Question

A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selected nine students. Their mean age was 19.1 years with a sample standard deviation of 1.5 years. What is the 99% confidence interval for the population mean?

A. [0.44,3.80]
B. [14.23,23.98]
C. [17.42,20.78]
D. [17.48,20.72]

Answer 1

`19.1-3.355(1.5)/(√(9))=17.42`

`19.1+3.355(1.5)/(√(9))=20.78`

And the best option would be:

C. [17.42,20.78]

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=19.1` represent the sample mean

`\mu` population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 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=9-1=8`

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(0.005,8)".And we see that
`t_(\alpha/2)=`

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

`19.1-3.355(1.5)/(√(9))=17.42`

`19.1+3.355(1.5)/(√(9))=20.78`

And the best option would be:

C. [17.42,20.78]