Answer :
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 [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].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
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{14.27}{\sqrt{31}} = 6.60[/tex]
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.
The confidence interval and margin of error for the given confidence level are;
CI = (99.24, 112.44)
E = 6.60
We are given;
Sample size; n = 31
sample mean; x' = 105.84
standard deviation; σ = 14.27
Now, formula for confidence interval is;
CI = x' ± z(σ/√n)
where z is the critical value at the given confidence level.
z at CL of 99% is 2.576
Thus;
CI = 105.84 ± 2.576(14.27/√31)
CI = 105.84 ± 2.576(2.56297)
CI = 105.84 ± 6.60
CI = (105.84 - 6.60), (105.84 + 6.60)
CI = (99.24, 112.44)
Formula for margin of error is;
E = z(σ/√n)
From what we have above;
E = 6.60
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