Question

If Upper X overbar equals 62​, Upper S equals 8​, and n equals 36​, and assuming that the population is normally​ distributed, construct a 99 % confidence interval estimate of the population​ mean, mu.

Answer 1

The 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

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=62` represent the sample mean

`\mu` population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size

Part a: Confidence interval

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=36-1=35`

Since the Confidence is 0.99 or 99%, the value of
`\alpha=0.01` and
`\alpha/2 =0.005`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,35)".And we see that
`t_(\alpha/2)=2.72`

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

`62-2.72(8)/(√(36))=58.373`

`62+2.72(8)/(√(36))=65.627`

So on this case the 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.