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 :

Answer:

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

Step-by-step explanation:

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

[tex]\alpha = \frac{1 - 0.98}{2} = 0.01[/tex]

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

That is z with a pvalue of [tex]1 - 0.01 = 0.99[/tex], so Z = 2.327.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 2.327\frac{4.2}{\sqrt{49}} = 1.4[/tex]

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.

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