Chapter 7
9.
a.
x  xi / n 
465
 93
5
b.
Totals
s
11. a.
b.
15. a.
( xi  x )
( xi  x ) 2
94
100
85
94
92
465
+1
+7
-8
+1
-1
0
1
49
64
1
1
116
( xi  x ) 2
116

 5.39
n 1
4
x  xi / n 
s
xi
$45,500
 $4,550
10
( xi  x )2
9,068,620

 $1003.80
n 1
10  1
The sampling distribution is normal with
E ( x ) =  = 200
 x   / n  50 / 100  5
For 5, 195  x  205
Using Standard Normal Probability Table:
At x = 205, z 
At x = 195, z 
x 
x
x 
x

5
1
P( z  1) = .8413
5

5
 1 P ( z  1) = .1587
5
P(195  x  205) = .8413 - .1587 = .6826
Using Excel: =NORMDIST(205,200,5,TRUE)-NORMDIST(195,200,5,TRUE) = .6827
b.
For  10, 190  x  210
Using Standard Normal Probability Table:
At x = 210, z 
x
x

10
2
5
P( z  2) = .9772
7-1
Chapter 7
At x = 190, z 
x 

x
10
 2 P ( z  2) = .0228
5
P(190  x  210) = .9772 - .0228 = .9544
Using Excel: =NORMDIST(210,200,5,TRUE)-NORMDIST(190,200,5,TRUE) = .9545
17. a.
b.
 x   / n  10 / 50  141
.
n / N = 50 / 50,000 = .001
Use  x   / n  10 / 50  141
.
c.
n / N = 50 / 5000 = .01
Use  x   / n  10 / 50  141
.
d.
n / N = 50 / 500 = .10
Use  x 
N n 

N 1 n
500  50 10
 134
.
500  1 50
Note: Only case (d) where n /N = .10 requires the use of the finite population correction factor.
 = 2.34  = .20
25.
a.
n = 30
z
x 
/ n

.03
.20 / 30
 .82
P(2.31  x  2.37) = P(-.82  z  .82) = .7939 - .2061 = .5878
Using Excel: =NORMDIST(2.37,2.34,.20/SQRT(30),TRUE)NORMDIST(2.31,2.34,.20/SQRT(30),TRUE)=.5887
b.
n = 50
z
x 
/ n

.03
.20 / 50
 1.06
P(2.31  x  2.37) = P(-1.06  z  1.06) = .8554 - .1446 = .7108
Using Excel: =NORMDIST(2.37,2.34,.20/SQRT(50),TRUE)NORMDIST(2.31,2.34,.20/SQRT(50),TRUE)=.7112
c.
n = 100
z
x 
/ n

.03
.20 / 100
 1.50
7-2
Sampling and Sampling Distributions
P(2.31  x  2.37) = P(-1.50  z  1.50) = .9332 - .0668 = .8664
Using Excel: =NORMDIST(2.37,2.34,.20/SQRT(100),TRUE)NORMDIST(2.31,2.34,.20/SQRT(100),TRUE)=.8664
d.
None of the sample sizes in parts (a), (b), and (c) are large enough. At z = 1.96 we find P(-1.96  z 
1.96) = .95. So, we must find the sample size corresponding to z = 1.96. Solve
.03
 1.96
.20 / n
 .20 
n  1.96 
  13.0667
 .03 
n  170.73
Rounding up, we see that a sample size of 171 will be needed to ensure a probability of .95 that the
sample mean will be within  $.03 of the population mean.
31. a.
p 
p(1  p)
.30(.70)

 .0458
n
100
p
.30
The normal distribution is appropriate because np = 100(.30) = 30 and n(1 - p) = 100(.70) = 70 are
both greater than 5.
b.
P (.20  p  .40) = ?
z
.40  .30
 2.18 P(z ≤ 2.18) = .9854
.0458
P(z < -2.18) = .0146
P(.20 ≤ p ≤ .40) = .9854 - .0146 = .9708
Using Excel:
=NORMDIST(.40,.30,.0458,TRUE)-NORMDIST(.20,.30,.0458,TRUE) = .9710
c.
P (.25  p  .35) = ?
7-3
Chapter 7
z
.35  .30
 1.09 P(z ≤ 1.09) = .8621
.0458
P(z < -1.09) = .1379
P(.25 ≤ p ≤ .35) = .8621 - .1379 = .7242
Using Excel:
=NORMDIST(.35,.30,.0458,TRUE)-NORMDIST(.25,.30,.0458,TRUE) = .7250
33. a.
Normal distribution
E ( p ) = .50
p 
b.
z
p (1  p)

n
p p
p

(.50)(1  .50)
 .0206
589
.04
 1.94
.0206
P(z ≤ 1.94) = .9738
P(z < -1.94) = .0262
P(.46 ≤ p ≤ .54) = .9738 - .0262 = .9476
Using Excel:
=NORMDIST(.54,.50,.0206,TRUE)-NORMDIST(.46,.50,.0206,TRUE) = .9478
c.
z
p p
p

.03
 1.46
.0206
P(z ≤ 1.46) = .9279
P(z < -1.46) = .0721
P(.47 ≤ p ≤ .53) = .9279 - .0721 = .8558
Using Excel:
=NORMDIST(.53,.50,.0206,TRUE)-NORMDIST(.47,.50,.0206,TRUE) = .8547
d.
z
p p
p

.02
 .97
.0206
P(z ≤ .97) = .8340
P(z < -.97) = .1660
P(.48 ≤ p ≤ .52) = .8340 - .1660 = .6680
Using Excel:
=NORMDIST(.52,.50,.0206,TRUE)-NORMDIST(.48,.50,.0206,TRUE) = .6684
7-4
Sampling and Sampling Distributions
35. a.
Normal distribution with E( p ) = p = .25 and
p(1  p)
.25(1  .25)

 .0137
n
1000
p 
z
b.
p p
p

.03
 2.19
.0137
P(z ≤ 2.19) = .9857
P(z < -2.19) = .0143
P(.22  p  .28) = P(-2.19  z  2.19) = .9857 - .0143 = .9714
Using Excel:
=NORMDIST(.28,.25,.0137,TRUE)-NORMDIST(.22,.25,.0137,TRUE) = .9715
z
c.
p p
.25(1  .25)
500

.03
 1.55
.0194
P(z ≤ 1.55) = .9394
P(z < -1.55) = .0606
P(.22  p  .28) = P(-1.55  z  1.55) = .9394 - .0606 = .8788
Using Excel:
=NORMDIST(.28,.25,.0194,TRUE)-NORMDIST(.22,.25,.0194,TRUE) = .8780
Chapter 8
13. a.
x
xi 80

 10
n
8
b.
s
xi
( xi  x )
( xi  x ) 2
10
8
12
15
13
11
6
5
0
-2
2
5
3
1
-4
-5
0
4
4
25
9
1
16
25
84
( xi  x )2
84

 3.464
n 1
7
7-5
Chapter 7
c.
t.025 (s / n )  2.365(3.464 / 8)  2.9
d.
x  t.025 (s / n )
10 ± 2.9 or 7.1 to 12.9
15.
x  t / 2 (s / n )
90% confidence
df = 64
t.05 = 1.669
19.5 ± 1.669 (5.2 / 65)
19.5 ± 1.08 or 18.42 to 20.58
95% confidence
df = 64
t.025 = 1.998
19.5 ± 1.998 (5.2 / 65)
19.5 ± 1.29 or 18.21 to 20.79
17.
For the Miami data set, the output obtained using Excel’s Descriptive Statistics tool follows:
Rating
Mean
Standard Error
Median
Mode
6.34
0.3059
6.5
8
Standard Deviation
2.1629
Sample Variance
4.6780
Kurtosis
-1.1806
Skewness
-0.1445
Range
8
Minimum
2
Maximum
10
Sum
317
Count
Confidence Level(95.0%)
50
0.6147
The 95% confidence interval is x  margin of error
21.
6.34
 0.6147 or 5.73 to 6.95
x
xi 2600

 130 liters of alcoholic beverages
n
20
7-6
Sampling and Sampling Distributions
s
( xi  x )2
81244

 65.39
n 1
20  1
t.025 = 2.093
df = 19
95% confidence interval: x  t.025 ( s / n )
130  2.093 (65.39 / 20)
130  30.60
35. a.
or 99.40 to 160.60 liters per year
p = 281/611 = .4599
(46%)
p (1  p )
.4599(1  .4599)
 1.645
 .0332
n
611
b.
z.05
c.
p ± .0332
.4599  .0332 or .4267 to .4931
43. a.
Margin of Error = z / 2
p (1  p )
(.53)(.47)
 1.96
 .0253
n
1500
95% Confidence Interval: .53  .0253 or .5047 to .5553
b.
Margin of Error = 1.96
(.31)(.69)
= .0234
1500
95% Confidence Interval: .31  .0234 or .2866 to .3334
c.
Margin of Error = 1.96
(.05)(.95)
= .0110
1500
95% Confidence Interval: .05  .0110 or .039 to .061
d.
The margin of error decreases as p gets smaller. If the margin of error for all of the interval estimates
must be less than a given value (say .03), an estimate of the largest proportion should be used as a
planning value. Using p*  .50 as a planning value guarantees that the margin of error for all the
interval estimates will be small enough.
Chapter 9
1.
a.
H0:   600
Manager’s claim.
Ha:  > 600
b.
We are not able to conclude that the manager’s claim is wrong.
c.
The manager’s claim can be rejected. We can conclude that  > 600.
7-7
Chapter 7
3.
5.
a.
H0:  = 32
Specified filling weight
Ha:   32
Overfilling or underfilling exists
b.
There is no evidence that the production line is not operating properly. Allow the production process
to continue.
c.
Conclude   32 and that overfilling or underfilling exists. Shut down and adjust the production
line.
a.
The Type I error is rejecting H0 when it is true. This error occurs if the researcher concludes that
young men in Germany spend more than 56.2 minutes per day watching prime-time TV when the
national average for Germans is not greater than 56.2 minutes.
b. The Type II error is accepting H0 when it is false. This error occurs if the researcher concludes that
the national average for German young men is  56.2 minutes when in fact it is greater than 56.2
minutes.
23. a.
b.
t
x  0
s/ n

14  12
4.32 / 25
 2.31
Degrees of freedom = n – 1 = 24
Upper tail p-value is the area to the right of the test statistic
Using t table: p-value is between .01 and .025
Using Excel: p-value = TDIST(2.31,24,1) = .0149
c.
p-value  .05, reject H0.
c.
With df = 24, t.05 = 1.711
Reject H0 if t  1.711
2.31 > 1.711, reject H0.
25. a.
t
x  0
s/ n

44  45
5.2 / 36
 1.15
Degrees of freedom = n – 1 = 35
Lower tail p-value is the area to the left of the test statistic
Using t table: p-value is between .10 and .20
Using Excel: p-value = TDIST(1.15,35,1) = .1290
p-value > .01, do not reject H0
7-8
Sampling and Sampling Distributions
b.
t
x  0
s/ n

43  45
 2.61
4.6 / 36
Lower tail p-value is the area to the left of the test statistic
Using t table: p-value is between .005 and .01
Using Excel: p-value = TDIST(2.61,35,1) = .0066
p-value  .01, reject H0
c.
t
x  0
s/ n

46  45
5 / 36
 1.20
Lower tail p-value is the area to the left of the test statistic
Using t table: p-value is between .80 and .90
Using Excel: p-value = TDIST(1.20,35,1) = .8809
p-value > .01, do not reject H0
H0:   47.50
31.
Ha:  > 47.50
t
x  0
s/ n

51  47.50
12 / 64
 2.33
Degrees of freedom = n - 1 = 63
Upper tail p-value is the area to the right of the test statistic
Using t table: p-value is between .01 and .025
Using Excel: p-value = TDIST(2.33,91,1) = .0110
Reject H0; Atlanta customers are paying a higher mean water bill.
37. a.
H0: p  .125
Ha: p > .125
b.
p
z
52
 .13
400
p  p0
p0 (1  p0 )
n

.13  .125
.125(1  .125)
400
 .30
Upper tail p-value is the area to the right of the test statistic
7-9
Chapter 7
Using normal table with z = .30: p-value = 1.0000 - .6179 = .3821
Using Excel: p-value = 1-NORMSDIST(.30) = .3821
c.
43. a.
p-value > .05; do not reject H0. We cannot conclude that there has been an increase in union
membership.
H0: p ≤ .10
Ha: p > .10
b.
There are 13 “Yes” responses in the Eagle data set.
p
c.
z
13
 .13
100
p  p0
p0 (1  p0 )
n

.13  .10
.10(1  .10)
100
 1.00
Upper tail p-value is the area to the right of the test statistic
Using normal table with z = 1.00: p-value = 1 - .8413 = .1587
Using Excel: p-value = 1-NORMSDIST(1) = .1587
p-value > .05; do not reject H0.
The statistical results do not allow us to conclude that p > .10. But, given that p = .13, management
may want to authorize a larger study before deciding not to go national.
Chapter 10
54
9
6
x2 
42
7
6
11. a.
x1 
b.
s1 
( xi  x1 ) 2
 2.28
n1  1
s2 
( xi  x2 ) 2
 1.79
n2  1
c.
x1  x2 = 9 - 7 = 2
d.
 s12 s22 
 2.282 1.792 






n1 n2 
6
6 


df 

 9.5
2
2
2
2
1  2.282  1  1.792 
1  s12 
1  s22 
  
 

  

5 6  5 6 
n1  1  n1  n2  1  n2 
2
2
7 - 10
Sampling and Sampling Distributions
Use df = 9, t.05 = 1.833
x1  x2  1.833
2 2.17
13. a.
x1 
s1 
x2 
s2 
b.
2.282 1.792

6
6
(-.17 to 4.17)
xi 111.6

 9.3
n1
12
( xi  x1 ) 2
71.12

 2.54
n1  1
12  1
xi 42

 4.2
n2
10
( xi  x2 ) 2
18.4

 1.43
n2
10.1
x1  x2 = 9.3 - 4.2 = 5.1 tons
Memphis is the higher volume airport and handled an average of 5.1 tons per day more than
Louisville. Memphis handles more than twice the volume of Louisville.
2
c.
2
 s12 s22 
 2.542 1.432 

  


10 
 n1 n2 
 12
df 

 17.8
2
2
2
2
1  2.542  1  1.432 
1  s12 
1  s22 
  
 

  

11  12  9  10 
n1  1  n1  n2  1  n2 
Use df = 17, t.025 = 2.110
( x1  x2 )  t.025
5.1  2.110
2.542 1.432

12
10
5.1  1.82
15.
s12 s22

n1 n2
(3.28 to 6.92)
1 for 2001 season
2 for 1992 season
H0: 1  2  0
Ha: 1  2  0
7 - 11
Chapter 7
b.
x1  x2 = 60 - 51 = 9 days
9/51(100) = 17.6% increase in number of days.
c.
t
 x1  x2   0
2
1
2
2

(60  51)  0
s
s

n1 n2
182 152

45 38
 2.48
2
2
 s12 s22 
 182 152 

  


n1 n2 
45 38 


df 

 81
2
2
2
2
1  182 
1  152 
1  s12 
1  s22 
  
 

  

44  45  37  38 
n1  1  n1  n2  1  n2 
Using t table, p-value is between .005 and .01.
Exact p-value corresponding to t = 2.48 is .0076
p-value  .01, reject H0. There is a greater mean number of days on the disabled list in 2001.
d.
31. a.
Management should be concerned. Players on the disabled list have increased 32% and time on the
list has increased by 17.6%. Both the increase in inquiries to players and the cost of lost playing time
need to be addressed.
p1 = 150/250 = .46 Republicans
p2 = 98/350 = .28 Democrats
b.
p1  p2 = .46 - .28 = .18
Republicans have a .18, 18%, higher participation rate than Democrats.
c.
d.
z.025
p1 (1  p2 ) p2 (1  p2 )

n1
n2
1.96
.46(1  .46) .28(1  .28)

 .0777
250
350
Yes, .18  .0777
(.1023 to .2577)
Republicans have a 10% to 26% higher participation rate in online surveys than Democrats. Biased
survey results of online political surveys are very likely.
33. a.
p1 = 256/320 = .80
b.
p2 = 165/250 = .66
c.
p1  p2 = .80 - .66 = .14
7 - 12
Sampling and Sampling Distributions
.14  z.025
p1 (1  p1 ) p2 (1  p2 )

n1
n2
.14  1.96
.80(1  .80) .66(1  .66)

320
250
.14  .0733
35. a.
(.0667 to .2133)
H0: p1 - p2 = 0
H a: p 1 - p 2  0
p1 = 63/150 = .42
p2 = 60/200 = .30
p
z
n1 p1  n2 p2
63  60

 .3514
n1  n2
150  200
p1  p2
1 1 
p 1  p    
 n1 n2 

.42  .30
1 
 1
.3514 1  .3514  


150
200


 2.33
p-value = 2(1.0000 - .9901) = .0198
p-value  .05, reject H0. There is a difference between the recall rates for the two commercials.
b.
p1  p2  z.025
p1 (1  p1 ) p2 (1  p2 )

n1
n2
.42  .30  1.96
.42(1  .42) .30(1  .30)

150
200
.12  .1014
(.0186 to .2214)
Commercial A has the better recall rate.
7 - 13