CHAPTER 6: Midterm Exam
In the following exercises, illustrate by sketching the graph whenever possible to indicate
your solution to the problem.
1. Excluding Monday mornings, a secretary can type an average of 50 words per
minute (with a standard deviation of 4 words). The secretary's boss will invest in a
new typewriter if the new model can be shown to increase productivity. Forty
separate tests are conducted on the new model with a resulting mean of 53 words
per minute. At the significance level of 0.01, test the claim that the new model
increases productivity.
Solution:
a.
H0 :
H1 :
  50
  50 (claim) words per minute
b. Since  = 0.01 and the test is a one-tailed right test, the critical
value is z = 2.33
c. z
= [( X -  )] / [ /
= [53 – 50] / [4/
= 3 / 0.63
= 4.74
n]
40 ]
d. Since the test value 4.74 is more than the critical value 2.33, the
decision is reject the null hypothesis that the mean is less than or
equal to 50. The new model does appear to increase production.
z = 4.74
 = 50 or z = 0
Words per minute
2. A local brand sugar is packaged in plastic bags labeled 5 kilos. One hundred
packages are randomly selected and measured. The sample mean and standard
deviation are found to be 4.95 kilos and 0.15 kilo, respectively. At the 0.01
significance level, test the claim that the sample comes from a population which
the mean weight is less than 5 kilos. Assume that the sample standard deviation
can be used for the population standard deviation.
Solution:
a.
H0 :
  5 kilos
H1 :
  5 kilos (claim)
b. Since  = 0.01 and the test is a one-tailed left test, the critical value
is z = - 2.33
c. x =  /

= 0.15 /
= 0.015
100
Z = ( X - x) / x
= (4.95 – 5) / 0.015
= - 0.05 / 0.015
= - 3.33
d. Since the test value – 3.33 is less than the critical value - 2.33, the
decision is reject the null hypothesis that the mean is greater than
or equal to 5 kilos. The sample data support the claim that the
mean weight is less than 5 kilos.
z = - 3.33
 = 5 or z = 0
Kilos of sugar per bag
3. At the significant level of 0.03, test the claim the following I.Q. scores come from
a special group in which the mean is above 100. Use the sample standard deviation as
an estimate for the population standard deviation.
Solution:
X
101
107
107
110
90
89
114
91
105
105
99
130
79
95
106
144
117
103
111
93
100
99
103
118
101
120
98
107
82
101
103
123
82
112
Total 3545
(X - X )
-3.3
2.7
2.7
5.7
-14.3
-15.3
9.7
-13.3
0.7
0.7
-5.3
25.7
-25.3
-9.3
1.7
39.7
12.7
-1.3
6.7
-11.3
-4.3
-5.3
-1.3
13.7
-3.3
15.7
-6.3
2.7
-22.3
-3.3
-1.3
18.7
-22.3
7.7
-1.2
(X - X )2
10.9
7.3
7.3
32.5
204.5
234.1
94.1
176.9
0.5
0.5
28.1
660.5
640.1
86.5
2.9
1576.1
161.3
1.7
44.9
127.7
18.5
28.1
1.7
187.7
10.9
246.5
39.7
7.3
497.3
10.9
1.7
349.7
497.3
59.3
6054.7
s2
= (X- X ) / (n-1)
= 6054.7 / [34 – 1]
= 6054.7 / 33
= 183.47
s
=
s2
=
183.47
= 13.54 or 13.5
a.
H0 :
H1 :
  100
  100 (claim) I.Q. scores
d. Since  = 0.03 and the test is a one-tailed right test, the critical
value is z = 1.88
e. x =  /

= 13.5 /
34
= 13.5 / 5.83
= 2.315
Z = ( X - x) / x
= (104.3 – 100) / 2.315
= 4.3 / 2.32
= 1.853 or 1.86
f. Since the test value 1.86 is less than the critical value 1.88, the
decision is not to reject the null hypothesis that the mean is less
than or equal to 100. The sample data do not support the claim
that the mean is above 100.
Critical and Noncritical
Regions for  = 0.03
One-Tailed Right Test
0.9700
 = 0.03
 = 5 or z = 0
1.88
z= 1.86
I. Q. scores
4. A sociologist designs a test to measure prejudicial attitudes and claims the
mean population score is 60. The test is then administered to 28 randomly
selected subjects and the results produce a mean and standard deviation of
69 and 12, respectively. At the 5% level of significance, test the
sociologist's claim.
a.
H0:  = 60 (claim) test score
H1:   60
b.
Since  = 0.05 and the test is a two-tailed test, the critical value
is t = - 0.025 and 0.025
c.
t = ( X -  ) / (s /
= [69 – 60] / [12/
= 9 / [12 / 5.29]
= 9 / 2.26
= 3.968 or 3.97
)
28 ]
d.
In this two-tailed test, the critical value of - 2.052 and 2.052 is
found from the table using DF = 28 – 1 = 27 and  = 0.05
e.
Since the test statistic of t = 3.97 does not fall in the critical
region, we reject the null hypothesis mean = 60.
f.
The mean appears to be different from 60 test score.
3.97 belongs to the
non-critical region
Finding the Critical
Values for  = 0.05
Two-Tailed Test
0.9500
0.025
0.025
0.475
 = 60 or z = 0
Test scores
5. Test the claim that the mean female reaction to a highway signal is less
than 0.70 second. Eighteen females are randomly selected and tested; their
mean is 0.668 second. Assume that the population standard deviation is
0.10 second and use a 5% level of significance.
a.
H0:  > 0.70 (claim) test score
H1:   0.70
b. Since  = 0.05 and the test is a left-tailed test
c. t = ( X -  ) / (s /  )
= [0.668 – 0.70] / [0.10 / 18 ]
= [0.668 – 0.70] / [0.10 / 4.24]
= - 0.032 / 0.023
= - 1.357 or - 1.36
d. In this left-tailed test, the critical value of 1. 645 is found from the
table.
e. Since the test statistic of t = - 1.36 does not fall in the critical
region, we do not reject the null hypothesis that the mean is
greater than or 0.70. The data do not support the claim that the
mean is less than 0.70 second.
Critical value and critical
region
0.05
0.4500
- 0.136
 = 0.70 or z = 0
Seconds