Part A: For questions 1-3 please refer to the following situation. The pulse rate per minute of the adult male population between 18 and 25 years of age in the United States is known to have a normal distribution with mean of 72 beats per minute and a standard deviation of 9.7.
1.
Suppose that a male between the age of 18 to 25 years is randomly selected. What is the probability that his pulse rate will exceed 100 beats per minute? [Such an event will disqualify him from getting into the Marine Corps (not “cool enough”)!]
Answer: P(X > 100) = P(Z > (100-72)/9.7) = P(Z > 2.87) = .5 - .4979 = .0021.
2.
What is the probability that his pulse rate will be between 70 and 82 beats per minute?
Answer: P(70 < X < 82) = P((70-72)/9.7 < Z < (82-72)/9.7) = P(-.21 < Z < 1.03) = .0832 +
.3485 = .4317.
3.
What is the lowest pulse rate that a male between 18 and 25 years of age could have but still belong to the highest 5% of all pulse rates?
Answer: For the Z-curve, to get a right-tail probability of .05, the z-score must be 1.645
[Look for .45 inside the table and find the associated z]. Therefore, if A is the lowest pulse rate to still belong to the highest 5%, then (A - 72)/9.7 = 1.645. Solving for A, we obtain A =
72 + (1.645)(9.7) = 89.9565.
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Part B: For questions 4-6 , refer to the information provided in Part A regarding the distribution of pulse rates of American males between 18 and 25 years old, and consider the situation where
36 of these males are randomly sampled. Let these 36 males.
X denote the sample mean of the pulse rates of
4.
Describe the sampling distribution of and its standard error.
X . That is, describe its shape and also state its mean
Answer: The sampling distribution of X is normal with mean equal to 72 (the population mean) and standard deviation (or standard error) equal to 9.7/sqrt(36) = 1.6167.
5.
Find the probability that X will either be between 70 and 74 beats per minute.
Answer: P{70 < X < 74} = P{(70 - 72)/1.6167 < Z < (74 - 72)/1.6167} = P{-1.24 < Z < 1.24} =
.3925 + .3925 = .7850.
6.
Find the value of A such that the probability that X will be at most A is 0.05.
Answer: The z-score associated with A is -1.645 since we want the left-tail probability.
Consequently, (A - 72)/1.6167 = -1.645 so solving for A we obtain A = 72 - (1.645)(1.6167) =
69.34.
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Part C: For questions 7-8 please refer to the following situation. Assume that when a person is afflicted with melanoma (a form of skin cancer), the probability that it will kill him/her is 0.45.
Suppose that we have 500 patients who are afflicted by melanoma. Let p denote the proportion of these 500 patients who will be killed by the melanoma.
7.
What is the approximate sampling distribution of
ˆ
, and what are the mean and standard error of p ?
Answer: The approximate sampling distribution of p is normal with mean of 0.45 (the population proportion) and standard error sqrt{(.45)(1-.45)/500} = .0222.
8.
Find the approximate probability that p is between 0.44 and 0.47?
Answer: P{.44 < p < .47} = P{(.44 - .45)/.0222 < Z < (.47 - .45)/.0222} = P{-.45 < Z < .90} =
.1736 + .3159 = .4895.
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Part D: For questions 9-11 please refer to the following situation. A researcher wants to estimate

, the mean academic year salary of female faculty members in US universities. She takes a random sample of n = 100 female faculty members from US universities and determines their academic year salaries. From this sample she observes a sample mean academic year salary of
X

$51,000 with a sample standard deviation of S = $10,000.
9.
Based on this information, construct a 95% confidence interval for the mean academic year salary of female faculty members in US universities.
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Answer: The 95% CI for the population mean academic year salary of female faculty members will be [51000 +/- (1.96)(10000)/sqrt(100)] = [51000 +/- 1960] = [$49040, $52960].
10.
What will happen to the length of your confidence interval if you change the confidence level from 95% to 99%?
Answer: The length of the interval will increase.
11.
Personally, which interval would you prefer in this situation: the 95% confidence interval or the 99% confidence interval. Explain your reasoning for your choice.
Answer: Preference here will depend on your level of risk-taking as well as the utility of the resulting CI based on its length. Conventionally, a 95% CI is used, but it is not written in stone that a 95% confidence level should always be used.
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Part E: For questions 12-13 , refer to the following situation. A researcher would like to determine the proportion, p, of South Carolinians who play the state lottery. Suppose that this researcher desires to have a 95% confidence interval for p whose margin of error is at most 0.03.
12. How many SC voters should this researcher sample [Assume that you don't have any prior information about the value of p.]?
Answer: Here, 1 -

= .95, B = .03, and we don't have any knowledge of p. Thus the appropriate formula is: n = z 2 / (4B 2 ) = (1.96) 2 /(4(.03) 2 ) = 1067.11. Thus, 1068 will be required.
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12.
Discuss what will happen to the desired sample size if
(a) the confidence coefficient 100(1 -

)% is changed from 95% to 99%;
Answer: It will increase since the appropriate z-score will increase. Here you want to be more certain about your conclusions, so you incur more cost since you need more information by sampling more units.
(b) the maximum tolerable margin of error (denoted by B) is changed from 0.03 to 0.02.
Answer: The sample size will increase since B appears in the denominator of the sample size formula. Here, you want to be more accurate, this entails additional cost.
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Part F: For questions 14-15 refer to the following information. Suppose that the researcher in
Part E decided to take a sample of size n = 1500 South Carolinians, and from this sample, the researcher finds that 400 play the state lottery.
14. Based on this information, construct a 99% confidence interval for the population proportion p?
Answer: For a 99% CI, the z-score is 2.575. The sample proportion is 400/1500 = .2667.
Therefore, the 99% CI for p is
[.2667 +/- (2.575)sqrt{(.2667)(1-.2667)/1500}] = [.2667 +/- .0294] = [.2373, .2961].
15. Based on the confidence interval in question 14, could the researcher conclude that the proportion (in percent) of all South Carolinians who play the state lottery is at least 30%?
Answer: Since the upper limit of the confidence interval obtained in problem 14 is smaller than .30, then the true population proportion is less than .30 (with 99% confidence).
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Part G: Consider a population whose probability function is described by: x 1 2 p(x) .5 .2
3
.3
16. Determine the population mean,

, and the population standard deviation,

.
Answers:

= (1)(.5) + (2)(.2) + (3)(.3) = 1.8

2 = (1-1.8) 2 (.5) + (2-1.8) 2 (.2) + (3-1.8) 2 (.3) = .76.

= sqrt(.76) = .8717.
17. Enumerate all possible samples of size n=2 together with their respective probabilities. Also, for each of these samples, compute the sample mean.
Answer:
Possible Samples
1, 1
1, 2
Probability
(.5)(.5) = .25
Sample Mean
1
1, 3
2, 1
2, 2
(.5)(.2) = .10
(.5)(.3) = .15
(.2)(.5) = .10
(.2)(.2) = .04
1.5
2
1.5
2
2, 3
3, 1
3, 2
(.2)(.3) = .06
(.3)(.5) = .15
(.3)(.2) = .06
2.5
2
2.5
3, 3 (.3)(.3) = .09 3
18. Determine the sampling distribution of the sample mean. Present this in tabular form.
Answer: Sampling Distribution of the Sample Mean
Value of the Sample Mean
1
1.5
2
2.5
3
Probability
.25
.20 (.10 + .10)
.34 (.15 + .04 + .15)
.12 (.06 + .06)
.09
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19. Based on the sampling distribution you constructed in (18), determine the mean of the sample mean. Is this equal to the population mean?
Answers:
Mean of Sample Mean = (1)(.25) + (1.5)(.20) + (2)(.34) + (2.5)(.12) + (3)(.09) = 1.8
This coincides with the population mean.
20. Determine the standard error of the sample mean. How does this relate to the population standard deviation?
Answers:
Variance of Sample Mean = (1 - 1.8) 2 (.25) + (1.5 - 1.8) 2 (.20) +(2 - 1.8) 2 (.34) + (2.5 - 1.8) 2 (.12)
+ (3 - 1.8) 2 (.09) = .38
Standard Error of Sample Mean = sqrt(.38) = .6164.
Notice that .6164 = .8717/sqrt(2) =

/sqrt(n).
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