Question 407784
1. Consider a normal population with µ = 25 and σ = 8.0.
(A)Calculate the standard score for a value x of 26.
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z(26) = (26-25)/8 = 1/8
That means 26 is 1/8 standard deviation above the mean.
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(B)Calculate the standard score for a randomly selected sample of 45 with x(bar) = 26.
z(26) = (26-25)/[8/sqrt(45)] = sqrt(45)/8 = 0.1491
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(C)Explain why the standard scores of 26 are different between A and B above. (Points : 6)
The Central Limit Theorem says "the mean of the sample means is the 
same as the mean of the population" but "the standard deviation of
the sample means is s/sqrt(n), where n is the sample size".
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2. Assume that the mean SAT score in Mathematics for 11th graders across the nation is 500, and that the standard deviation is 100 points. Find the probability that the mean SAT score for a randomly selected group of 150 11th graders is between 475 and 525.
z(475) = (475-500)/[100/sqrt(150)] = -25*sqrt(150)/100 = -3.0619
z(525) = (475-500)/[100/sqrt(150)] = +3.0619 
P(475 < x < 525) = P(-3.0619 < z < 3.0619) = 0.9978
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3. Assume that a sample is drawn and z(&#945;/2) = 1.96 and &#963; = 30. Answer the following questions:

(A)If the Maximum Error of Estimate is 0.04 for this sample, what would be the sample size?
n = [zs/E]^2
n = [1.96*30/0.04]^2 = 2160900
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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?
E = zs/sqrt(n)
E = 1.96*30/sqrt(400)
etc.
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(C)What happens to the Maximum Error of Estimate as the sample size gets smaller?
ME = zs/sqrt(n)
This mean Me and z are directly related
Me and s are directly related
Me and sqrt(n) are indirectly related
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As sample gets smaller Me gets larger
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(D)What effect does the answer to C above have on the size of the confidence interval?
The confidence interval is always 2*ME wide
As sample gets smaller Me gets larger so CI would get wider
(Points : 8)
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4. By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the standard deviation is 3.78 seconds. 
Answer each of the following (show all work):
(A) How many measurements should be made in order to be 99% certain that the maximum error of estimation will not exceed 2.0 seconds? 
Use the formula:
n = [zs/E]^2 and round up to the next whole number.
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(B) What sample size is required for a maximum error of 1.5 seconds?
Same formula as above.
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5. A 95% confidence interval estimate for a population mean was computed to be (34.6, 49.4). Determine the mean of the sample, which was used to determine the interval estimate (show all work).
x-bar - ME = 34.6
x-bar + ME = 49.4
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Add to get
2*x-bar = 84
x-bar = 42
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6. A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having two children. A sample of 195 was taken, and the mean amount spent was $210.97. Assuming a standard deviation equal to $41.63, find the 90% confidence interval for &#61549;, the mean for all such families (show all work).
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x-bar = 210.97
E = 1.645*41.63/sqrt(195) = 4.904
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90%CI: 210.97-4.904 < u < 210.97+4.904

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7. A confidence interval estimate for the population mean is given to be (36.28, 44.76). If the standard deviation is 17.803 and the sample size is 48, answer each of the following (show all work): 

(A) Determine the maximum error of the estimate, E.
Method same as #6
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(B) Determine the confidence level used for the given confidence interval.
Once you determine ME
solve ME = zs/sqrt(n) for z.
Then determine twice the size of the tail for that z.
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Get back to me if some of this doesn't make any sense.
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Cheers,
Stan H.
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