Question

A group of 49 randomly selected students has a mean age of 22.4 years with a standarddeviation of 3.8. Construct a 98% confidence interval for the population mean knowing thatthe population standard deviation is 4.2 years.

Answer 1

The 98% confidence interval for the population mean is between 21 and 23.8 years.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.98)/(2) = 0.01`

Now, we have to find z in the Z-table as such z has a p-value of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.01 = 0.99`, so Z = 2.327.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.327(4.2)/(√(49)) = 1.4`

The lower end of the interval is the sample mean subtracted by M. So it is 22.4 - 1.4 = 21 years.

The upper end of the interval is the sample mean added to M. So it is 22.4 + 1.4 = 23.8 years.

The 98% confidence interval for the population mean is between 21 and 23.8 years.