1. Consider a population with mean of 59.6 and st.dev 3.45(A) Calculate the z-score for
x/bar 60.7 from a sample of size 15. (B) Could this z-score be used in calculating
probabilities using Table 3 in Appendix B of the text? Why or why not? (Points : 4)
1(A) z = (xbar-59.6)/3.45 = (60.7-59.6)/3.45 =0.3188
1(B) Yes. Because z follows standard normal distribution. P(xbar < 60.7) can be
calculated.
P(xbar < 60.7) = P(z < 0.3188) = 0.6251
2. Given a level of confidence of 99% and a population standard deviation of 8, answer the
following: (A) What other information is necessary to find the sample size (n)? (B) Find the
Maximum Error of Estimate (E) if n = 94. Show all work. (Points : 4)
2(A)
If E is given, n can be found.
2(B)
E = z*σ/√n = 2.5758*8/√94 = 2.1254
3. A sample of 142 golfers showed that their average score on a particular golf course was
90.46 with a standard deviation of 3.84. Answer each of the following (show all work and
state the final answer to at least two decimal places.): (A) Find the 95% confidence interval
of the mean score for all 142 golfers. (B) Find the 95% confidence interval of the mean score
for all golfers if this is a sample of 120 golfers instead of a sample of 142. (C) Which
confidence interval is larger and why? (Points : 6)
3(A)
n = 142
xbar = 90.46
s = 3.84
df = 142-1 =141
E = t*s/√n = 1.9769*3.84/√142 =0.6371
95% confidence interval for population mean μ is
(xbar-E, xbar+E) = (89.82, 91.10)
3(B)
n = 120
xbar = 90.46
s = 3.84
df = 120-1 =119
E = t*s/√n = 1.9769*3.84/√120 =0.6941
95% confidence interval for population mean μ is
(xbar-E, xbar+E) = ( 89.77, 91.15)
3(C)
The confidence interval in (B) is larger as it is based on a smaller sample size.
4. Assume that the population of heights of male college students is approximately normally
distributed with mean m of 73.16 inches and standard deviation s of 6.39 inches. A random
sample of 70 heights is obtained. Show all work. (A) Find the mean and standard error of the
x/bar distribution P x /bar <74.50 (B) Find (Points : 6)
4(A)
m = 73.16
σ = 6.39
n = 70
Mean of xbar = m = 73.16
Standard error of xbar = σ/√n = 6.39/√70 = 0.76
4(B)
P(xbar < 74.50) = P(z < (74.50 – 73.16)/0.76) =P(z < 1.7545) = 0.9603
5. The diameters of oranges in a certain orchard are normally distributed with a mean of 4.08
inches and a standard deviation of 0.67 inches. Show all work. (A) What percentage of the
oranges in this orchard is larger than 4.17 inches? (B) A random sample of 100 oranges is
gathered and the mean diameter is calculated. What is the probability that the sample mean is
greater than 4.17 inches? (Points : 6)
5(A)
μ = 4.08
σ = 0.67
x = 4.17
z = (4.17-4.08)/0.67 = 0.1343
P(x > 4.17) = P(z > 0.1343) = 1 – P(z ≤ 0.1343) = 1-0.5534 = 0.4466
5(B)
n = 100
xbar = 4.17
Standard error of xbar = σ/√n = 0.67/√100 = 0.067
z = (4.17 – 4.08)/0.067 = 1.3433
P(xbar > 4.17) = P(z > 1.3433) = 1- P(z ≤ 1.3433) = 1-0.9104 = 0.0896
6. A researcher is interested in estimating the noise levels in decibels at area urban hospitals.
She wants to be 90% confident that her estimate is correct. If the standard deviation is 5.31,
how large a sample is needed to get the desired information to be accurate within 0.63
decibels? Show all work. (Points : 6)
Q6
Confidence coefficient = 0.90
z = 1.6449
σ = 5.31
E = 0.63
n = (z*σ/E)^2 = (1.6449*5.31/0.63)^2 = 192.204
The minimum sample size is 193
7. Consider a normal population with µ = 25 and σ = 7.0. (A) Calculate the standard score for
a value x of 22. (B) Calculate the standard score for a randomly selected sample of 30 with =
22. (C) Explain why the standard scores of 22 are different between A and B above. (Points :
6)
7(A)
µ = 25
σ = 7.0
x = 22
z = (22-25)/7.0 = -0.4286
7(B)
xbar = 22
n = 30
Standard error of xbar = σ/√n = 7.0/√30 = 1.2780
z = (22-25)/1.2780 = -2.3474
7(C)
The standard error of xbar is σ/√n which is less than σ, the standard deviation of x. So the z –
score in (B) is numerically greater than that in (A)
8. Assume that the mean score on a certain aptitude test across the nation is 100, and that the
standard deviation is 20 points. Find the probability that the mean aptitude test score for a
randomly selected group of 150 8th graders is between 95 and 105. (Points : 3)
Q8
μ = 100
σ = 20
n = 150
x1 = 95
x2 = 105
z1 = √150*(95-100)/20 = -3.0619
z1 = √150*(105-100)/20 = 3.0619
P(95 < x < 105) = P(-3.0619 < z < 3.0619) = 0.9989 – 0.0011 = 0.9978
9. Assume that a sample is drawn and z(α/2) = 1.96 and σ = 35. Answer the following
questions: (A)If the Maximum Error of Estimate is 0.02 for this sample, what would be the
sample size? (B)Given that the sample Size is 400 with this same z(α/2) and σ, what would
be the Maximum Error of Estimate? (C)What happens to the Maximum Error of Estimate as
the sample size gets smaller? (D)What effect does the answer to C above have to the size of
the confidence interval? (Points : 8)
9(A)
z = 1.96
σ = 35
E = 0.02
n = (z*σ/E)^2 = (1.96*35/0.02)^2 = 11764900
9(B)
n = 400
E = z*σ/√n = 1.96*35/√400 = 3.43
9(C)
As the sample size becomes smaller the maximum error E increases.
9(D)
The width of the confidence interval is 2E and it increases as the sample size decreases.
10. 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 3.63 seconds.
Answer each of the following (show all work): (A) How many measurements should be made
in order to be 95% certain that the maximum error of estimation will not exceed 2.0 seconds?
(B) What sample size is required for a maximum error of 0.5 seconds? (Points : 6)
10(A)
σ = 3.63
E = 2.0
n = (z*σ/E)^2 = (1.96*3.63/2.0)^2 = 12.65
The minimum sample size is 13
10(B)
σ = 3.63
E = 0.5
n = (z*σ/E)^2 = (1.96*3.63/0.5)^2 = 202.47
The minimum sample size is 203
11. A 90% confidence interval estimate for a population mean was computed to be (38.5,
58.3). Determine the mean of the sample, which was used to determine the interval estimate
(show all work). (Points : 4)
Q11
Mean of the sample = xbar = (58.3+38.5)/2 = 48.4
12. 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
$182.25. Assuming a standard , thedeviation equal to $42.16, find the 95% confidence
interval for mean for all such families (show all work). (Points : 4)
n = 195
xbar = 182.25
σ = 42.16
E = z*σ/√n = 1.96*42.16/√195 = 5.92
The confidence interval is (xbar-E, xbar+E) = (182.25-5.92, 182.25+5.92)= (176.33, 188.17)
13. A confidence interval estimate for the population mean is given to be (35.71, 44.14). If
the standard deviation is 13.045 and the sample size is 52, answer each of the following
(show all work): (A) Determine the maximum error of the estimate, E. (B) Determine the
confidence level used for the given confidence interval. (Points : 4)
13(A)
E =(44.14-35.71)/2 = 4.215
z = √n E/σ = √52*4.215/13.045 = 2.33
P(|z| ≤ 2.33) = 0.98
The confidence is 98%