Question

1. Assume the age of night school students is normally distributed. A simple random sample of 24 night school students had an average age of 27.3 years. Use this information and a sample standard deviation of 5.6 to find a a. 90% confidence interval estimate for the population variance b. 90% confidence interval estimate for the population standard deviation c. Explain the meaning of the confidence interval that you found in b. What does the 90% mean

Answer 1

a)
`27.3-1.714(5.6)/(√(24))=25.34`

`27.3+1.714(5.6)/(√(24))=29.26`

So on this case the 90% confidence interval would be given by (25.34;29.26)

b)
`((23)(5.6)^2)/(35.172) \leq \sigma^2 \leq ((23)(5.6)^2)/(13.091)`

`20.507 \leq \sigma^2 \leq 55.097`

Now we just take square root on both sides of the interval and we got:

`4.528 \leq \sigma \leq 7.423`

So the best option would be:

4.528<σ<7.423

c) For this case the 90% means that we have 90% of confidence that the true parameter
`\mu, \sigma` is between the limits calculated on the intervals

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=27.3` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s=5.6 represent the sample standard deviation

n=24 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)

Part a

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=24-1=23`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a talel to find the critical value. The excel command would be: "=-T.INV(0.05,23)".And we see that
`t_(\alpha/2)=1.714`

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

`27.3-1.714(5.6)/(√(24))=25.34`

`27.3+1.714(5.6)/(√(24))=29.26`

So on this case the 90% confidence interval would be given by (25.34;29.26)

Part b

The confidence interval for the population variance is given by the following formula:

`((n-1)s^2)/(\chi^2_(\alpha/2)) \leq \sigma^2 \leq ((n-1)s^2)/(\chi^2_(1-\alpha/2))`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,23)" "=CHISQ.INV(0.95,23)". so for this case the critical values are:

`\chi^2_(\alpha/2)=35.172`

`\chi^2_(1- \alpha/2)=13.091`

And replacing into the formula for the interval we got:

`((23)(5.6)^2)/(35.172) \leq \sigma^2 \leq ((23)(5.6)^2)/(13.091)`

`20.507 \leq \sigma^2 \leq 55.097`

Now we just take square root on both sides of the interval and we got:

`4.528 \leq \sigma \leq 7.423`

So the best option would be:

4.528<σ<7.423

Part c

For this case the 90% means that we have 90% of confidence that the true parameter
`\mu, \sigma` is between the limits calculated on the intervals