Question

A student wanted to quickly construct a 95% confidence interval for the average age of students in her statistics class. She randomly selected 9 students. From the sample she learned their average age was 19.1 years with a sample standard deviation of 1.5 years. What is the 95% confidence interval for the population mean

Answer 1

The 95% confidence interval for the population mean is between 15.641 years and 22.559 years.

Explanation:

We are in posession of the sample's standard deviation, so we use the Students t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 9 - 1 = 8

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.95)/(2) = 0.975([tex]t_(975)`). So we have T = 2.3060

The margin of error is:

M = T*s = 2.3060*1.5 = 3.459

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 19.1 - 3.459 = 15.641 years

The upper end of the interval is the sample mean added to M. So it is 19.1 + 3.459 = 22.559 years

The 95% confidence interval for the population mean is between 15.641 years and 22.559 years.

Answer 2

`19.1-2.306(1.5)/(√(9))=17.947`

`19.1+2.306(1.5)/(√(9))=20.253`

So on this case the 95% confidence interval would be given by (17.947;20.253)

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 0.95 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(0.025,8)".And we see that
`t_(\alpha/2)=2.306`

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

`19.1-2.306(1.5)/(√(9))=17.947`

`19.1+2.306(1.5)/(√(9))=20.253`

So on this case the 95% confidence interval would be given by (17.947;20.253)