7.11)
A simple random sample of n = 300 full-time employees is selected from a company list containing the names of all N =
5,000 full-time employees in order to evaluate job satisfaction.
A) Give an example of possible coverage error.
C) Give an example of possible sampling error.
(A) A possible coverage error occurs if none of the employees from a particularly large department are included in the
sample
(B) A possible sampling error can occur if a particular department whose employees are known to be consistently overperforming has been over-represented in the sample
7.15)
Given a normal distribution with µ = 100 and σ = 10, if you select a sample of n - 25, what is the probability that X is
A) Less than 95?
C) Above 102.2?
(A) z = (x - μ)/(σ/√n) = (95 - 100)/(10/√25) = -2.5
P(x-bar < 95) = P(z < -2.5) = 0.0062
(C) z = (x-bar - μ)/(σ/√n) = (102.2 - 100)/(10/√25) = 1.10
P(x-bar > 102.2) = P(z > 1.10) = 0.1357
7.27)
You plan to conduct a marketing experiment in which students are to taste one of two different brands of soft drink.
Their task is to correctly identify the brand tasted. You select a random sample of 200 students and assume that the
students have no ability to distinguish between the two brands. (Hint : If an individual has no ability to distinguish
between the two drinks, then each brand is equally likely to be selected).
A) What is the probability that the sample will have between 50% and 60% of the identifications correct?
C) What is the probability that the sample percentage of correct identifications is greater than 65%?
(A) p = 0.5, n = 200
q = 1 - p = 0.5
z = (p’ - p)/√(pq/n) = (0.5 - 0.5)/√(0.5 * 0.5/200) = 0 and x = (0.6 - 0.5)/(√(0.5 * 0.5/200) = 2.8284
P(0.5 < p’ < 0.6) = P(0 < z < 2.8284) = 0.4977
(C) z = (p’ - p)/(√(pq/n) = (0.65 - 0.5)/(√(0.5 * 0.5/200) = 4.2426
P(p’ > 0.65) = P(z > 4.2426) = 0
8.1)
If X = 85, σ = 8, and n = 64, construct a 95% confidence interval estimate of the population mean, µ.
n = 64
x-bar = 85
s= 8
% = 95
Standard Error, SE = s/√n = 8 /√64 = 1
z- score = 1.959963985
Width of the confidence interval = z * SE =
1.95996398454005 * 1 = 1.959963985
Lower Limit of the confidence interval = x-bar - width =
85 - 1.95996398454005 = 83.04003602
Upper Limit of the confidence interval = x-bar + width =
85 + 1.95996398454005 = 86.95996398
The 95% confidence interval is [83.04, 86.96]
8.9)
Assuming that the population is normally distributed, construct a 95% confidence interval estimate for the population
mean for each of the following samples:
Sample A: 1,1,1,1,8,8,8,8
Sample B: 1,2,3,4,5,6,7,8
Explain why these two samples produce different confidence intervals even though they have the same mean and range.
(A) x-bar = 4.5, s = 3.74
n= 8
x-bar = 4.5
s = 3.74
% = 95
Standard Error, SE = s/n = 3.74/√8 = 1.322289681
Degrees of freedom = n - 1 = 8 -1 = 7
t- score = 2.364624251
Width of the confidence interval = t * SE =
2.36462425094932 * 1.32228968081884 = 3.126718246
Lower Limit of the confidence interval = x-bar - width =
4.5 - 3.12671824604427 = 1.373281754
Upper Limit of the confidence interval = x-bar + width =
4.5 + 3.12671824604427 = 7.626718246
The 95% confidence interval is [1.373, 7.627]
(B) x-bar = 4.5, s = 2.45
n= 8
x-bar = 4.5
s = 2.45
% = 95
Standard Error, SE = s/n = 2.45/√8 = 0.866205807
Degrees of freedom = n - 1 = 8 -1 = 7
t- score = 2.364624251
Width of the confidence interval = t * SE =
2.36462425094932 * 0.866205806953521 = 2.048251257
Lower Limit of the confidence interval = x-bar - width =
4.5 - 2.04825125743542 = 2.451748743
Upper Limit of the confidence interval = x-bar + width =
4.5 + 2.04825125743542 = 6.548251257
The 95% confidence interval is [2.452, 6.548]
We can clearly see that sample A has more variation in the scores than sample B. (In fact, standard deviation is 3.74 for
A and 2.45 for B). Therefore, they produce different confidence intervals even though they have the same mean and
range.
8.11)
A stationery store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample
of 100 greeting cards indicates a mean value of $2.55 and a standard deviation of $0.44.
A) Assuming a normal distribution, construct a 95% confidence interval estimate of the mean value of all greeting cards
in the store's inventory.
B) Suppose there were 2,500 greeting cards in the store's inventory. How are the results in (a) useful in assisting the
store owner to estimate the total value of her inventory?
(A) n = 100
x-bar = 2.55
0.44
% = 95
Standard Error, SE = σ/n = 0.44 /√100 = 0.044
z- score = 1.959963985
Width of the confidence interval = z * SE =
1.95996398454005 * 0.044 = 0.086238415
Lower Limit of the confidence interval = x-bar - width =
2.55 - 0.0862384153197624 = 2.463761585
Upper Limit of the confidence interval = x-bar + width =
2.55 + 0.0862384153197624 = 2.636238415
The 95% confidence interval is [$2.46, $2.64]
(B) Using the above confidence interval, the store owner can know how much money is locked in the inventory. In this
case, the money locked up in the inventory lies in the range [2500 * $2.46, 2500 * $2.64] = [$6150, $6600]
8.13)
The U.S. Department of Transportation requires tire manufacturers to provide tire performance information on the
sidewall of the tire to better inform prospective customers when making purchasing decisions. One very important
measure of tire performance is the tread wear index, which indicates the tire's resistance to tread wear compared with a
tire graded with a base of 100. This means that a tire with a grade of 200 should last twice as long, on average, as a tire
graded with a base of 100. A consumer organization wants to estimate the actual tread wear index of a brand name of
tires that claims "graded 200" on the sidewall of the tire. A random sample of n = 18 indicates a sample mean tread
wear index of 195.3 and a sample standard deviation of 21.4.
A) Assuming that the population of tread wear indexes is normally distributed, construct a 95% confidence interval
estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name.
B) Do you think that the consumer organization should accuse the manufacturer of producing tires that do not meet the
performance information provided on the sidewall of the tire? Explain.
(A) n = 18
x-bar = 195.3
s = 21.4
% = 95
Standard Error, SE = s/n = 21.4/√18 = 5.044028372
Degrees of freedom = n - 1 = 18 -1 = 17
t- score = 2.109815559
Width of the confidence interval = t * SE =
2.10981555859266 * 5.04402837246404 = 10.64196954
Lower Limit of the confidence interval = x-bar - width =
195.3 - 10.6419695382074 = 184.6580305
Upper Limit of the confidence interval = x-bar + width =
195.3 + 10.6419695382074 = 205.9419695
The 95% confidence interval is [184.66, 205.94]
(B) Since 200 lies within the above confidence interval, the consumer organization can’t accuse the manufacturer of
producing tires that do not meet the performance information provided on the sidewall of the tire
8.27)
The start of the twenty-first century saw many corporate scandals and many individuals lost faith in business. In a 2007
poll conducted by the New York City-based Edelman Public Relations firm, 57% of respondents say they trust business to
"do what is right" This percentage was the highest in the annual survey since 2001 (data extracted from G. Colvin,
"Business is Back" Fortune, May 14, 2007, pp. 40-48).
A) Construct a 95% confidence interval estimate of the population proportion of individuals who trust business to "do
what is right" assuming that the poll surveyed:
1. 100 individuals
2. 200 individuals
3. 300 individuals
B) Discuss the effect that sample size has on the width of confidence intervals.
(A)
(1) n = 100
p = 0.57
% = 95
Standard Error, SE = p(1 - p)/n} = √(0.57(1 - 0.57))/100 = 0.049507575
z- score = 1.959963985
Width of the confidence interval = z * SE =
1.95996398454005 * 0.0495075751779463 = 0.09703306
Lower Limit of the confidence interval = P - width =
0.57 - 0.0970330643106838 = 0.47296694
Upper Limit of the confidence interval = P + width =
0.57 + 0.0970330643106838 = 0.66703306
The 95% confidence interval is [0.473, 0.667]
(2) n = 200
p = 0.57
% = 95
Standard Error, SE = p(1 - p)/n} = √(0.57(1 - 0.57))/200 = 0.035007142
z- score = 1.959963985
Width of the confidence interval = z * SE =
1.95996398454005 * 0.0350071421284286 = 0.06861274
Lower Limit of the confidence interval = P - width =
0.57 - 0.0686127377733949 = 0.50138726
Upper Limit of the confidence interval = P + width =
0.57 + 0.0686127377733949 = 0.63861274
The 95% confidence interval is [0.501, 0.639]
(3) n = 300
p = 0.57
% = 95
Standard Error, SE = p(1 - p)/n} = √(0.57(1 - 0.57))/300 = 0.028583212
z- score = 1.959963985
Width of the confidence interval = z * SE =
1.95996398454005 * 0.0285832118559129 = 0.05602207
Lower Limit of the confidence interval = P - width =
0.57 - 0.0560220658000676 = 0.51397793
Upper Limit of the confidence interval = P + width =
0.57 + 0.0560220658000676 = 0.62602207
The 95% confidence interval s [0.514, 0.626]
(B) As the sample size increases, the width of the confidence interval decreases.
8.33)
If you want to be 95% confident of estimating the population proportion to within a sampling error of ± 0.02 and there is
historical evidence that the population proportion is approximately 0.40, what sample size is needed?
N = (z/E)^2 * p(1 - p) = (1.96/0.02)^2 * 0.40(1 - 0.40) = 2305
N = 2305
8.35)
If the manager of a paint supply store wants to estimate the mean amount of paint in a 1 - gallon can to within ± 0.004
gallon with 95% confidence and also assumes that the standard deviation is 0.02 gallon, what sample size is needed?
N = (z * σ/E)^2 = (1.96 * 0.02/0.004)^2 = 97
N = 97