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

Answer:

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

Step-by-step 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:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

The confidence interval on this case is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (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:

$12.15 - 2.58\frac{0.15}{\sqrt{60}}=12.10$  

$12.15 + 2.58\frac{0.15}{\sqrt{60}}=12.20$  

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