1
Stat501 Hw#5 Solution Hints
9.10
a If the airline is to determine whether or not the flight is unprofitable, they are interested in finding out
whether or not   60 (since a flight is profitable if  is at least 60). Hence, the alternative hypothesis is
Ha :   60 and the null hypothesis is H0 :   60 . Formally,
H0 :   60
vs.
Ha :   60
b Since only small values of x (and hence, negative values of z) would tend to disprove H0 in favor of Ha,
this is a one-tailed test.
c
For this exercise, n  120, x  58, and s  11 .
One-Sample T
Test of mu = 0.6 vs < 0.6
N
120
Mean
0.5800
StDev
0.1100
SE Mean
0.0100
95% Upper
Bound
0.5966
T
-1.99
P
0.024
Since the p-value of 0.024 is small (<0.05) we can reject the
null hypothesis and conclude the alternative that this flight is
unprofitable.
---Period------------ This section can be omitted ----Hence, the test statistic is
z
x  0

n

x  0
s
n

58  60
11 120
 1.992
The rejection region with   .05 is determined by a critical value of z such that P  z  z0   .05 . This
value is z0  1.645 and H0 will be rejected if z  1.645 (compare the right-tailed rejection region in
Exercise 9.6). The observed value of z falls in the rejection region and H0 is rejected. The flight is
unprofitable.
2
-------------------------------------------------------------------------------------9.14
The hypothesis to be tested is
H0 :   7 versus Ha :   7
----Insert MINITAB output and make conclusions
-------------------------------- end-----and the test statistic is
z
x  0

n

x  0
s
n

6.7  7
2.7
80
 .992
The rejection region with   .05 is z  1.645 (similar to Exercise 9.10). The observed value, z  .992 ,
does not fall in the rejection region and H0 is not rejected. The data do not provide sufficient evidence to
indicate that   7 .
9.15
a The hypothesis to be tested is
H0 :   7.4 versus Ha :   7.4
----------------------------------------------Insert MINITAB output and make conclusions
--------------------------------
----- Optional------and the test statistic is
z
x  0

n

x  0
s
n

7.9  7.4
1.9
100
 2.63
with p-value  P  z  2.63  1  .9957  .0043 . To draw a conclusion from the p-value, use the guidelines
for statistical significance in Section 9.3. Since the p-value is less than .01, the test results are highly
significant. We can reject H0 at both the 1% and 5% levels of significance.
-------------------------------------------
3
b You could claim that you work significantly fewer hours than those without a college education.
c If you were not a college graduate, you might just report that you work an average of more than 7.4
hours per week..
9.16
a The hypothesis to be tested is
H0 :   98.6 versus Ha :   98.6
----------------------------------------------Insert MINITAB output and make conclusions
To replace the following.
and the test statistic is
z
x  0

n

x  0
s
n

98.25  98.6
.73 130
 5.47
with p-value  P  z  5.47   P( z  5.47)  2(0)  0 . Alternatively, we could write
p-value  2P  z  5.47   2(.0002)  .0004 With   .05 , the p-value is less than  and H0 is rejected.
There is sufficient evidence to indicate that the average body temperature for healthy humans is different
from 98.6.
-------------------------------------------------------------------------b-c
Using the critical value approach, we set the null and alternative hypotheses and calculate the
test statistic as in part a. The rejection region with   .05 is | z | 1.96 . The observed value, z  5.47 ,
does fall in the rejection region and H0 is rejected. The conclusion is the same is in part a.
d How did the doctor record 1 million temperatures in 1868? The technology available at that time
makes this a difficult if not impossible task. It may also have been that the instruments used for this
research were not entirely accurate.
9.17
The hypothesis to be tested is
H0 :   5.97 versus Ha :   5.97
----------------------------------------------Insert MINITAB output and make conclusions
4
To replace the following:
and the test statistic is
z
x  0

n

x  0
s
n

9.8  5.97
1.95
31
 10.94
with p-value  P  z  10.94  1  .9998  .0002 (or p-value  0 ). Since the p-value is less than .05, the
null hypothesis is rejected . There is sufficient evidence to indicate that the average diameter of the
tendon for patients with AT is greater than 5.97 mm.
-----------------------9.18
a-b The hypothesis of interest is one-tailed:
H0 : 1  2  0 versus Ha : 1  2  0
c
The test statistic, calculated under the assumption that 1  2  0 , is
-----------------------------------------------
Insert MINITAB output and make conclusions
To replace:
z
 x1  x2    1  2 
 12
n1

 22
n2
with  12 and  22 known, or estimated by s12 and s22 , respectively. For this exercise,
z
 x1  x2   0
2
1
2
2
s
s

n1 n2

11.6  9.7
27.9 38.4

80
80
 2.09
a value which lies slightly more than two standard deviations from the hypothesized difference of zero.
This would be a somewhat unlikely observation, if H0 is true.
d The p-value for this one-tailed test is
5
Since the p-value is not less than   .01 , the null hypothesis cannot be rejected at the 1% level. There is
insufficient evidence to conclude that 1  2  0 .
e Using the critical value approach, the rejection region, with   .01 , is z  2.33 (see Exercise 9.3a).
Since the observed value of z does not fall in the rejection region, H0 is not rejected. There is insufficient
evidence to indicate that 1  2  0 , or 1  2 .
9.21
a The hypothesis of interest is one-tailed:
H0 : 1  2  0 versus Ha : 1  2  0
b The test statistic, calculated under the assumption that 1  2  0 , is
z
 x1  x2   0
2
1
2
2
s
s

n1 n2

6.9  5.8
 2.9 
35
2
1.2 

2
 2.074
35
The rejection region with   .05 , is z  1.645 and H0 is rejected. There is evidence to indicate that
1  2  0 , or 1  2 . That is, there is reason to believe that Vitamin C reduces the mean time to
recover.
9.22
a The hypothesis of interest is one-tailed:
H0 : 1  2  0 versus Ha : 1  2  0
MINITAB output ……(use P-value)
The rejection region with   .01, is z  2.33 and H0 is rejected. There is evidence to indicate that
1  2  0 , or 1  2 . The average per-capita beef consumption has decreased in the last ten years.
(Alternatively, the p-value for this test is the area to the right of z  5.33 which is very close to zero and
less than   .01 .)
b For the difference 1  2 in the population means this year and ten years ago, the 99% lower
confidence bound uses z.01  2.33 and is calculated as
(The lower bound can be found in MINITAB output)
6
Since the difference in the means is positive, you can again conclude that there has been a decrease in
average per-capita beef consumption over the last ten years. In addition, it is likely that the average
consumption has decreased by more than 5.63 pounds per year.
9.24
The hypothesis of interest is two-tailed:
H0 : 1  2  0 versus Ha : 1  2  0
(Get a Minitab output and make conclusions)
b .
9.27
a The hypothesis of interest is two-tailed:
H0 : 1  2  0 versus Ha : 1  2  0
(Get a Minitab output and make conclusions)
b The 95% confidence interval for 1  2 is approximately
(From MINITAB output)
Since the value 1  2  0 does not fall in the interval in part b, it is not likely that 1  2 . There is
evidence to indicate that the means are different, confirming the conclusion in part a.
9.29
a The hypothesis of interest is two-tailed:
H0 : 1  2  0 versus Ha : 1  2  0
(Get a Minitab output and make conclusions)
b Since the p-value = …., we can reject H0 at the 5% level (p-value < .05), but not at the 1% level (p-value
> .01). Using the guidelines for significance given in Section 9.3 of the text, we declare the results
statistically significant, but not highly significant.
9.34
a The hypothesis of interest is two-tailed:
H0 : p  .75 versus Ha : p  .75
b (Get a Minitab output and make conclusions in stead of using following)
With x  58 and n  100 , so that pˆ 
x 58

 .58 , the test statistic is
n 100
7
z
pˆ  p0
p0 q0
n

.58  .75
.75 .25
100
 3.93
with p-value  P  z  3.93  2 .0002   .0004 or p-value  0 . Since this p-value is less than .01, H0 is
rejected at the 1% level of significance and the results are declared highly significant. There is evidence
that the proportion of red flowered plants is not .75.
9.35
a-b Since the survival rate without screening is p  2 3 , the survival rate with an effective program may
be greater than 2/3. Hence, the hypothesis to be tested is
H0 : p  2 3 versus Ha : p  2 3
c
With pˆ 
x 164

 .82 , the test statistic is
n 200
z
pˆ  p0
p0 q0
n

.82  2 3
 2 31 3
 4.6
200
The rejection region is one-tailed, with   .05 or z  1.645 and H0 is rejected. The screening program
seems to increase the survival rate.
d For the one-tailed test,
p-value  P  z  4.6  1  .9998  .0002
That is, H0 can be rejected for any value of   .0002 . The results are highly significant.
9.38
a The hypothesis of interest is
H0 : p  .25 versus Ha : p  .25
With pˆ 
x 22

 .275 , the test statistic is
n 80
z
pˆ  p0
p0 q0
n

.275  .25
.25 .75 
80
 .52
The rejection region is two-tailed   .05 , or z  1.96 and H0 is not rejected. There is insufficient
evidence to indicate that the claim is incorrect.
8
b The hypothesis of interest is
H0 : p  .24 versus Ha : p  .24
With pˆ 
x 15

 .1875 , the test statistic is
n 80
z
pˆ  p0
p0 q0
n

.1875  .24
 1.10
.24 .76 
80
The rejection region is two-tailed   .05 , or z  1.96 and H0 is not rejected. There is insufficient
evidence to indicate that the claim is incorrect.
c Unless the experimenter had some preconceived idea that the proportion might be greater or less
than claimed, there would be no reason to run a one-tailed test.
9.40
The hypothesis of interest is
H0 : p  .35 versus Ha : p  .35
with pˆ 
x 123

 .41 , the test statistic is
n 300
z
pˆ  p0
p0 q0
n

.41  .35
.35 .65 
300
 2.17
The rejection region with  =.01 is | z | 2.58 and the null hypothesis is not rejected. (Alternatively, we
could calculate p-value  2P  z  2.17   2(.0150)  .0300 . Since this p-value is greater than .01, the null
hypothesis is not rejected.) There is insufficient evidence to indicate that the percentage of adults who
say that they always vote is different from the percentage reported in Time.
9.45
a The hypothesis of interest is:
H0 : p1  p2  0 versus Ha : p1  p2  0
Calculate pˆ1  .36 , pˆ 2  .60 and pˆ 
z
n1 pˆ1  n2 pˆ 2 18  30

 .48 . The test statistic is then
n1  n2
50  50
pˆ1  pˆ 2
1 1
ˆ ˆ  
pq
 n1 n2 

.36  .60
.48 .52 1 50  1 50 
 2.40
9
The rejection region, with   .05 , is z  1.645 and H0 is rejected. There is evidence of a difference in
the proportion of survivors for the two groups.
b From Section 8.7, the approximate 95% confidence interval is
 pˆ1  pˆ 2   1.96
pˆ1qˆ1 pˆ 2 qˆ2

n1
n2
.36 .64  .60 .40 

50
50
.24  .19 or  .43   p1  p2   .05
.36  .60   1.96
9.46
a The hypothesis of interest is
H0 : p1  p2  0 versus Ha : p1  p2  0
Calculate pˆ1 
x x
123  145
123
145
 .280 , pˆ 2 
 .259 , and pˆ  1 2 
 .268 . The test statistic is then
440
560
n1  n2 440  560
z
pˆ1  pˆ 2
1 1 
ˆ ˆ  
pq
 n1 n2 

.280  .259
.268 .732 1 440  1 560 
 0.74
The rejection region, with   .01 , is z  2.58 and H0 is not rejected. There is no evidence of a difference
in the proportion of frequent moviegoers in the two demographic groups.
b A difference in the proportions might mean that the advertisers would choose different products to
advertise before this movie.
9.50
a Since the two treatments were randomly assigned, the randomization procedure can be implemented
as each patient becomes available for treatment. Choose a random number between 0 and 9 for each
patient. If the patient receives a number between 0 and 4, the assigned drug is aspirin. If the patient
receives a number between 5 and 9, the assigned drug is clopidogrel.
b Assume that n1  7720 and n2  7780 . It is given that pˆ1  .054 , pˆ 2  .038 , so that
pˆ 
n1 pˆ1  n2 pˆ 2 7720 .054   7780 .038 

 .046 .
n1  n2
15,500
The test statistic is then
z
pˆ1  pˆ 2
1 1
ˆ ˆ  
pq
 n1 n2 

.054  .038
.046 .954 1 7720  1 7780 
 4.75
10
with p-value  P  z  4.75  2(.0002)  .0004 . Since the p-value is less than .01, the results are
statistically significant. There is sufficient evidence to indicate a difference in the proportions for the two
treatment groups.
c
9.51
Clopidogrel would be the preferred treatment, as long as there are no dangerous side effects.
The hypothesis of interest is
H0 : p1  p2  0 versus Ha : p1  p2  0
Calculate pˆ1 
x x
93
93  119
119
 .598 , and pˆ  1 2 
 .6625 . The test statistic is then
 .769 , pˆ 2 
199
n1  n2 121  199
121
pˆ1  pˆ 2
z
1 1
ˆ ˆ  
pq
 n1 n2 

.769  .598
.6625 .3375 1 121  1 199 
 3.14
with p-value  P  z  3.14  1  .9992  .0008 . Since the p-value is less than .01, the results are reported
as highly significant at the 1% level of significance. There is evidence to confirm the researcher’s
conclusion.
9.60
a The hypothesis of interest is
H0 : p  .5 versus Ha : p  .5
with pˆ 
x 6

 .222 , and the test statistic is
n 27
z
pˆ  p0
p0 q0
n

.222  .5
.5 .5
27
 2.89
Since no value of  is specified in advance, we calculate p-value  P  z  2.89  .5  .4981  .0019 .
Since this p-value is less than .10, you can reject H0 at the 1% level (highly significant) and reject the DA’s
claim of 50% or greater.
b If you take x  455 and n  503 with pˆ 
x 455

 .905 , the 95% confidence interval for p is
n 503
approximately
pˆ  1.96
.905 .095 
ˆˆ
pq
 .905  1.96
 .905  .026
n
503
11
or .879  p  .931
c Even with the conservative value of x in part b, you can see that all the possible values for p are
greater than p  .5 . You cannot conclude that the DA’s claim of 50% or greater is wrong – in fact, it
appears that the DA is correct!
9.65
a-b The hypothesis to be tested is
H0 : p1  p2  0 versus Ha : p1  p2  0
Calculate pˆ1 
x x
42
48
42  48
 .636 , pˆ 2 
 .623 , and pˆ  1 2 
 .629 . The test statistic is
66
77
n1  n2 66  77
then
z
pˆ1  pˆ 2
1 1
ˆ ˆ  
pq
 n1 n2 

.636  .623
.629 .3711 66  1 77 
 0.16
and the p-value is P  z  0.16  1  .5636  .4364 .
c Since the observed p-value, .4364, is greater than   .05 , H0 cannot be rejected. There is insufficient
evidence to support the researcher’s belief.
9.66
Refer to the figure below, which represents the two probability distributions, one assuming that
p1  p2  0 and one assuming that p1  p2  .1 .
12
The right curve is the true distribution of the random variable pˆ1  pˆ 2 and consequently any probabilities
that we wish to calculate concerning the random variable must be calculated as areas under the curve to
the right. The objective of this exercise is to find a common sample size so that
  P reject H0 when H0 is true  .05 and   P accept H0 when H0 is false  .20 . For
  .05 consider the critical value of pˆ1  pˆ 2 that separates the rejection and acceptance regions. This
value will be denoted by  pˆ1  pˆ 2 C . Recall that the random variable z   x     measures the
distance from a particular value x to the mean (in units of standard deviation). Since the z-value
corresponding to  pˆ1  pˆ 2 C is z  1.645 , we have
1.645 
 pˆ1  pˆ 2 C  0
p1q1 p2 q2

n1
n2
or
 pˆ1  pˆ 2 C
Now
 1.645
p1q1 p2 q2

n1
n2
  P accept H0 when p1  p2  .1 which is the area under the right hand curve to the left of
 pˆ1  pˆ 2 C .
Since it is required that   .20 we must find the z-value corresponding to A  .2 , which is
z  .84 (see Table 3). Then
.84 
where  pˆ1  pˆ 2 C  1.645
 pˆ1  pˆ 2 C  .1
p1q1 p2 q2

n1
n2
p1q1 p2 q2
. Substituting for  pˆ1  pˆ 2 C ,

n1
n2
1.645
.84 
2.485 
p1q1 p2 q2

 .1
n1
n2
p1q1 p2 q2

n1
n2
.1
p1q1 p2 q2

n1
n2
The following two assumptions will allow us to calculate the appropriate sample size:
1
n1  n2  n .
13
2
The maximum value of p 1  p  will occur when p  1  p  .5 . Since values of p1 and p2 are
unknown, the use of p  .5 will provide a valid sample size, although it may be slightly larger than
necessary.
Then, solving for n, we obtain
2.485 
.1
1 1
.5 .5    
n n
 2.485 
.1
.5 2 n
n  17.57 or n  308.76 . Hence, a common sample size for the researcher’s test will be n  309 .
9.71
a The hypothesis to be tested is
H0 : 1  2  0 versus Ha : 1  2  0
and the test statistic is
z
 x1  x2   0
2
1
2
2

s
s

n1 n2
240  227
980 820

200 200
 4.33
The rejection region, with   .05, is one-tailed or z  1.645 and the null hypothesis is rejected. There is a
difference in mean yield for the two types of spray.
b An approximate 95% confidence interval for 1  2 is
 x1  x2   1.96
s12 s22

n1 n2
980 820

200 200
13  5.88 or 7.12   1  2   18.88
 240  227   1.96
9.72
a Let p1 be the proportion of cells in which RNA developed normally when treated with a .6 micrograms
per millimeter concentration of Actinomysin-D, and p2 be the proportion of normal cells treated with the
higher concentration of Actinomysin-D. The hypothesis to be tested is
H0 : p1  p2  0 versus Ha : p1  p2  0
14
Calculate pˆ1 
x x
55  23
55
23
 .786 , pˆ 2 
 .329 , and pˆ  1 2 
 .557 . The test statistic is
70
70
n1  n2 70  70
then
z
pˆ1  pˆ 2
1 1 
ˆ ˆ  
pq
 n1 n2 

.786  .329
.557 .4431 70  1 70 
 5.44
and the p-value is P  z  5.44   2 .0002   .0004 .
b Since the observed p-value < .0004, is must be less than   .05 , and H0 is rejected. We can conclude
that there is a difference in the rate of normal RNA synthesis for cells exposed to the two different
concentrations of Actinomysin-D.
9.73
a The hypothesis to be tested is
H0 :   508 versus Ha :   508 .
The test statistic is
z
x 
s
n

499  508
98 100
 .92
and the p-value is
p-value  P  z  .92  P  z  .92  2(.1788)  .3576
Since the p-value, .3576, is greater than   .05 , and H0 cannot be rejected and we cannot conclude that
the average verbal score for California students in 2005 is different from the national average.
b The hypothesis to be tested is
H0 :   520 versus Ha :   520 .
The test statistic is
z
and the p-value is
x 
s
n

516  520
96
100
 .42
15
p-value  P  z  .42  P  z  .42  2(.3372)  .6744
Since the p-value, .6744, is greater than   .05 , and H0 cannot be rejected and we cannot conclude that
the average math score for California students in 2005 is different from the national average.
c Since the same students are used to measured verbal and math scores, there would not be two
independent samples, and the two sample z-test would not be appropriate.
9.75
The hypothesis to be tested is
H0 :   5 versus Ha :   5
and the test statistic is
z
x  0

n

x  0
s
n

7.2  5
6.2
38
 2.19
The rejection region with   .01 is z  2.33 . Since the observed value, z  2.19 , does not fall in the
rejection region and H0 is not rejected. The data do not provide sufficient evidence to indicate that the
mean ppm of PCBs in the population of game birds exceeds the FDA’s recommended limit of 5 ppm.
9.76
Refer to Exercise 9.75, in which the rejection region was given as z  2.33 where
z
x  0
s
n

x  2.3
.29
35
Solving for x we obtain the critical value of x necessary for rejection of H0.
x 5
6.2
38
 2.33  x  2.33
6.2
38
 5  7.34
The probability of a Type II error is defined as
  P  accept H0 when H0 is false 
Since the acceptance region is x  7.34 from part a,  can be rewritten as
  P  x  7.34 when H0 is false   P  x  7.34 when   5
16
Several alternative values of  are given in this exercise.
(Get a Minitab output using power calculation and make conclusions to do the followings)
Insert your output
a For   6 ,(difference = 6-5=1 in Minitab dialog)

  P  x  7.34 when   6   P  z 
 P  z  1.33  .9082

7.34  6 

6.2 38 
and 1    1  .9082  .0918 .
b For   7 ,

  P  x  7.34 when   7   P  z 
 P  z  .34   .6331

and 1    1  .6331  .3669 .
c
For   8 ,
1    1  P  x  7.34 when   8 

7.34  8 
 1  P  z 

6.2 38 

 1  P  z  .66   .7454
For   9 ,
1    1  P  x  7.34 when   9 

7.34  9 
 1  P  z 

6.2 38 

 1  P  z  1.65   .9505
For   10 ,
7.34  7 

6.2 38 
17
1    1  P  x  7.34 when   10 

7.34  10 
 1  P  z 

6.2 38 

 1  P  z  2.64   .9959
For   12 ,
1    1  P  x  7.34 when   12 

7.34  12 
 1  P  z 

6.2 38 

 1  P  z  4.63  1
d The power curve is shown on the next page.
1.0
Power
0.8
0.6
0.4
0.2
0.0
6
7
8
9
Mean
10
11
12
You can see that the power becomes greater than or equal to .90 for a value of  a little smaller than
  9 . To find the exact value, we need to solve for  in the equation:

7.34   
1    1  P  x  7.34   1  P  z 
  .90
6.2 38 


7.34   
or P  z 
  .10
6.2 38 

From Table 3, the value of z that cuts off .10 in the lower tail of the z-distribution is z  1.28 , so that
7.34  
6.2 / 38
 1.28
  7.34  1.28
6.2
38
 8.63.
18
9.78
a The hypothesis to be tested is
H0 : 1  2  0 versus Ha : 1  2  0
and the test statistic is
z
 x1  x2   0
2
1
2
2

s
s

n1 n2
69.58  64.43
2.622 2.582

48
77
 10.75
The rejection region, with   .01, is one-tailed or z  2.33 and the null hypothesis is rejected. There is
sufficient evidence to indicate that the average height for males is greater than females.
b An approximate 99% one-sided confidence bound for 1  2 is approximately
 x1  x2   2.33
s12 s22

n1 n2
2.622 2.582

48
77
5.15  1.12  4.03 or  1   2   4.03
 69.58  64.43  2.33
Males are at least 4.03 inches taller than females on average.
10.23
a If you check the ratio of the two variances using the rule of thumb given in this section you will find:
 4.67 
larger s 2

 1.36
2
2
smaller s
 4.00
2
which is less than three. Therefore, it is reasonable to assume that the two population variances are
equal.
b
From the Minitab printout, the test statistic is t  .06 with p-value  .95 .
c
The value of s  4.38 is labeled “Pooled StDev” in the printout, so that s 2   4.38   19.1844 .
2
d Since the p-value  .95 is greater than .10, the results are not significant. There is insufficient
evidence to indicate a difference in the two population means.
e A 95% confidence interval for  1  2  is given as
19
1 1
s2   
 n1 n2 
 x1  x2   t.025
1 1
19.1844   
6 7
.14  5.363 or  5.223   1   2   5.503
 29  28.86   2.201
Since the value 1  2  0 falls in the confidence interval, it is possible that the two population means
are the same. There insufficient evidence to indicate a difference in the two population means.
10.25
a The hypothesis to be tested is
H0 : 1  2  0 versus Ha : 1  2  0
From the Minitab printout, the following information is available:
x1  .896
s12  .400 
2
n1  14
x2  1.147
s  .679 
2
n2  11
2
2
and the test statistic is
t
 x1  x2   0
1 1
s2   
 n1 n2 
 1.16
The rejection region is two-tailed, based on n1  n2  2  23 degrees of freedom. With   .05 , from
Table 4, the rejection region is t  t.025  2.069 and H0 is not rejected. There is not enough evidence to
indicate a difference in the population means.
b It is not necessary to bound the p-value using Table 4, since the exact p-value is given on the printout
as P-Value = .260.
c
If you check the ratio of the two variances using the rule of thumb given in this section you will find:
.679 
larger s 2

 2.88
2
2
smaller s
.400 
2
which is less than three. Therefore, it is reasonable to assume that the two population variances are
equal.
10.27
a Check the ratio of the two variances using the rule of thumb given in this section:
20
larger s 2
2.78095

 16.22
2
.17143
smaller s
which is greater than three. Therefore, it is not reasonable to assume that the two population variances
are equal.
b You should use the unpooled t test with Satterthwaite’s approximation to the degrees of freedom for
testing
H0 : 1  2  0 versus Ha : 1  2  0
The test statistic is
t
 x1  x2   0
2
1
2
2

s
s

n1 n2
3.73  4.8
2.78095 .17143

15
15
 2.412
with
2
 s12 s22 
2
  
n n
.185397  .0114287 
df   12 2  2 
 15.7
.002455137  .00000933
 s12   s22 
   
 n1    n2 
n1  1 n2  1
With df  15 , the p-value for this test is bounded between .02 and .05 so that H 0 can be rejected at the
5% level of significance. There is evidence of a difference in the mean number of uncontaminated
eggplants for the two disinfectants.
10.40
a-b The table of differences, along with the calculation of d and sd2 , is presented below.
d 
sd2 
Week
1
2
3
4
Totals
di
–1.77
–15.03
–23.22
127.05
–67.07
 d i 67.07

 16.7675
n
4
 di2 
  di 
n 1
n
2

1499.9047 
The hypothesis of interest is
3
 67.07 
4
2
 125.102825 and sd  11.1849
21
H 0 : 1  2  0 or H 0 : d  0
H a : 1  2  0 or H a : d  0
and the test statistic is
t
d  d
sd
n

16.7675  0
 3.00
11.1849
4
Since t  3.00 with df  n  1  3 falls between the two tabled values, t.025 and t.05 ,
.05  p-value  .10
for this two tailed test and H0 is not rejected. We cannot conclude that the means are different.
c
The 99% confidence interval for 1  2  d is
d  t.005
sd
11.1849
  16.7675  5.841
  16.7675  32.666
n
4
or 49.433   1  2   15.899 .
10.43
a A paired-difference test is used, since the two samples are not random and independent (at any
location, the ground and air temperatures are related). The hypothesis of interest is
H0 : 1  2  0 Ha : 1  2  0
The table of differences, along with the calculation of d and sd2 , is presented below.
d 
sd2 
Location
1
2
3
4
5
Total
di
–.4
–2.7
–1.6
–1.7
–1.5
–7.9
 di 7.9

 1.58
n
5
 di2 
  di 
n 1
2
n
and the test statistic is

15.15 
 7.9 
4
5
2
 .667 and sd  .8167
22
t
d  d
sd
1.58  0
 4.326
.8167
5

n
A rejection region with   .05 and df  n  1  4 is t  t.025  2.776 , and H0 is rejected at the 5% level of
significance. We conclude that the air-based temperature readings are biased.
b The 95% confidence interval for 1  2  d is
d  t.025
sd
  1.58  2.776
n
.8167
5
  1.58  1.014
or 2.594   1  2   .566 .
c
The inequality to be solved is
t 2 SE  B
We need to estimate the difference in mean temperatures between ground-based and air-based sensors
to within .2 degrees centigrade with 95% confidence. Since this is a paired experiment, the inequality
becomes
sd
t.025
n
 .2
With sd  .8167 and n represents the number of pairs of observations, consider the sample size obtained
by replacing t.025 by z.025  1.96 .
1.96
.8167
n
 .2
n  8.0019  n  64.03 or n  65
Since the value of n is greater than 30, the use of z 2 for t 2 is justified.
10.47
A paired-difference analysis must be used. The hypothesis of interest is
H 0 : 1  2  0 or H 0 : d  0
H a : 1  2  0 or H a : d  0
The table of differences is presented below. Use your scientific calculator to find d and sd ,
di
3
3
2
1
1
3
1
23
Calculate d  .857 , sd  2.193 , and the test statistic is
t
d  d
sd
n

.857  0
 1.03
2.193
7
Since t  1.03 with df  n  1  6 is smaller than the smallest tabled value t.10 ,
p-value  .10
for this one-tailed test and H0 is not rejected. We cannot conclude that the average time outside the
office is less when music is piped in.