Question

You research commute times to work and find that the population standard deviation is 9.3 minutes. Repeat Exercise, using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results. In a random sample of eight people, the mean commute time to work was 35.5 minutes and the standard deviation was 7.2 minutes.

Answer 1

Case
`s =7.2`

`35.5-2.36(7.2)/(√(8))=29.49`

`35.5+2.36(7.2)/(√(8))=41.51`

So on this case the 95% confidence interval would be given by (29.49;41.51)

Case
`\sigma =9.3`

`35.5-1.96(9.3)/(√(8))=29.06`

`35.5+1.96(9.3)/(√(8))=41.94`

So on this case the 95% confidence interval would be given by (29.06;41.94)

And we conclude that the intervals are very similar.

Explanation:

If we assume that for this question we need to find a confidence interval for the population mean. We have the following procedure:

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= 35.5` represent the sample mean

`\mu` population mean (variable of interest)

s=7.2 represent the sample standard deviation

n=8 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=8-1=7`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that
`t_(\alpha/2)=2.36`

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

`35.5-2.36(7.2)/(√(8))=29.49`

`35.5+2.36(7.2)/(√(8))=41.51`

So on this case the 95% confidence interval would be given by (29.49;41.51)

If we assume that the real population standard deviation is
`\sigma =9.3` the confidence interval is given by:

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

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that
`z_(\alpha/2)=1.96`

`35.5-1.96(9.3)/(√(8))=29.06`

`35.5+1.96(9.3)/(√(8))=41.94`

So on this case the 95% confidence interval would be given by (29.06;41.94)

And we conclude that the intervals are very similar.