Question 80068: Question 1
A. A large public university want to estimate the average (mean) teaching experience of it's faculty. A preliminary random sample of 20 faculty yields a standard deviation of s=2.5 years. The dean wants to be 95% confident that the sample does not have a maximum error of estimate of more than 1.6 years. (Hint: this is maximum error of estimate E). How large of a sample should she use? Use the preliminary sample standard deviation as an approximation of the population standard deviation. Round up to next whole number.
B. If she decides on a maximum error of 1.8 years and a level of significance of 95% how large of a sample should she use? Round up to next whole number.
Question 2
Suppose you are given a sample set of data consisting of 17 body temperatures, with s= 1.02. Assume that body temperatures are normally distributed. Find the 95% confidence interval for the population variance. Round off 2 decimal places.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Question 1
A. A large public university want to estimate the average (mean) teaching experience of it's faculty. A preliminary random sample of 20 faculty yields a standard deviation of s=2.5 years. The dean wants to be 95% confident that the sample does not have a maximum error of estimate of more than 1.6 years. (Hint: this is maximum error of estimate E). How large of a sample should she use? Use the preliminary sample standard deviation as an approximation of the population standard deviation. Round up to next whole number.
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E=z*(sigma/sqrt(n))
sqrt(n)=z*sigma/E
Your problem:
sqrt(n)=1.96[2.5]/1.6
sqrt(n)=3.0625
n=9.4
Rounding up gives n=10
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B. If she decides on a maximum error of 1.8 years and a level of significance of 95% how large of a sample should she use? Round up to next whole number.
sqrt(n)=1.96[2.5]/1.8
sqrtr(n)= 2.72222
n=7.7
Rounding gives n=8
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Question 2
Suppose you are given a sample set of data consisting of 17 body temperatures, with s= 1.02. Assume that body temperatures are normally distributed. Find the 95% confidence interval for the population variance. Round off 2 decimal places.
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[(n-1)s^2/X(r)]< sigma^2 < [(n-1)s^2/X(l)]
[(17-1)1.02^2/28.845] < sigma^2 < [(17-1)1.02^2/6.90766
0.5771 < sigma^2 <2.4098
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Cheers,
Stan H.
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