1. In recent years the interest rate on home mortgages has declined to less than 6.0
percent. However, according to a study by Federal Reserve Board the rate charge on
credit card debit is more than 14 percent. Listed below is the interest rate charged on a
sample of 10 credit cards.
14.6
16.7
17.4
17.0
17.8
15.4
13.1
15.8
14.3
14.5
Is it reasonable to conclude the mean rate charged is greater than 14 percent? Use the
0.01 significance level.
Solution
H 0 :   14
H a :  > 14
x =15.66 s=1.544
t 101 =
15.66  14
=3.4
1.544
10
Critical value t 9 , 0.01 =2.821
Since 3.4>2.821, reject
H0
at  =0.01 significance level and we conclude
the mean rate charged is greater than 14 percent.
2. The Watch Corporation of Switzerland claims that their watches on average will
neither gain nor lose time during a week. A sample of 18 watches provided the
following gains(+) or losses(-) in seconds per week.
-0.38
-0.20
-0.38
-0.32 +0.32
-0.23
+0.30
+0.25
-0.10
-0.37
-0.61 -0.48
-0.47 -0.64
-0.04
-0.20
-0.68
+0.05
Is it reasonable to conclude the mean gain or loss in time for the watches is 0? Use the
0.05 significance level.
Solution
H 0 :  =0
x =-0.232
t 181 =
Ha :   0
s=0.312
 0.232  0
=-3.155
0.312
18
critical value:  t 17, 0.05 / 2 =  2.11
Since -3.155<-2.11, reject
H 0 at 0.05 significance level and conclude the mean gain
or loss in time for the watches is not 0.
3. An urban planner claims that, nationally, 20 percent of all families renting
condominiums move during a given year. A random sample of 200 families renting
condominiums in Dallas Metroplex reveal that 56 had moved during the past year. At
the 0.1 significance level, does this evidence suggest that more than 20% of
condominium owners moved in the Dallas area?
Solution
H 0 :   0.2
H a :   0.2
n  40  5 and (1  p)  160  5
p=
z=
56
=0.28
200
0.28  0.2
=2.828
0.2(1  0.2)
200
critical value: z 0.1 =1.285
Since 2.828>1.285, reject H 0 at 0.1 significance level and suggest that 20% of
condominium owners moved in the Dallas area.
4. A coffee manufacture is interested in whether the mean daily consumption of
regular coffee drinker is less than that of decaffeinated-coffee drinkers. A random
sample of 50 regular-coffee drinkers showed a mean of 4.35 cups per day, with a
standard deviation of 1.20 cups per day. A sample of 40 decaffeinated-coffee drinkers
showed a mean of 5.84 cups per day, with a standard of 1.36 cups per day. Use the
0.01 significance level.
Solution: n1  50, X1  4.35, s1  1.20 , n2  40, X 2  5.84, s1  1.36
(1) H 0 : 1  2 ; H1 : 1  2
(2)   0.01
(3)Since  1 and  2 are unknown, assumed they are equal. This is a t-test.
df  n1  n2  2  88
(4) S p2 
t
(n1  1) s12  (n2  1) s22
 1.6215
n1  n2  2
X1  X 2  0
4.35  5.84

 5.516
0.27013
1
2 1
Sp(  )
n1 n2
(5) df=88>29, t0.01  z0.01  2.326 , the critical value is -2.326
(6) Since -5.516<-2.326, we can reject H 0 at 0.01 significance level and suggest that
the mean daily consumption of regular coffee drinker is less than that of
decaffeinated-coffee drinkers.
5. An Ohio university wishes to demonstrate that car ownership is detrimental to
academic achievement. A random sample of 100 students who do not own cars had a
mean grade point average (GPA) of 2.68, while a random sample of 100 students who
own cars had a mean GPA of 2.55.
a. Set up the null and alternative hypotheses that should be used to attempt to justify
that the mean GPA for non-car owners is higher than the mean GPA for car owners.
b. Test the hypotheses that you set up in part a with 0.05 level of significance. Again
assume that  1  0.7 and  2  0.6 . Interpret the results of this test. That is, what do
your results say about whether the university can statistically justify that car
ownership hurts academic achievement?
Solution:
n1  100 , x1  2.68 n2  100 , x 2  2.55
H 0 : 1  2 H1 : 1  2
z
x1  x2  ( 1   2 )
 12
n1
For

 22
 1.41
n2
  0.05 , the critical value is 1.645
Since 1.41<1.645, we fail to reject H 0 .
The university can not statistically justify that car ownership hurts academic achievement.
6. The Computer Anxiety Rating Scale (CARS) measures an individual’s level of
computer anxiety on a scale from 20 (no anxiety) to 100 (highest level of anxiety).
Researchers at Miami University administered CARS to 172 business students. One
of the objectives of the study was to determine if there is a difference between the
level of computer anxiety experienced by female and male business students. We
assume that variances of two populations are equal.
Males
Females
40.26
36.85
x
s
13.35
9.42
n
100
72
At the 0.05 level of significance, is there evidence of a difference in the mean
computer anxiety experienced by female and male business students?
Solution: a. n1  100 , x1  40.26 , s1  13.35
n2  72 , x 2  36.85 , s2  9.41
H 0 : 1  2 H1 : 1  2
n1  n2
s 2p 
t
(n1  1) s12  (n2  1) s22
 140.85
n1  1  n2  1
x1  x2  ( 1   2 )
 1.859
1 1
sp

n1 n2
df=100+72-2=170>30, the critical value is approximately equal to z0.025  1.96
Since 1.859<1.969, we fail to reject H 0 .
There is no evidence of a difference in the mean computer anxiety experienced by female and
male business students.
7. Multiple myeloma or blood plasma cancer is characterized by increased blood
vessel formulation (angiogenesis) in the bone marrow that is a prognostic factor in
survival. One treatment approach used for multiple myeloma is stem cell
transplantation with the patient’s own stem cells. The following data MYELOMA
represent the bone marrow microvessel density for patients who had a complete
response to the stem cell transplant as measured by blood and urine tests. The
measurements were taken immediately prior to the stem cell transplant and at the time
of the complete response.
Patient
Before
After
1
158
284
2
189
214
3
202
101
4
353
227
5
416
290
6
426
176
7
441
290
At the 0.05 level of significance, is there evidence that the mean bone marrow
microvessel density is higher before the stem cell transplant than after the stem cell
transplant?
Solution:
d1 =-126, d 2 =-25, d 3 =101, d 4 =126, d 5 =126, d 6 =250, d 7 =151
H 0 : d  0
d =86.14
t 71 =
H a :  d >0
sd =123.7
d   d 86.14  0
=
=1.842
sd
123.7
7
n
Critical value: t 6, 0.05 =1.943
Since 1.842<1.943, we can’t reject H 0 at 0.05 level of significance and there is
not enough evidence that the mean bone marrow microvessel density is higher
before the stem cell transplant than after that.
8.A number of minor automobile accidents occur at various high-risk intersections in
Teton County despite traffic lights. The Traffic Department claims that a modification
in the type of light will reduce these accidents. The county commissioners have
agreed to a proposed experiment. Eight intersections were chosen at random, and the
lights at those intersections were modified. The numbers of minor accidents during a
six-month period before and after the modifications were:
Number of Accidents
A
B
C
D
E
F
G
H
Before
5
7
6
4
8
9
8
10
After
3
7
7
0
4
6
8
2
At the 0.05 level of significance, is there evidence that the modification in the type of
light will reduce the accidents?
9. Two hospital emergency rooms use different procedures for triage of their patients.
We want to test the claim that the mean waiting time of patients is the same for both
hospitals. The 40 randomly selected subjects from one hospital produce a mean of
18.3 minutes. The 50 randomly selected patients from the other hospital produce a
mean of 25.31 minutes. Assume a a=2.1 minutes and b=2.92 minutes. Setup the null
and alternative hypothesis to determine if there is a difference in the mean waiting
time between the two hospitals. Calculate the test statistic for testing these hypothesis and
then test the hypothesis at the 0.01level of significance.
Ho: µa-µb=0 Ha: µa-µb0
Reject Ho
10. A test of spelling ability is given to a random sample of 10 students before and
after they completed a spelling course. The mean score before the course was 119.60
and after the course the mean score was 130.80. The standard deviation of the
difference was 16.061. Setup the null and alternative hypothesis to test the claim that
the test score is not affected by the course. Test the hypothesis at
=.05
H 0 : d  0, H a : d  0
Cannot reject the Ho
11. The registrar at a state college is interested in determining whether there is a
difference of more than one credit hour between male and female students in the
average number of credit hours taken during a term. She selected a random sample of
60 male and 60 female students and observed the following sample information.
Setup the null and alternative hypothesis to test the claim. Test the claim at =.05
Ho: µf-µm1 Ha: µf-µm>1,
For
  0.05 , the critical value is 1.645
Since 1.73>1.645, we can reject H 0 .
We have strong evidence that there is a difference of more than one credit hour between male and
female students in the average number of credit hours taken during a term.(z=1.73,p-value<0.05)
12. The production department of NDB Electronics wants to explore the relationship
between the number of employees who assemble a subassembly and the number
produced. As an experiment, two employees were assigned to assemble the
Production
subassemblies. They produced 15 during a one-hour period. Then four employees
assembled them. They produced 25 during a one-hour period. The complete set of
paired observations follows.
Number of
One-Hour Production
Assemblers
(units)
2
15
4
25
1
10
5
40
3
30
The dependent variable is production; that is, it is assumed that the level of production
depends upon the number of employee.
a. Based on scatter diagram, does there appear to be any linear relationship between
the number of assemblers and production? Explain.
b. Determine the regression equation.
c. Compute the coefficient of correlation.
d. Predict the production of 6 employee.
40
30
20
10
0
0
1
2
3
4
5
6
Asse mble rs
Yes, as the number of assemblers increases, so does the production.
b.
x =3, y =24
X
 X  X  Y  Y   X  X  Y  Y   X  X Y  Y 
2
Y
2 15
4 25
1 10
5 40
3 30
15 120
b1 =
ss xy
ss xx
=
-1
1
-2
2
0
-9
1
-14
16
6
1
1
4
4
0
10
 ( x  x )( y  y ) = 70 =7
10
(x  x)
i
i
2
i
b0 = y  b1 x =24-7  3=3
So the regression equation is: ŷ =3+7x.
c. the coefficient of correlation: r=0.9272
2
81
1
196
256
36
570
9
1
28
32
0
70
r
SS xy
SS xx SS yy
 0.9272
d. x=6, ŷ =3+7*6=45.