Question

Use the given information to construct a 99% confidence interval estimate of the mean of the population. n=39, standard deviation= 3.86, sample mean = 41.7

Answer 1

`41.7-2.71(3.86)/(√(39))=40.02`

`41.7+2.71(3.86)/(√(39))=43.38`

So on this case the 99% confidence interval would be given by (40.02;43.38)

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=41.7` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s=3.86 represent the sample standard deviation

n=39 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=39-1=38`

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(0.005,38)".And we see that
`t_(\alpha/2)=2.71`

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

`41.7-2.71(3.86)/(√(39))=40.02`

`41.7+2.71(3.86)/(√(39))=43.38`

So on this case the 99% confidence interval would be given by (40.02;43.38)