Question 633821
1. A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having two children. A sample of 180 was taken, and the mean amount spent was $219.76. Assuming a standard deviation equal to $41.77, find the 99% confidence interval for m, the mean for all such families (show all work).
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x-bar = 219.76
E = 1.645*(41.77/sqrt(180)) = 5.1215
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90% CI: 219.76-5.12 < x < 219.76+5.12
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2. Assume that a sample is drawn and z(&#945;/2) = 1.65 and &#963; = 15. Answer the following questions:

(A)If the Maximum Error of Estimate is 0.05 for this sample, what would be the sample size?
E = z*s/sqrt(n)
0.05 = 1.65*15/sqrt(n)
sqrt(n) = 1.65*15/0.05
sqrt(n) = 495
n = 22.25 = 22 when rounded down
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(B)Given that the sample Size is 400 with this same z(&#945;/2) and &#963;, what would be
the Maximum Error of Estimate?
Substitute for "n" and "z" and sigma.
Then solve for "E".
E = 1.2375
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(C) What happens to the Maximum Error of Estimate as the sample size gets larger?
E and sqrt(n) are indirectly related.
AS n get larger, E get smaller.
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(D) What effect does the answer to C above have to the size of the confidence interval?
The size of the CI is ALWAYS 2*E.
If n get larger, E get smaller, and the CI gets smaller (narrower).
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Cheers,
Stan H.