however, still not enough to offset the lower hours in the remaining sample to the
degree that the null hypothesis can be rejected.
9.53
(a)
0
1
:
2
:
2
H
H
Decision rule: Reject
0
H
if
|t|
> 2.0096
d.f.
= 49
Test statistic:
–
2.0007-2
0.1143
0.0446
50
X
t
S
n
Decision: Since
|t
| < 2.0096, do not reject
0
H
.
There is not enough evidence to
conclude that the mean amount of soft drink filled is different from 2.0 liters.

Solutions to End-of-Section and Chapter Review Problems
177
9.53
(b)
p
-value = 0.9095.
The probability of observing a sample of 50 soft drinks that will
cont.
result in a sample average amount of fill more different from 2.0 liters is 0.9095 if
the population average amount of soft drink filled is indeed 2.0 liters.
(c)
In order for the
t
test to be valid, the data are assumed to be independently drawn
from a population that is normally distributed.
Since the sample size is 50, which is
considered fairly large, the
t
distribution will provide a good approximation to the
sampling distribution of the mean as long as the population distribution is not very
skewed.
(d)
Box-and-whisker Plot
Amount
1.8
1.85
1.9
1.95
2
2.05
2.1
2.15
2.2
The box-and-whisker plot suggests that the data are rather symmetrically distributed.
Hence, the results in (a) are valid in terms of the normality assumption.
(e)
Time Series Plot
1.75
1.8
1.85
1.9
1.95
2
2.05
2.1
2.15
1
4
7
10
13
16
19
22
25
28
31
34
37
40
43
46
49
Amount
The time series plot of the data reveals that there is a downward trend in the amount
of soft drink filled.
This violates the assumption that data are drawn independently
from a normal population distribution because the amount of fill in consecutive
bottles appears to be closely related.
As a result, the
t
test in (a) becomes invalid.

178
Chapter 9: Fundamentals of Hypothesis Testing: One-Sample Tests
9.54
(a)
0
1
:
0
:
0
H
H
Decision rule: Reject
0
H
if
|t|
> 1.9842
d.f.
= 99
Test statistic:
–
-0.00023
-1.3563
0.00170
100
X
t
S
n
Decision: Since
|t
| < 1.9842, do not reject
0
H
.
There is not enough evidence to
conclude that the average difference is different than 0.0 inches.
(b)
p
-value = 0.1781.
The probability of observing a sample of 100 steel parts that will
result in a sample average difference further away from the hypothesized value than
this sample is 0.9095 if the population average difference is indeed 0.0 inches.
(c)
In order for the
t
test to be valid, the data are assumed to be independently drawn
from a population that is normally distributed.
Since the sample size is 100, which is
considered quite large, the
t
distribution will provide a good approximation to the
sampling distribution of the mean as long as the population distribution is not very
skewed.
(d)
Box-and-whisker Plot
Error
-0.006
-0.004
-0.002
0
0.002
0.004
0.006
The box-and-whisker plot suggests that the data has a distribution that is skewed
slightly to the right.
Given the relatively large sample size of 100 observations, the
t
distribution should still provide a good approximation to the sampling distribution of
the mean.
9.55
(a)
0
1
:
5.5
:
5.5
H
H
Decision rule: Reject
0
H
if
|t|
> 2.680
d.f.
= 49
Test statistic:
–
5.5014-5.5
0.0935
0.1058
50
X
t
S
n
Decision: Since
|t
| < 2.680, do not reject
0
H
.
There is not enough evidence to