<PRE><B>Question 550860</B>
Since E = z*s/sqrt(n) to find &quot;margin of error&quot;
n = [z*s/E]^2 to find sample size.

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1. Given a level of confidence of 99% and a population standard deviation of 6, answer the following:
(A) What other information is necessary to find the sample size (n)?
Ans: maximum error or E
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(B) Find the maximum error of estimate (E) if n = 87. show all work.
E = 2.5758*6/sqrt(87) = 1.6569
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2. A sample of 89 golfers showed that their average score on a particular golf course was 90.98 with a standard deviation of 6.47. 
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Answer each of the following (show all work and state the final answer to at least two decimal places.): 
(A) Find the 98% confidence interval of the mean score for all 89 golfers. 
x-bar = 90.89
ME = 2.3263*6.47/sqrt(89) = 1.5955
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98% CI: (x-bar)-ME &lt; u &lt; (x-bar)+ME
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(B) Find the 98% confidence interval of the mean score for all golfers if this is a sample of 135 golfers instead of a sample of 89. 
x-bar = 90.89
ME = 2.3263*6.47/sqrt(135) = 1.2954
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Same formula as above.

(C) Which confidence interval is larger and why?
The 1st is larger.  ME and n are inversely related;
as sample size increases (more information) the 
interval of confidence can get smaller, and does.
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3. Assume that the population of heights of male college students is approximately normally distributed with a mean M of 70.63 inches and standard deviation s of 6.47 inches. 
A random sample of 88 heights is obtained. show all work. 
(A) Find the mean and standard error of the x-bar distribution.
mean of the sample means = 70.63
std of the sample means = 6.47/sqrt(88) = 0.6897
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(B) Find P(x-bar &gt; 72.25)
t(72.25) = (72.25-70.63)/(0.6897) = 2.3488
P(x-bar &gt; 72.25) = P(t &gt; 2.3488 when df = 87) = tcdf(2.3488,100,87) = 0.0105 
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4. The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.97 inches and a standard deviation of 0.69 inches. show all work. 
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(A) What percentage of the grapefruits in this orchard is larger than 6.01 inches?
z(6.01) = (6.01-5.97)/0.69 = 0.0580
P(fruit above 6.01) = P(z &gt; 0.0580) = normalcdf(0.0580,100) = 47.69%
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(B) A random sample of 100 grapefruits is gathered and the mean diameter is calculated. 
What is the probability that the sample mean is greater than 6.01 inches?
z(6.01) = (6.01-4.97)/[0.69/sqrt(100)] = 0.5797
P(sample mean greater than 6.01) = P(z &gt; 0.5797 ) = 28.11%
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5. A researchers is interested in estimating the noise levels in decibels at area urban hospitals. She wants to be 99% confident that her estimate is correct. If the standard deviation is 4.92, how large a sample is needed to get the desired information to be accurate within 0.71 decibels? show all work.
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n = [z*s/E]^2
n = [2.5758*4.92/0.71]^2 = 319 when rounded up
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Cheers,
Stan H.
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