1. Exhibit 10-1
Salary information regarding male and female employees of a large company is shown
below.
Male Female
Sample Size 64 36
Sample Mean Salary (in $1,000) 44 41
Population Variance (σ²) 128 72
Solution: margin of error is: z(0.025)√(128/45+72/36) = 1.96(2) = 3.92
Refer to Exhibit 10-1. At 95% confidence, the margin of error is
a. 1.645
b. 1.96
c. 2.000
d. 3.920
_____________
2. Refer to Exhibit 10-1. If you are interested in testing whether or not the average salary
of males is significantly greater than that of females, the test statistic is
Solution: statistic = (44-41)/ √(128/45+72/36) = 3/2 = 1.5
a. 1.96
b. 2.0
c. 1.5
d. 1.645
________________________________
4. Exhibit 10-3
Today Five Years Ago
x̄(xbar) 82 88
σ² 112.5 54
n 45 36
Refer to Exhibit 10-3. The p-value for the difference between the two population
means is
Solution: statistic = (82-88)/√112.5/45+54/36) = -6/2 = -3
p-value = 2P(z<-3) = 2(0.0013) = 0.0026
a. .0026
b. .4987
c. .9987
d. .0013
_____________________________________________________
5.
Exhibit 12-1
When individuals in a sample of 150 were asked whether or not they supported
capital punishment, the following information was obtained.
Do you support capital punishment? # of individuals
Yes 40
No 60
No Opinion 50
We are interested in determining whether or not the opinions of the individuals (as to
Yes, No, and No opinion) are uniformly distributed.
The conclusion of the test (at 95% confidence) is that the
Solution:
Ho: Opinions are uniformly distributed
Ha: Opinions are not uniformly distributed
Fo
Fe
(Fo-Fe)2/Fe
Yes
40
50
2
No
60
50
2
No opinión Total
50
150
50
150
0
4
Statistical value = 4
Df = 2, p-value = P(chi2 > 4) = 0.1353 > 0.05
We fail to reject Ho
a. test is inconclusive
b. distribution is not uniform
c. distribution is uniform
d. none of these alternatives is correct
_______________________________
6. Exhibit 12-2
Last school year, the student body of a local university consisted of 30% freshmen,
24% sophomores, 26% juniors, and 20% seniors. A sample of 300 students taken from
this year’s student body showed the following number of students in each
classification
Freshmen 83
Sophomores 68
Juniors 85
Seniors 64
We are interested in determining whether or not there has been a significant change
in the classifications between the last school year and this school year.
Solution:
Ho: Opinions are uniformly distributed
Ha: Opinions are not uniformly distributed
Freshmen Sophomores Juniors
Total
Fo
Fe
(Fo-Fe)2/Fe
83
90
0.5444
68
72
0.2222
85
78
0.6282
Seniors
64
60
0.2667
Total
300
300
1.6615
Statistical value = 1.6615
Refer to Exhibit 12-2. The calculated value for the test statistic equals
a. 300
b. 1.6615
c. 6.6615
d. 0.5444
_________________________________
7. In a sample of 100 Republicans, 60 favored the President's anti-drug program. While
in a sample of 150 Democrats, 84 favored his program. At 95% confidence, test to see
if there is a significant difference in the proportions of the Democrats and the
Republicans who favored the President's anti-drug program.
a.State the hypotheses involved in this test.
Η 0: P1 - P2 - Select your answer -greater thangreater than or equal to equal to less
than or equal toless thannot equal toItem 1 0
Η a: P1 - P2 - Select your answer -greater thangreater than or equal toequal toless
than or equal toless than not equal to Item 2 0
b.Compute the value of the z test statistic (to 2 decimals).
Solution:
p1-hat = 60/100 = 0.6
p2-hat = 84/150 = 0.56
p-hat = (60+84)/(100+150) = 0.576
Statistic = z = (0.6-0.56)/√[0.56(1-0.56)(1/100+1/150)] = 0.6269623
Answer: 0.63
c.What is the p-value (to 4 decimals)?
Solution: p-value = 2P(z>0.63) = 0.5287
Answer: 0.5287
d.What is your conclusion?
- Select your answer -Conclude there is a significant difference between the proportions of
Democrats and Republicans who favor the president's anti-drug plan Do not conclude there is a
significant difference between the proportions of Democrats and Republicans who favor the
president's anti-drug plan
_______________________________________________
8.
For a one-tailed test (upper tail), a sample size of 26 at 90% confidence, t =
a. 1.316
b. 1.740
c. -1.740
d. -1.316
__________________________________________________
9. The average hourly wage of computer programmers with 2 years of experience has been
$21.80. Because of high demand for computer programmers, it is believed there has been a
significant increase in the average wage of computer programmers. To test whether or not
there has been an increase, the correct hypotheses to be tested are
a. H0: μ > 21.80 Ha: μ ≤ 21.80
b. H0: μ = 21.80 Ha: μ ≡ 21.80
c. H0: μ ≤ 21.80 Ha: μ > 21.80
d. H0: μ < 21.80 Ha: μ ≥ 21.80
10. A weatherman stated that the average temperature during July in Chattanooga is 80
degrees or less. A sample of 32 Julys is taken. The correct set of hypotheses is
a. H0: μ ≤ 80 Ha: μ > 80
b. H0: μ ≡ 80 Ha: μ = 80
c. H0: μ ≥ 80 Ha: μ < 80
d. H0: μ < 80 Ha: μ > 80
11. A group of young businesswomen wish to open a high fashion boutique in a vacant
store but only if the average income of households in the area is at least $25,000. A random
sample of 9 households showed the following results.
$28,000 $24,000 $26,000 $25,000
$23,000 $27,000 $26,000 $22,000
$24,000
Assume the population of incomes is normally distributed.



1. Compute the sample mean and the standard deviation (to 2 decimals, if
necessary).
x ¯ = 25,000 and σ = 1936.49
2. State the hypotheses for this problem.
Η 0: μ - Select your answer -greater than greater than or equal to equal toless than or
equal toless thannot equal to 25,000
Η a: μ - Select your answer -greater thangreater than or equal toequal toless than or
equal to less than not equal to 25,000
3. Compute the test statistic.
Answer: 0
12. The average starting salary of students who graduated from colleges of Business in
2009 was $48,400. A sample of 100 graduates of 2010 showed an average starting salary of
$50,000. Assume the standard deviation of the population is known to be $8,000. We want
to determine whether or not there has been a significant increase in the starting salaries.

1. State the null and alternative hypotheses to be tested.
Ho: μ ≤ 48,400
Ha: μ > 48,400

2. Compute the test statistic.
Answer: 2

3. The null hypothesis is to be tested at 95% confidence. Determine the critical
value for this test (to 2 decimals).
Answer: 1.65

4. What do you conclude?
We reject Ho (there has been a significant increase in the starting salaries)

5. Compute the p-value (to 4 decimals).
%
Answer: 0.0228
13. A department store believes that telephone calls come into the switchboard at 10-minute
intervals, according to a Poisson distribution. Before ordering new equipment, the store
wishes to determine whether the Poisson model is a valid assumption. Records on the
number of calls received were kept for a random selection of 150 ten-minute intervals. The
results are shown below.
Number of Calls
Frequency
0
5
1
18
2
24
3
30
4
32
5
13
6
20
7
8
___
150

1. What is the average number of calls during these ten-minute intervals (to 1
decimal)?
Answer: 3.5

2. Generate the expected number of calls using a Poisson probability table (to 3
decimals).
Number of Calls
0 or 1
2
3
4
5
6
7 or more



ei
20.383
27.744
32.368
28.322
19.825
11.565
9.793
3.
4. Give the null and alternative hypotheses for the appropriate test.
Η 0: - Select your answer -The number of telephone calls during a 10 minute
interval follows a Poisson distributionThe number of telephone calls during a 10
minute interval does not follow a Poisson distribution
Η a: - Select your answer -The number of telephone calls during a 10 minute
interval follows a Poisson distributionThe number of telephone calls during a 10
minute interval does not follow a Poisson distribution
5. Determine the number of degrees of freedom for this test.
Let k = number of categories or classes remaining after combining classes
k = 7, Df = k-1 = 7-1 = 6
Answer: df = 6

6. Calculate the value of the test statistic (to 2 decimals).
χ2 =
(23-20.343)2/20.343 +…..+(8-9.793)2/9.793 = 9.99
Answer: 9.99

7. Based on the p-value what is your conclusion?
If we use = 0.05
p-value = P(chi>9.99)= 0.1251 > 0.05
Answer: we fail to reject Ho
- Select your answer -The Poisson distribution is not a valid model
The Poisson distribution is a valid model