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Confidence interval, t distribution, sample size, error

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(1) For a t distribution with 12 degrees of freedom, find the area, or probability, that is in each region:
(A)To the left of 1.782
(B)To the right of -1.356
(C)To the right of 2.681
(D)To the left of -1.782
(E)Between -2.179 and +2.179
(F)Between -1.356 and +1.782

(2) Due to constituents' complaints to their representatives about rising prescription drug costs, the U.S. congress considered creating a law that would force drug companies to offer prescription discounts to Senior citizens. The house government reform committee provided data for the drug costs for some of the most widely used drugs. Assume the following data shows a sample of the drug costs for Zocor (a cholesterol drug):
$110 $112 $115 $99 $100 $98 $104 $126
Assume a normal distribution and develop a 95% confidence interval estimate of the population mean cost for a prescription of Zocor.

(3)Annual starting salaries for college graduates with a degree in business administration degrees are believed to have a standard deviation of approximately $2000.00. Assume that a 95% confidence interval estimate of the mean annual starting salary is requested. How large of a sample should be taken if the margin of error is:

(A)$500?
(B)$200?
(C)$100?

(4) The travel to work time for residents of the 15 largest cities within the United States is reported in a magazine. Suppose that a preliminary simple random sample of residents in New York City is used to develop a planning value of 6.25 minutes for the population standard deviation.

(A)If we want to estimate the population mean travel to work time for New York City residents with a margin of error of 2 minutes what sample size should be used? Assume a 95% confidence interval.
(B)If we want to estimate the population mean travel to work time for New York City residents with a margin of error of 1 minute, what sample size should be used if you assume a 95% confidence interval?

(5) Audience profile data collected from the espn website showed that 26% of the users were women. Assume this percentage is based on a sample of 400 users.
(A) What is the margin of error associated with a 95% confidence interval for the group of users who are women?
(B) What is the 95% confidence interval for the population proportion of espn website users who are women?
(C) How large a sample should be taken if the desired margin of error is 0.03?

(6) In a survey, 200 people were asked to identify their major source of news. One hundred and ten (110) people stated they mainly got their news information from a tv.
(A) Construct a 95% confidence interval for the group of people in the population who get their major source of news from a tv set.
(B) How large of a sample would be needed to estimate the population proportion with a margin of error of 0.05 at a 95% confidence level?

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Solution Summary

6 problems on probability calculations for t distribution, confidence interval estimate, sample size, margin of error etc have been answered.

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(1)For a t distribution with 12 degrees of freedom, find the area, or probability, that is in each region:
(A)To the left of 1.782
(B)To the right of -1.356
(C)To the right of 2.681
(D)To the left of -1.782
(E)Between -2.179 and +2.179
(F)Between -1.356 and +1.782

(A)To the left of 1.782

t= 1.782
No of tails= 1
degrees of freedom= 12
Prob-value corresponding to t= 1.782 is 5.00%
Thus area to the right of 1.782 is 5.00%
Hence area to the left of 1.782 is 95.00% =100% - 5.%
Answer: 95.00% 0r 0.9500

(B)To the right of -1.356

t= -1.356
No of tails= 1
degrees of freedom= 12
Prob-value corresponding to t= -1.356 is 10.00%
Thus area to the left of -1.356 is 10.00%
Hence area to the right of -1.356 is 90.00% =100% - 10.%
Answer: 90.00% 0r 0.9000

(C)To the right of 2.681

t= 2.681
No of tails= 1
degrees of freedom= 12
Prob-value corresponding to t= 2.681 is 1.00%
Thus area to the right of 2.681 is 1.00%
Answer: 1.00% 0r 0.0100

(D)To the left of -1.782

t= -1.782
No of tails= 1
degrees of freedom= 12
Prob-value corresponding to t= -1.782 is 5.00%
Thus area to the left of -1.782 is 5.00%
Answer: 5.00% 0r 0.0500

(E)Between -2.179 and +2.179

t= -2.179
No of tails= 1
degrees of freedom= 12
Prob-value corresponding to t= -2.179 is 2.50%
Thus area to the left of -2.179 is 2.50%

t= 2.179
No of tails= 1
degrees of freedom= 12
Prob-value corresponding to t= 2.179 is 2.50%
Thus area to the right of 2.179 is 2.50%

Thus the area between -2.179 and 2.179 is 95.00% =100% - (2.5% + 2.5%)

Answer: 95.00% 0r 0.9500

(F)Between -1.356 and +1.782

t= -1.356
No of tails= 1
degrees of freedom= 12
Prob-value corresponding to t= -1.356 is 10.00%
Thus area to the left of -1.356 is 10.00%

t= 1.782
No of tails= 1
degrees of freedom= 12
Prob-value corresponding to t= 1.782 is 5.00%
Thus area to the right of 1.782 is 5.00%

Thus the area between -1.356 and 1.782 is 85.00% =100% - (10.% + 5.%)

Answer: 85.00% 0r 0.8500

(2)Due to constituents' complaints to their representatives about rising prescription drug costs, the U.S. congress considered creating a law  that would force drug companies to offer prescription discounts to Senior citizens. The house government reform committee provided data for the drug costs for some of the most widely used drugs. Assume the following data shows a sample of the drug costs for Zocor (a cholesterol drug):
$110 $112 $115 $99 $100 $98 $104 $126
Assume a normal distribution and develop a 95% confidence interval estimate of the population mean cost for a prescription of Zocor.

First find the mean and standard deviation of the cost

X= X 2 =
$110 12100
$112 12544
$115 13225
$99 9801
$100 10000
$98 9604
$104 10816
$126 15876
Total= 864 93966
n=no of observations= 8
Mean= 108 =864/8

variance={summation of X 2 - n(Mean) 2 }/(n-1)= 93.4286 =(93966-8*108^2)/(8-1)
standard deviation =square root of Variance= 9.67 =square root of 93.4286

95% Confidence limits

Mean=M = $108
Standard deviation =s= $9.67
sample size=n= 8
sx=standard error of mean=s/square root of n= 3.4189 = ( 9.67 /square root of 8) ...

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