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LenKa [72]
3 years ago
5

We have IQ test scores of 31 seventh-grade girls in a Midwest school district. We have calculated that sample mean is 105.84 and

the standard deviation is 14.27.
Give a 99% confidence interval for the average score in the population. What is the margin of error?
Mathematics
1 answer:
yuradex [85]3 years ago
4 0

Answer:

The 99% confidence interval for the average score in the population is between 99.24 and 112.44.

The margin of error is 6.60.

Step-by-step explanation:

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

\alpha = \frac{1-0.99}{2} = 0.005

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

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

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

M = 2.575*\frac{14.27}{\sqrt{31}} = 6.60

The lower end of the interval is the mean subtracted by M. So it is 105.84 - 6.60 = 99.24

The upper end of the interval is the mean added to M. So it is 105.84 + 6.60 = 112.44

The 99% confidence interval for the average score in the population is between 99.24 and 112.44.

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