Solution 1
Step 1: State the null and alternate hypotheses :
The null hypothesis refers to the “no change” situation. That is, there has been no change in the Clipper's efficiency; it is 6.25 or less litres per 100 km.
It is written: H
0
:

≤ 6.25 and is read that the population mean is less than or equal to 6.25.
The alternate hypothesis written: H
1
:

> 6.25 and is read that the population mean is greater than 6.25.
If the null hypothesis is rejected, then the alternate hypothesis is accepted. It would be concluded that the
Clipper uses more than 6.25 litres per 100 km.
Step 2: S elect the level of significance:
The testing agency decided on the 0.01significance level. This is the probability that the null hypothesis will be rejected when in fact it is true.
Step 3: Decide on a test statistic:
The use of the standard normal distribution requires that the population standard deviation

be known.
When it is not known, as in this problem, the sample standard deviation designated by s , is used as an estimate of

. When the sample standard deviation is based on a large sample, the standard normal distribution is still an appropriate test statistic. “Large” is usually defined as being more than 30.
To determine z we use formula [9-2]. z

X
  s / n
Where:

X is the sample mean.
is the population mean. s is the standard deviation computed from the sample. n is the sample size.
Step 4: Develop the decision rule:
The decision rule is a statement of the conditions under which the null hypothesis is rejected. The decision rule is shown in the following diagram. If the computed value of z is to the left of

2.33, the null hypothesis is rejected. The

2.33 is the critical value. How is it determined?
Remember that the significance level stated in the problem is 0.01. This indicates that the area to the left of the critical value under the normal curve is 0.01. For the standard normal distribution the total area to the left of 0 is 0.5000. Therefore, the area between the critical value and 0 is 0.4900, found by 0.5000

0.0100. Now refer to Appendix D and search the body of the table for a value close to 0.4900. The closest value is 0.4901. Read 2.3 in the left margin and 0.03 in the column containing 0.4901. Thus the z value corresponding to 0.4901 is 2.33.
153 Chapter 9
One-Sample Tests of A Hypothesis

Recall from Step 1 that the alternate hypothesis is H
1
:

> 6.25. The inequality sign points in the negative direction. Thus the critical value is

2.33 and the rejection region is all in the lower left tail.
Step 5: Compute the value of the test statistic, make a decision regarding the null hypothesis, and interpret the results :
Since the standard deviation of the population is not known, the sample standard deviation is used as its estimate. Repeating the formula for z : z

X
  s / n
Recall that the manufacturer claims 6.25 l / 100 km and the mean of the sample is 6.8 l / 100 km. Solving for z: z

1.5 / 36
 
2.20
The computed value of

2.20 is to the right of

2.33, so the null hypothesis is not rejected. We do not reject the claim of the manufacturer that the Clipper uses at most 6.25 litres per 100 km. It is reasonable that the 0.55 litres per 100 km between (6.25 and 6.8) could be due to chance.
We do observe, however, that

2.20 is fairly close to the critical value of

2.33. What is the likelihood of a z value to the left of

2.20? It is 0.0139, found by 0.5000

0.4861, where 0.4861 is the likelihood of a z value between 0 and 2.20. (The probabilities are found in Appendix D.)
The 0.0139 is referred to as the pvalue. It is the probability of getting a value of the test statistic ( z in this case) more extreme than that actually observed, if the null hypothesis is true. Had the significance level been set at 0.02 instead of .0.0l, the null hypothesis would have been rejected. By reporting the pvalue we give information on the strength of the decision regarding the null hypothesis.
Solution 2
Step 1: State the null and alternate hypotheses :
The null hypothesis is that there is no change in the construction time. That is, the construction time is at least 3.5 days. The alternate hypothesis is that the construction time is less than 3.5 days.
154 Chapter 9
One-Sample Tests of A Hypothesis

Symbolically, these statements are written as follows:
H
0
:
 
3.5
H
1
:

< 3.5
Step 2: Select the level of significance:
We decided on the 0.05 significance level. This is the probability that the null hypothesis will be rejected when in fact it is true.
Step 3: Select the test statistic:
The test static in this situation is the t distribution. The distribution of construction times follows the normal distribution, however we do not know the value of the population standard deviation. Also, we have a small sample.
Step 4: Develop the decision rule:
The decision rule is a statement of the conditions under which the null hypothesis is rejected. If the computed value of t is to the left of

1.796, the null hypothesis is rejected. The

1.796 is the critical value. How is it determined?
Remember that the significance level stated in the problem is 0.05. The critical values of t are given in
Appendix F. The number of degrees of freedom is ( n

1) = (12

1) = 11. We have a one-tailed test, so we find the portion of the table labeled “one-tailed.” Locate the column for the 0.05 significance level.
Read down the column until it intersects the row with 11 degrees of freedom. The value is 1.796.
Since this is a one-tailed test and the rejection region is in the left tail, the critical value is negative.
The decision rule is to reject H
0
if the value of t is less than

1.796.
Step 5: Make a decision regarding the null hypothesis and interpret the results :
We use text Formula [9-5]: t =
X
  s / n
Where: t is the value of the test statistic.
X is the sample mean. (3.0 days)
 
is the population mean. (3.5 days) s is the standard deviation of the sample. (0.9 days) n is the sample size. (12)
Thus: t =
X
  s / n

0.9 / 12


0.5
0.2598
 
1.925
155 Chapter 9
One-Sample Tests of A Hypothesis

Because

1.925 lies to the left of the critical value

1.796, the null hypothesis is rejected at the 0.05 significance level. This indicates that the use of the “quick setting concrete” has reduced the mean construction time to less than 3.5 days.
Solution 3
Step 1: State the null and alternate hypotheses :
The null hypothesis is that the machine is not out of adjustment. That is, the mean diameter of the discs is
6.125 inches. The alternate hypothesis is that the mean diameter is not 6.125 inches. .
Symbolically, these statements are written as follows:
H
0
H
1
:
:
 
6.125
 
6.125
Step 2: Select the level of significance:
We decided on the 0.01 significance level. This is the probability that the null hypothesis will be rejected when in fact it is true.
Step 3: Select the test statistic:
The test statistic in this situation is the t distribution. The disc diameters follow the normal distribution, however we do not know the value of the population standard deviation. Also, we have a small sample.
Step 4: Develop the decision rule:
The decision rule is a statement of the conditions under which the null hypothesis is rejected. The alternate hypothesis does not state a direction, so this is a two-tailed test.
Remember that the significance level stated in the problem is 0.01. The critical values of t are given in
Appendix F. The number of degrees of freedom is ( n

1) = (8

1) = 7. We have a two-tailed test, so we find the portion of the table labeled “two-tailed.” Locate the column for the 0.01 significance level. Read down the column until it intersects the row with 7 degrees of freedom. The value is 3.499.
The decision rule is: Reject the null hypothesis if the computed value of t is to the left of

3.449, or to the right of 3.449
Step 5: Make a decision regarding the null hypothesis, and interpret the results :
We use text Formula [9-5]: t =
X
  s / n
156 Chapter 9
One-Sample Tests of A Hypothesis

Where: t is the value of the test statistic.
X is the sample mean.
 
is the population mean. (6.125 inches) s is the standard deviation of the sample. n is the sample size. (8)
We need to calculate the mean and standard deviation of the sample. The standard deviation of the sample can be determined using either Formula [3-10] or [3-11].
Computing the sample standard deviation both ways:
X
6.115
X

X ( X

X )
2
X
2

0.008 0.000064 37.393225
6.127
6.129
6.113
0.004 0.000016 37.540129
0.006 0.000036 37.564641

0.010 0.000100 37.368769
6.124
6.121
6.131
6.124

48.984

0.001
0.002
0.000001
0.000004
37.503376
37.466641
0.008 0.000064 37.589161
0.001 0.000001 37.503376
0.000 0.000286 299.929318 s
X

 n
X

48.984

6.123
8



( X

X )
2 n

1

0.000286
0.000040857

0.0063919

0.0064
or s




X
2 
(

X n

1 n
)
2
299.929318

(48.984)
2
8
7
0.000040857

0.0064
The value of t is computed using Formula [9-5]: t =
X
  s / n

0.0064 / 8


0.002
0.0022627
 
0.8839
The computed value of

0.8829 lies between the two critical values:

3.449 and 3.449. The null hypothesis is not rejected at the 0.01 significance level. The foreman’s suspicion that the machine is out of adjustment cannot be substantiated with this sample.
Solution 4
As usual, the first step is to state the null and alternate hypotheses. The null hypothesis is that there is no change in the percent employed. That is, the population proportion is at least 0.30. The alternate hypothesis is that the population proportion is less than 0.30. This is the statement we are trying to test empirically. Symbolically, these statements are written as follows:
The 0.01 significance level is to be used. The assumptions of the binomial distribution are met in the problem. That is
H
0
: π 
0.30
H
1
:
π
< 0.30
157 Chapter 9
One-Sample Tests of A Hypothesis

1. There are only two outcomes for each trial--the student is either employed or isn't employed.
2. The number of trials is fixed

100 students.
3. Each trial is independent, meaning the employment of one student selected does not affect another.
4. The probability that any randomly selected student is employed is 0.30.
The normal approximation to the binomial is used because both n

and n (l
 
) exceed 5. That is: [ n

=
100(0.30)] = 30 and n (l
 
) = 100(0.70)= 70. The standard normal distribution, z is the test statistic. To formulate the decision rule, we need the critical value of z . Using the 0.01 significance level, the area is
0.4900, (0.5000

0.0100).
Search the body of Appendix D for a value as close to 0.4900 as possible. It is 0.4901. The z value associated with 0.4901 is 2.33. The alternate hypothesis points in the negative direction, hence the rejection region is in the left tail and the critical value of z is

2.33.
The decision rule is to reject the null hypothesis if the computed value of the test statistic lies in the rejection region to the left of

2.33.
Recall that the sample of 100 Scandia
Tech students revealed that 25 were employed. The question is whether the sample proportion of 0.25, found by
25/100, is significantly less than 0.30.
H
0
:

H
1
:


0.30

0.30
z

 p

1




.
.
.
.
g
 
.
n 100
The computed value of z falls in the region between 0 and

2.33. H
0
is not rejected. There is a difference between the Dean's hypothesized proportion (0.30) and the sample proportion (0.25), but this difference of 0.05 is not sufficient to reject the null hypothesis. The 0.05 can be attributed to sampling (chance). The
Dean's claim cannot be refuted.
The p -value is the probability of a z value to the left of

1.09. It is 0.1379, found by (0.5000

0.3621).
The p -value is larger than the significance level of 0.01, which is consistent with our decision not to reject the null hypothesis.
158 Chapter 9
One-Sample Tests of A Hypothesis