# chap08

Document Sample

```					150   Chapter 8: Confidence Interval Estimation

CHAPTER 8

                       8
8.1     X  Z         = 85 1.96                                                   83.04    86.96
n                      64

                       24
8.2     X  Z         = 125  2.58                                              114.68    135.32
n                       36

8.3    If all possible samples of the same size n are taken, 95% of them include the true population
average monthly sales of the product within the interval developed. Thus we are 95 percent
confident that this sample is one that does correctly estimate the true average amount.

8.4    Since the results of only one sample are used to indicate whether something has gone wrong
in the production process, the manufacturer can never know with 100% certainty that the
specific interval obtained from the sample includes the true population mean. In order to
have every possible interval estimate of the true mean, the entire population (sample size N)
would have to be selected.

8.5    To the extent that the sampling distribution of sample means is approximately normal, it is
true that approximately 95% of all possible sample means taken from samples of that same
size will fall within 1.96 times the standard error away from the true population mean. But
the population mean is not known with certainty. Since the manufacturer estimated the mean
would fall between 10.99408 and 11.00192 inches based on a single sample, it is not
necessarily true that 95% of all sample means will fall within those same bounds.

8.6    Approximately 5% of the intervals will not include the true population mean somewhere in
the interval. Since the true population mean is not known, we do not know for certain
whether it is in the one interval we have developed, between 10.99408 and 11.00192 inches.

                      0.02
8.7    (a)       X  Z          0.995  2.58                                    0.9877    1.0023
n                       50
(b)       Since the value of 1.0 is included in the interval, there is no reason to believe that the
average is different from 1.0 gallon.
(c)       No. Since  is known and n = 50, from the central limit theorem, we may assume that
the sampling distribution of X is approximately normal.
(d)       An individual can with an observed value of 0.98 gallon of paint yields a Z-value of
– 1.0. The confidence interval represents bounds on the estimate of the average of a
sample of 50 cans of paint, not an individual value.
(e)       The reduced confidence level narrows the width of the confidence interval.
                       0.02
(a)      X  Z           0.995 1.96                           0.9895    1.0005
n                       50
(b)      Since the value of 1.0 is still included in the interval, there is no reason to
believe that the average is different from 1.0 gallon.
Solutions to End-of-Section and Chapter Review Problems        151

8.8    (a)

(b)   No. The manufacturer cannot support a claim that the bulbs last an average 400
hours. Based on the data from the sample, a mean of 400 hours would represent a
distance of 4 standard deviations above the sample mean of 350 hours.
(c)   No. Since  is known and n = 64, from the central limit theorem, we may assume that
the sampling distribution of X is approximately normal.
(d)   An individual value of 320 is only 0.30 standard deviations below the sample mean
of 350. The confidence interval represents bounds on the estimate of a sample of 64,
not an individual value.
(e)   The confidence interval is narrower based on a process standard deviation of 80
hours rather than the original assumption of 100 hours.
                    80
(a)      X  Z         350 1.96                             330.4    369.6
n                    64
(b)      Based on the smaller standard deviation, a mean of 400 hours would
represent a distance of 5 standard deviations above the sample mean of 350
hours. No, the manufacturer cannot support a claim that the bulbs last an
average of 400 hours.

                    0.05
8.9    (a)   X  Z         1.99 1.96                                  1.9802    1.9998
n                    100
(b)   No. Since  is known and n = 100, from the central limit theorem, we may assume that
the sampling distribution of X is approximately normal.
(c)   An individual value of 2.02 is only 0.60 standard deviation above the sample mean
of 1.99. The confidence interval represents bounds on the estimate of a sample of
100, not an individual value.
(d)   A shift of 0.02 units in the sample average shifts the confidence interval by the same
distance without affecting the width of the resulting interval.
                    0.05
(a)      X  Z        1.97 1.96                          1.9602    1.9798
n                    100

8.10   (a)   t9 = 2.2622
(b)   t9 = 3.2498
(c)   t31 = 2.0395
(d)   t64 = 1.9977
(e)   t15 = 1.7531
152    Chapter 8: Confidence Interval Estimation

s                 24
8.11     X t       75  2.0301                                           66.8796    83.1204
n                 36

s                 15
8.12     X t       50  2.9467                                           38.9499    61.0501
n                 16

3.7417
8.13    Set 1: 4.5  2.3646                                                    1.3719    7.6281
8
2.4495
Set 2: 4.5  2.3646                                                    2.4522    6.5478
8
The data sets have different confidence interval widths because they have different values for
the standard deviation.

6.4660
8.14    Original data: 5.8571  2.4469                                      – 0.1229    11.8371
7
2.1602
Altered data: 4.00  2.4469                                            2.0022    5.9978
7
The presence of an outlier in the original data increases the value of the sample mean and
greatly inflates the sample standard deviation.

s                    21.4
8.15    (a)      X t       195.3  2.1098                              184.6581    205.9419
n                    18
(b)      No, a grade of 200 is in the interval.
(c)      It is not unusual. A tread-wear index of 210 for a particular tire is only 0.69 standard
deviation above the sample mean of 195.3.

s                    0.32
8.16    (a)      X t        1.67  2.0930                                     \$1.52    \$1.82
n                     20
(b)      The store owner can be 95% confidence that the population mean retail value of
greeting cards that it has in its inventory is somewhere in between \$1.52 and \$1.82.

8.17    (a)

(b)      The population of dental expenses must be approximately normally distributed.
Solutions to End-of-Section and Chapter Review Problems           153

s                      138.80
8.17    (d)     X t        261.40  2.2622                             \$162.11    \$360.69
n                       10
cont.   (e)     The additional \$500 in dental expenses, divided across the sample of 10, raises the
mean by \$50 and increases the standard deviation by nearly \$20. The interval half-
width increases over \$11 in the process. The new interval is:
s                    157.056
X t        311.40 1.8331                              \$220.36    \$402.22
n                      10

s                          194645.95
8.18    (a)      X t       2153666.667  2.145                    107575.24<  323158.09
n                             15
(b)     The number of shares traded is approximately normally distributed.

s                      1.8154
8.19    (a)      X t     3.8415  2.0227                                       3.26<  4.42
n                        40
s                      1.8154
(b)      X t     3.8415  2.7079                                       3.06<  4.62
n                        40
(c)     The 99% confidence interval is wider because a higher percentage of all the intervals
that could be constructed will have to include the true population mean.

s                      5.9005
8.20    (a)      X t     4.3171  2.0322                                        2.29<  6.34
n                         35
s                    16.9253
(b)      X t     11.4  2.0322                                         5.59<  17.21
n                       35
s                       0.6143
(c)      X t     0.6954  2.0322                                        0.48<  0.91
n                         35

s                      53.5766
8.21    (a)      X t     164.85  2.0930                                  139.78<  189.92
n                        20
s                     8.8572
(b)      X t     45.65  2.0930                                      41.50<  49.80
n                       20

(c)     We need to assume that the population has a normal distribution since the sample size
of 20 is not large enough to invoke the central limit theorem for distributions that are
not symmetric.

X 50                             ps (1 – ps )               0.25(0.75)
8.22    ps         = 0.25          ps  Z                  0.25  1.96
n 200                                 n                        200
0.19  p  0.31

X 25                             ps (1 – ps )               0.0625(0.9375)
8.23    ps         = 0.0625        ps  Z                  0.25  2.58
n 400                                 n                           400
0.0313  p  0.0937
154    Chapter 8: Confidence Interval Estimation

ps (1 – ps )                0.41(0.59)
8.24    (a)    ps = 0.41        ps  Z                  0.41 1.96 
n                         200
0.3418  p  0.4782
(b)    The dealer can infer that the proportion of all customers who still own the cars they
purchased at the dealership 5 years earlier is somewhere between 03418 and 0.4782
with a 95% level of confidence.

8.25    (a)

(b)    The manager in charge of promotional programs concerning residential customers
can infer that the proportion of households that would purchase an additional
telephone line if it were made available at a substantially reduced installation cost is
somewhere between 0.22 and 0.32 with a 99% level of confidence.

X 453                               ps (1 – ps )                0.53(1  0.53)
8.26    (a)     ps          0.5304           ps  Z                  0.53  1.96
n 854                                    n                           854
0.497  p  0.564
X 453
(b)     ps              0.5304
n 854
ps (1 – ps )                 0.53(1  0.53)
ps  Z                  0.53  2.5758                            0.486  p  0.574
n                            854
(c)    The 99% confidence interval is wider because a higher percentage of all the intervals
that could be constructed will have to include the true population proportion of
personnel officials that believe job seekers sometimes falsify past salaries.
Solutions to End-of-Section and Chapter Review Problems        155

X 229
8.27   (a)   ps              0.3001
n 763
ps (1 – ps )                   0.3001(1  0.3001)
ps  Z                  0.3001  1.6449
n                                763
0.2728  p  0.3274
X      99
(b)   ps              0.1298
n 763
ps (1 – ps )                   0.1298(1  0.1298)
ps  Z                  0.1298  1.6449
n                                763
0.1097  p  0.1498

X 1123
8.28   (a)   ps               0.9027
n 1244
ps (1 – ps )                 0.9027(1  0.9027)
ps  Z                  0.9027  1.96
n                             1244
0.8863  p  0.9192
X (1244  684)
(b)   ps                        0.4502
n         1244
ps (1 – ps )                 0.4502(1  0.4502)
ps  Z                  0.4502  1.96
n                             1244
0.4225  p  0.4778

ps (1 – ps )                 0.66(0.34)
8.29   (a)   ps  Z                  0.66  1.96                         0.6311  p  0.6896
n                          1007
ps (1 – ps )                  0.301(1  0.301)
(b)   ps  Z                  0.301  1.96                          0.266  p  0.336
n                              665
(c)   With this information, the Tupperware Corporation can better predict the demand for
their lunch boxes.

ps (1 – ps )                 0.67(0.33)
8.30   (a)   ps  Z                  0.67  1.96                           0.611  p  0.735
n                          220
With 95% confidence, between 61.1% and 73.5% of all child-care providers have at
least one safety hazard.
ps (1 – ps )                   0.3818(1  0.3818)
(b)   ps  Z                  0.3818  1.96                         0.3176  p  0.4460
n                                220
With 95% confidence, between 31.76% and 44.6% of all child-care providers have
children wearing clothing with neck drawstrings.
156     Chapter 8: Confidence Interval Estimation

ps (1 – ps )                  0.191(1  0.191)
8.30     (c)         ps  Z                  0.191  1.96                      0.139  p  0.243
n                              220
cont.               With 95% confidence, between 13.9% and 24.3% of all child-care providers have cribs
with soft bedding that can lead to suffocation.

ps (1 – ps )                  0.383(1  0.383)
8.31     (a)         ps  Z                  0.383  1.96                      0.365  p  0.401
n                             2943
With 95% confidence, between 36.5% and 40.1% of the population of U.S. Internet
households plan to buy a computer during the year.
ps (1 – ps )                   0.383(1  0.383)
(b)         ps  Z                  0.383  2.576                     0.360  p  0.406
n                              2943
With 99% confidence, between 36.0% and 40.6% of the population of U.S. Internet
households plan to buy a computer during the year.
ps (1 – ps )                  0.231(1  0.231)
(c)         ps  Z                  0.231  1.96                      0.216  p  0.246
n                             2943
With 95% confidence, between 21.6% and 24.6% of the population of U.S. Internet
households plan to buy a new car or truck during the year.
ps (1 – ps )                   0.231(1  0.231)
(d)         ps  Z                  0.231  2.576                     0.211  p  0.251
n                              2943
With 99% confidence, between 21.1% and 25.1% of the population of U.S. Internet
households plan to buy a new car or truck during the year.

Z 2  2 1.962 152
8.32           n 2           2    = 34.57                                             Use n = 35
e          5

Z 2  2 2.582 100 2
8.33           n       2         2     = 166.41                                     Use n = 167
e          20

Z 2 p(1– p) 2.582 (0.5)(0.5)
8.34           n          2              2     = 1,040.06                          Use n = 1,041
e          (0.04)

Z 2 p(1– p) 1.962 (0.4)(0.6)
8.35           n          2              2     = 2,304.96                          Use n = 2,305
e          (0.02)

Z 2  2 1.962  4002
8.36     (a)        n 2            2     = 245.86                                   Use n = 246
e          50
Z 2 2 1.962  4002
(b)        n 2                  = 983.41                                   Use n = 984
e         252

Z 2  2 1.962 (0.02)2
8.37     n 2               2    = 96.04                                               Use n = 97
e         (0.004)
Solutions to End-of-Section and Chapter Review Problems            157

Z 2  2 1.962 (100)2
8.38   n        2           2    = 96.04                                               Use n = 97
e          (20)

Z 2  2 1.962 (0.05) 2
8.39   n        2            2     = 96.04                                             Use n = 97
e         (0.01)

Z 2  2 2.582 252
8.40   (a)      n    2        2     = 166.41                                          Use n = 167
e          5
Z 2  2 1.962  252
(b)      n 2           2     = 96.04                                            Use n = 97
e          5

Z 2  2 1.6452 45 2
8.41   (a)      n 2            2      = 219.19                                        Use n = 220
e          5
Z 2  2 2.582 45 2
(b)      n 2           2      = 539.17                                         Use n = 540
e          5

Z 2  2 1.962  202
8.42   n        2        2     = 61.47                                                 Use n = 62
e          5

s                        108947.59
8.43   (a)      X t    134088.89  2.110                           79910.45    188267.33
n                           18
Z
2 2
1.96 108947.6
2          2
(b)      n 2                       114
e           200002
(c)      The sample size is computed using the sample standard deviation of the shares traded
for the 18 companies as an estimate for the population standard deviation. This is not
likely to be an accurate estimate and, hence, the result will not likely by valid if the
study is performed today.

Z 2 p(1– p) 1.6452 (0.5)(0.5)
8.44   (a)      n        2                    2    = 422.82                           Use n = 423
e              (0.04)
Z 2 p(1– p) 1.962 (0.5)(0.5)
(b)      n        2                  2     = 600.25                            Use n = 601
e             (0.04)
Z 2 p (1 – p ) 1.96 2 (0.5)(0.5)
(c)      n                                 = 1067.07                          Use n = 1068
e2            (0.03) 2
(d)      In general, larger sample size is needed if a higher level of confidence is required
holding everything else fixed. This increase in sample size reflects the higher price for
the increased in level of confidence. If a higher precision is required, which is reflected
in lower acceptable sampling error, the sample size will need to be raised to reflect the
higher associated cost holding everything else fixed.
158    Chapter 8: Confidence Interval Estimation

Z 2 p(1– p) 2.582 (0.10)(0.90)
8.45    (a)       n        2               2      = 122.26                         Use n = 123
e            (0.07)
Z 2 p(1– p) 1.962 (0.10)(0.90)
(b)       n        2                2     = 96.04                           Use n = 97
e            (0.06)

X      94
8.46    (a)       ps              0.1604
n 586
ps (1 – ps )                    0.1604(1  0.1604)
ps  Z                 0.1604  1.96                          0.1307  p  0.1901
n                                  586
Z 2 p (1 – p ) 1.96 2 (0.1604)(1  0.1604)
(b)       n                                            1293.34        Use n = 1294
e2                   (0.02) 2
Z 2 p (1 – p ) 2.57582 (0.1604)(1  0.1604)
(c)       n                                               2233.84        Use n = 2234
e2                   (0.02) 2

X      707
8.47    (a)       ps               0.6805
n 1039
ps (1 – ps )                   0.6805(1  0.6805)
ps  Z                  0.6805  1.96 
n                               1039
0.6521  p  0.7088
Z 2 p (1 – p ) 1.96 2 (0.6805)(1  0.6805)
(b)       n                                              2088.02         Use n = 2089
e2                   (0.02) 2
Z 2 p (1 – p ) 1.96 2 (0.6805)(1  0.6805)
(c)       n                                              8352.07         Use n = 8353
e2                   (0.01) 2

Z 2 p (1 – p ) 1.962 (0.50)(0.50)
8.48     n                                      2400.91                          Use n = 2401
e2              (0.02) 2

ps (1 – ps )                0.3(0.7)
8.49    (a)       ps  Z                0.3  1.96                          0.273  p  0.327
n                       1100
Z p (1 – p ) 2.576 (0.30)(0.70)
2                    2
(b)       n                                      13933.34               Use n = 13934
e2                 (0.01) 2

s    N –n                            7.8   500 – 25
8.50     (b)      N  X  N t               500 25.7  500 2.7969     
n   N –1                             25    500 –1

\$10,721.53  Population Total  \$14,978.47
Solutions to End-of-Section and Chapter Review Problems       159

8.51   Using Excel,
Di       (Di – D )2
13.76         164.719256
42.87        1759.3243
34.65        1137.32841
11.09         103.312994
14.54         185.349164
22.87         481.552302
25.52         604.879592
9.81          78.9307865
10.03          82.8882785
15.49         209.868823
Mean = 1.00315
Sum, first 10 items =     4780.82023
191*(0 – 1.00315)2 =       192.205195
(Di – D )2              4973.025425
St Dev =         4.99900765
sD     N –n                                      4.9990 10, 000 – 200
N  D  N t                 10, 000 1.00315  10, 000 1.96        
n     N –1                                        200   10, 000 –1
3,172.53  Total Difference in the Population  16,890.47

ps (1 – ps )                 0.04(1  0.04)
8.52   (a)        ps  Z                   0.04+1.2816                               p  0.0545
n                            300
ps (1 – ps )                0.04(1  0.04)
(b)        ps  Z                   0.04+1.645                                p  0.0586
n                           300
ps (1 – ps )                 0.04(1  0.04)
(c)        ps  Z                   0.04+2.3263                               p  0.0663
n                            300

s     N –n                                \$0.32 300 – 20
8.53   N  X  N t                 300  \$1.67  300  2.0930       
n    N –1                                  20   300 –1
\$457.52  Population Total  \$544.48

s     N –n                                  \$138.8046 3000 –10
8.54   N  X  N t                 3000  \$261.40  3000 1.8331          
n    N –1                                      10     3000 –1

\$543,176.96  Population Total  \$1,025,223.04

s     N –n                                    \$93.67 1546 – 50
8.55   N  X  N t                 1546  \$252.28  1546  2.0096        
n    N –1                                      50    1546 –1
\$349,526.64  Population Total  \$430,523.12
160    Chapter 8: Confidence Interval Estimation

sD    N –n                                     \$29.5523 4000 –150
8.56     N  D  N t                4000  \$7.45907  4000  2.6092          
n    N –1                                        150    4000 –1
\$5,125.99  Total Difference in the Population  \$54,546.57
Note:   The t-value of 2.6092 for 95% confidence and df = 149 was derived on Excel.

sD    N –n                                    \$25.2792 1200 –120
8.57     N  D  N t                1200  (–\$0.9583)  1200 1.98          
n    N –1                                       120    1200 –1
– \$6,354.02  Total Difference in the Population  \$4,054.10

ps (1 – ps )                  0.0367(1  0.0367)
8.58    (a)      ps  Z                     0.0367+1.645                                p  0.0545
n                               300
(b)     Since the upper bound is higher than the tolerable exception rate of 0.04, the auditor
should request a larger sample.

ps (1 – ps )                 0.024(1  0.024)
8.59    (a)      ps  Z                     0.024+1.645                                 p  0.0353
n                             500
(b)     With 95% level of confidence, the auditor can conclude that the rate of
noncompliance is less than 0.0353, which is less than the 0.05 tolerable exception
rate for internal control, and, hence, the internal control compliance is adequate.

8 60    The only way to have 100% confidence is to obtain the parameter of interest, rather than a
sample statistic. From another perspective, the range of the normal and t distribution is
infinite, so a Z or t value that contains 100% of the area cannot be obtained.

8.61    The t distribution is used for obtaining a confidence interval for the mean when  is
unknown.

8.62    If the confidence level is increased, a greater area under the normal or t distribution needs to
be included. This leads to an increased value of Z or t, and thus a wider interval.

8.63    The confidence interval will be narrower and the sample size will be smaller.

8.64    In some applications such as auditing, interest is primarily on the total amount of a variable
rather than the average amount.

8.65    Difference estimation involves determining the difference between two amounts, rather than
a single amount.
Solutions to End-of-Section and Chapter Review Problems            161

ps (1 – ps )                 0.43(0.57)
8.66   (a)   ps  Z                  0.43  1.96                          0.416  p  0.444
n                          5000
The sampling error is 0.014.
ps (1 – ps )                  0.159(1  0.159)
(b)   ps  Z                  0.159  1.96                         0.149  p  0.169
n                             5000
The sampling error is 0.010
(c)   The sampling error is the largest when ps  0.5 . Hence, the sampling error of (a)
with ps  0.43 is larger than that of (b) when ps  0.159

8.67   (a)   The population from which this sample was drawn was the collection of all
subscribers to the Redbook magazine who visited the magazine's web site.
(b)   The sample is not a random sample from this population. The sample consisted of
only those subscribers to the magazine who visited the magazine's web site and
chose to fill up the survey.
(c)   This is not a statistically valid study. There was selection bias since only those who
visited the magazine's web site and chose to answer the survey were represented.
There was possibly nonresponse bias as well. Visitors to the web site who chose to
fill up the survey might not answer all questions and there was no way for the
magazine to get back to them to follow-up on the nonresponses if this was an
anonymous survey.
(d)   To avoid the above potential pitfalls, the magazine could drawn a random sample
from the list of all subscribers to the magazine and offer them the option of filling up
the survey over the Internet or on the survey form that is mailed to the subscribers.
The magazine should also keep track of the subscribers who are invited to fill in the
survey and follow up on the nonresponses after a specified period of time with mail
or telephone to encourage them to participate in the survey.
Z 2  p  (1 – p ) 1.96 2  (0.6195)  (0.6195)
The sample size needed is n                                                      2264
e2                     (0.02) 2

ps (1 – ps )                  0.657(1  0.657)
8.68   (a)   ps  Z                0.657  1.96                            0.608  p  0.705
n                             364
Z  p  (1 – p ) 1.96  (0.657)  (0.657)
2                     2
(b)   n                                           2165
e2                   (0.02) 2
ps (1 – ps )                  0.870(1  0.870)
(c)   ps  Z                  0.870  1.96                          0.828  p  0.913
n                              239
ps (1 – ps )                 0.46(1  0.460)
(d)   ps  Z                  0.46  1.96                          0.397  p  0.523
n                            239
162    Chapter 8: Confidence Interval Estimation

s                      0.1058
8.69    (a)     X t        5.5014  2.6800                                       5.46<  5.54
n                        50
(b)    Since 5.5 gram is within the 99% confidence interval, the company can claim that the
average weight of tea in a bag is 5.5 gram with a 99% level of confidence.

s                       25.2835
8.70    (a)     X t        43.8889  2.0555                                  33.89<  53.89
n                         27
(b)    With 95% confidence, the average approval process takes somewhere between 33.89 to
53.89 days.

s                      0.0461
8.71    (a)     X t        8.4209  2.0106                                       8.41<  8.43
n                        49
(b)    With 95% confidence, the population average width of troughs is somewhere between
8.41 and 8.43 inches. Hence, the company's requirement of troughs being between 8.31
and 8.61 is being met with a 95% level of confidence.

s                      89.5508
8.72    (a)     X t        1723.4  2.0452                               1689.96<  1756.84
n                        30
(b)    With 95% confidence, the population average force is somewhere between 1689.96
and 1756.84 pounds. Hence, the company's requirement of strength of the insulators
being at least 1500 pounds is being met with a 95% level of confidence.

s                      3.8
8.73    (a)     X t       15.3  2.0227                                    14.085    16.515
n                      40
p (1 – ps )                 0.675(0.325)
(b)     ps  Z  s               0.675 1.96                            0.530  p  0.820
n                            40
Z 
2     2
1.96 5
2   2
(c)     n       2           2     = 24.01                                       Use n = 25
e            2

Z 2  p(1– p) 1.96 2 (0.5)(0.5)
(d)     n            2                 2      = 784                            Use n = 784
e            (0.035)
(e)    If a single sample were to be selected for both purposes, the larger of the two sample
sizes (n = 784) should be used.

s                     380
8.74    (a)     X t       1759  2.6490                                 1,638.69    1,879.31
n                      70
p (1– ps )                 0.60(0.40)
(b)     p s  Z s             0.60  1.96                              0.485  p  0.715
n                          70
Solutions to End-of-Section and Chapter Review Problems             163

s                       4
8.75   (a)   X t        9.7  2.0639                                            8.049    11.351
n                      25
ps (1 – ps )                0.48(0.52)
(b)   ps  Z                   0.48 1.96                                  0.284  p  0.676
n                          25
Z 2  2 1.96 2 4.5 2
(c)   n       2              2     = 34.57                                           Use n = 35
e             1.5
Z 2  p(1– p) 1.6452 (0.5)(0.5)
(d)   n           2                       2    = 120.268                           Use n = 121
e                  (0.075)
(e)   If a single sample were to be selected for both purposes, the larger of the two sample
sizes (n = 121) should be used.

Z 2  2 2.582 182
8.76   (a)   n        2        2    = 86.26                                                 Use n = 87
e         5
Note:       If the t-value used is carried out to 2.5758, the value of n is 85.986 and only
86 women would be sampled.
Z 2  p(1– p) 1.6452 (0.5)(0.5)
(b)   n            2                 2      = 334.07                               Use n = 335
e           (0.045)
(c)   If a single sample were to be selected for both purposes, the larger of the two sample
sizes (n = 335) should be used.

s                          \$11.39
8.77   (a)   X t         \$28.52  1.9949                                       \$25.80    \$31.24
n                             70
p (1 – ps )                    0.40(0.60)
(b)   ps  Z  s                 0.40  1.645                            0.3037  p  0.4963
n                             70
Z 
2     2
1.96 10
2      2
(c)   n       2             2      = 96.04                                           Use n = 97
e             2
Z 2  p(1– p) 1.6452 (0.5)(0.5)
(d)   n           2                       2       = 422.82                         Use n = 423
e                   (0.04)
(e)   If a single sample were to be selected for both purposes, the larger of the two sample
sizes (n = 423) should be used.

s                        \$9.22
8.78   (a)   X t         \$21.34  1.9949                                       \$19.14    \$23.54
n                          70
ps (1 – ps )                   0.3714(0.6286)
(b)   ps  Z                   0.3714  1.645
n                               70
0.2764  p  0.4664
Z  2      2
1.96 102    2
(c)   n      2             2 = 170.74                                              Use n = 171
e            1.5
Z 2  p(1– p) 1.6452 (0.5)(0.5)
(d)   n          2                  2     = 334.08                                 Use n = 335
e              (0.045)
(e)   If a single sample were to be selected for both purposes, the larger of the two sample
sizes (n = 335) should be used.
164    Chapter 8: Confidence Interval Estimation

s                        \$7.26
8.79    (a)      X t         \$38.54  2.0010                               \$36.66    \$40.42
n                          60
p (1 – ps )                 0.30(0.70)
(b)      ps  Z  s               0.30  1.645                        0.2027  p  0.3973
n                           60
Z 2  2 1.96 2 82
(c)     n       2            2    = 109.27                                     Use n = 110
e           1.5
Z 2  p(1– p) 1.6452 (0.5)(0.5)
(d)     n            2                     2     = 422.82                      Use n = 423
e                 (0.04)
(e)    If a single sample were to be selected for both purposes, the larger of the two sample
sizes (n = 423) should be used.

Z 2  p  (1 – p ) 1.96 2  (0.5)  (0.5)
8.80    (a)     n                                            = 384.16                  Use n = 385
e2                (0.05) 2
If we assume that the population proportion is only 0.50, then a sample of 385 would
be required. If the population proportion is 0.90, the sample size required is cut to
103.
ps (1 – ps )                 0.84(0.16)
(b)     ps  Z                  0.84  1.96 
n                           50
0.7384  p  0.94161
(c)    The representative can be 95% confidence that the actual proportion of bags that
will do the job is between 74.5% and 93.5%. He/she can according perform a
cost-benefit analysis to decide if he/she want to sell the Ice Melt product.
7
8.81    (a)     ps       0.14
50
ps (1 – ps )   N –n                     0.14(0.86) 1000 – 50
ps  Z                       0.14  1.2816                
n         N –1                          50        1000 – 1
p  0.2013
(b)    Since the upper bound is higher than the tolerable exception rate of 0.15, the auditor
should request a larger sample.

Z 2   2 1.962  302
8.82    (a)     n0                          = 138.292
e2             52
n0 N             138.292  25056
n                                       = 137.53                    Use n = 138
n0  ( N –1) 138.292  (25056 –1)
12
(b)     ps         0.087
138
ps (1 – ps ) N – n
ps  Z                 
n         N –1
0.087(0.913) 25, 056 –138
 0.087  1.645                 
138       25, 056 –1
0.0476  p  0.1263
Solutions to End-of-Section and Chapter Review Problems     165

8.82    (c)   Using Excel, we find the t-value for 95% confidence and 137 degrees
cont.         of freedom is t = 1.9774.

s    N –n                    \$34.55 25,056 –138
X t              \$93.70 1.9774        
n   N –1                      138   25,056 – 1

\$87.90  X  \$99.50
s    N –n
(d)   N  X  N t       
n   N –1
\$34.55 25, 056 –138
 25, 056  \$93.70  25, 056 1.9774          
138   25, 056 –1

\$2,202,427.61  Population Total  \$2,493,066.79

D 241                                   (D – D) 2   6432.12
(e)   D           1.7463768           sD                           6.85199
n    138                                  n–1          137
s      N –n
ND  Nt D 
n    N –1
\$6.85199 25,056 –138
 25,056 \$1.7463768  25,0561.9774               
138    25,056 –1

\$14,937.30  Total Difference in the Population  \$72,577.14

Z 2   2 2.582  4002
8.83    (a)   n                        = 26.6256                               Use n = 27
e2         2002
s     N –n                                   \$384.62 258 – 27
(b)   N  X  N t              258  \$3,054  258  2.77872         
n    N –1                                      27     258 –1

\$737,655.50  Population Total  \$838,275.58
166   Chapter 8: Confidence Interval Estimation

The Springville Herald Case

SH8.1 (a)      The advantages of random digit dialing is that this method allows the survey to reach
households that have unlisted telephone numbers. One of the disadvantages is that
additional telephone calls would be made due to calls to commercial and other non-
residential telephone numbers as well as calls to households outside the Springville
Herald distribution area. The other disadvantage would be the inability to reach
individuals who might be interested in purchasing the Springville Herald , but do not
have a telephone in their household.
(b)     An alternative approach would involve the use of a mail survey of households with a
telephone follow-up to contact non-responding households. The advantage of this
approach is that the mail survey would be less expensive to conduct and would be
able to reach some households without phones. However, the rate of response would
undoubtedly be lower than for a telephone survey.

SH8.2 Note: 95% confidence intervals used
Question 1: Yes: ps = .8670 .834 to .900           Question 2: Yes: ps = .3864 .3355 to.4373
Question 3: Monday - Saturday: ps = .1323 .0754 to .1894
Question 3: Sunday only: ps = .1838 .1187 to .2489
Question 3: Every day: ps = .6838 .6056 to .7620
Question 4: Every day: ps = .3611 .2970 to .4252
Question 4: Most days: ps = .4398 .3736 to .5060
Question 4: Occasionally: ps = .1991 .1458 to .2524
Question 5: Every Sunday: ps = .6389 .5757 to .7039
Question 5: 2-3 Sundays a month: ps = .25 .1923 to .3077
Question 5: Once a month: ps = .1111 .0692 to .1530
Question 6: Convenience store/Delicatessen: ps = .3426 .2793 to .4059
Question 6: Stationery/Candy store: ps = .4398 .3736 to .5060
Question 6: Vending machine: ps = .0972 .0577 to .1367
Question 6: Supermarket: ps = .0602 .0285 to .0919
Question 6: Other: ps = .0602 .0285 to .0919       Question 7: Yes: ps = .2130 .1584 to .2676
Question 8: X = 3.7348 s = .3602 3.6278 to 3.8418
Question 9: Yes: ps = .3920 .3410 to .4430
Question 10: Yes : ps = .1875 .1467 to .2283
A large proportion of households read the paper (estimated to be between 83.4 and 90%) but
it is estimated that only between 33.55% and 43.73% receive a home subscription. The bulk
of those who have a subscription receive the newspaper every day. Those who purchase the
newspaper are most likely to do so either every day or most days and most Sundays. They
are most likely to purchase the newspaper in a convenience store/delicatessen or stationery/
candy store. However, we estimate that only between 15.84% and 26.76% would consider a
discounted trial subscription. We should in the future investigate the reasons why readers do
not wish to receive a trial subscription. Of those who would consider a discounted trial
subscription, we estimate the average price that they would be willing to pay to be between
\$3.63 and \$3.84. We also estimate that only between 34.1% and 44.3% of readers read
another daily newspaper. Finally, between 14.67 and 22.83% of the readers would consider
prepaying for six months in order to receive a restaurant discount card.

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 12 posted: 4/9/2011 language: English pages: 17