A psychologist is interested in the mean IQ score of a given group of children. It is known that the IQ scores of the group have a standard deviation of 10. The psychologist randomly selects 100 children from this group and finds that their mean IQ score is 100. Based on this sample, find a 99% confidence interval for the true mean IQ score for all children of this group. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)

The 99% confidence interval for the true mean IQ score for all children of this group is (97.4, 102.6). The lower limit is 97.4 and the upper limit is 102.6.

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 - 0.99}{2} = 0.005[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575.

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 = 2.575\frac{10}{\sqrt{100}} = 2.575[/tex]

Rounding to one decimal place, the margin of error is 2.6.

The lower end of the interval is the sample mean subtracted by M. So it is 100 - 2.6 = 97.4.

The upper end of the interval is the sample mean added to M. So it is 100 + 2.6 = 102.6.

The 99% confidence interval for the true mean IQ score for all children of this group is (97.4, 102.6). The lower limit is 97.4 and the upper limit is 102.6.