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Sagot :
Answer:
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.
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