standard deviation in the cutting length is 0.15 feet. A sample of 60 cut tubes yields a mean length of 12.15 feet. This sample will be used to obtain a 99% confidence interval for the mean length cut by machine. Develop the 99% confidence interval for population mean.

Answer 1

The 99% confidence interval would be given by (12.10;12.20)

Explanation:

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".

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".

`\bar X=12.15` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma=0.15` represent the population standard deviation

n=60 represent the sample size

Assuming the X follows a normal distribution

`X \sim N(\mu, \sigma=0.15`

The sample mean
`\bar X` is distributed on this way:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

The confidence interval on this case is given by:

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

The next step would be find the value of
`\z_(\alpha/2)`,
`\alpha=1-0.99=0.01` and
`\alpha/2=0.005`

Using the normal standard table, excel or a calculator we see that:

`z_(\alpha/2)=2.58`

Since we have all the values we can replace:

`12.15 - 2.58(0.15)/(√(60))=12.10`

`12.15 + 2.58(0.15)/(√(60))=12.20`

So on this case the 99% confidence interval would be given by (12.10;12.20)