Statistics for managers using microsoft excel global 8th edition levine solutions manual

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CHAPTER 8 8.1

A (1 −  )  100% confidence interval estimate of the mean with σ known, where Z 2 is the value corresponding to an upper tail of  2 from the standardized normal distribution (that is, a cumulative area of 1 −  2 ).

X  Z 2

 n

or

X − Z 2

 n

   X + Z 2

 n

99% confidence corresponds to an  value of 0.01.

 2 1−

=



0.01 = 0.005 2

2

= 1 − 0.005 = 0.995

Z 2 = Z 0.005 = 2.58

Now students will need to construct a 99% confidence interval estimate of the populationmean, μ.

X − Z 2

 n

   X + Z 2

 n

= 66 − 2.58

6 64

   66 + 2.58

6 64

= 64.07    67.94

8.2

105.02    122.98

8.3

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 100% confidence, the entire population (sample size N) would have to be selected.

8.4

Yes, it is true since 5% of intervals will not include the population mean.

8.5

If all possible samples of the same size n=100 are taken, 95% of them will include the true population mean time spent on the site per day. Thus you are 95 percent confident that this sample is one that does correctly estimate the true mean time spent on the site per day.

8.6

(a)

(b)

8.7

You would compute the mean first because you need the mean to compute the standard deviation. If you had a sample, you would compute the sample mean. If you had the population mean, you would compute the population standard deviation.

If you have a sample, you are computing the sample standard deviation not the population standard deviation needed in Equation 8.1. If you have a population, and have computed the population mean and population standard deviation, you don't need a confidence interval estimate of the population mean since you already have computed it.

Note that if the population mean, μ, is in the interval indicated by the confidence interval estimate, then the confidence interval estimate is correct. If the population mean is not in the interval, then the confidence interval estimate is not correct.

Thus, the answer should be no. This is because the value of μ is not included within the confidence interval estimate.

8.8

Equation (8.1) assumes that you know the population standard deviation. Because you are selecting a sample of 100 from the population, you are computing a sample standard deviation, not the population standard deviation.

8.9

(a)

The upper and lower estimates of the confidence interval can be calculated using technology or the formula below, where X is the sample mean,  is the population standard deviation, n is the sample size, and Z 2 is the value corresponding to an uppertail probability of  2 from the standardized normal distribution. For the purpose of this exercise, use technology.

X  Z 2

 n

or X − Z 2

 n

   X  Z 2

 n

X = 0.927

 = 0.04 n = 50

Thus, the 95% confidence interval is 0.91591    0.93809. (b)

(c)

(d)

The distributor can be (1 −  )  100% confident that the population mean is within the confidence interval. Note that a 1-gallon bottle containing exactly 1-gallon of water is not the real population mean; it is what the distributor wants the population mean to be. So students should say no, because 1 falls outside the interval 0.91591 and 0.93809. Thus, if the population mean is within the confidence interval, the distributor does not have a right to complain. If the population mean is outside the confidence interval, the distributor does have a right to complain. Use the information above to determine whether the distributor has a right to complain based on the results of the confidence interval.

The sampling distribution of the sample mean, X , is normally distributed if the population for your characteristic of interest, X, follows a normal distribution. And, if the population of X does not follow a normal distribution, the Central Limit Theorem almost always ensures that X is normally distributed when n is large. However, when dealing with a small sample size and a population that does not follow a normal distribution, the sampling distribution of X is not normally distributed, and therefore the confidence interval discussed in this section is inappropriate. In practice, however, as long as the sample size is large enough and the population is not very skewed, you can use the confidence interval to estimate the population mean when σ is known. Use this information to determine if the population must be assumed to be normally distributed.  Using X  Z 2 , the 90% confidence interval is 0.91770    0.93630. Thus, a 1n gallon bottle containing exactly 1-gallon of water does not fall within the 90% confidence interval. Recall that if the population mean is within the confidence interval, the distributor does not have a right to complain. If the population mean is outside the confidence interval, the distributor does have a right to complain. Use the information above to determine whether the distributor has a right to complain based on the results of the confidence interval. Compare this result to the result found in part (b).

8.10

(a)

Confidence Interval Estimate for the Mean Data Population Standard Deviation Sample Mean Sample Size Confidence Level

1000 7500 64 95%

Intermediate Calculations Standard Error of the Mean 125 Z Value -1.9600 Interval Half Width 244.9955 Confidence Interval Interval Lower Limit Interval Upper Limit

X Z (b)

(c) (d)



n

= 7500  1.96 

7255.00 7745.00

1000 64

7255.00    7745.00

No. The manufacturer cannot support a claim that the bulbs last an average 8,000 hours with a 95% level of confidence because 8,000 does not fall inside the 95% confidence interval.

No. Since  is known and n = 64, from the Central Limit Theorem, we may assume that the sampling distribution of X is approximately normal.

The confidence interval is narrower based on a process standard deviation of 800 hours rather than the original assumption of 1000 hours.



7304.00    7696.00

X Z

(b)

No. The manufacturer cannot support a claim that the bulbs last an average 8,000 hours with a 95% level of confidence because 8,000 still does not fall inside the 95% confidence interval.

n

= 7500  1.96 

800 64

(a)

8.11

The equation below defines the (1 −  )  100% confidence interval estimate for the mean with  unknown, where t 2 is the critical value corresponding to an upper-tail probability  2 from the t distribution with n − 1 degrees of freedom.

S S X − t 2    X  t 2 n n  = 0.05  0.05 = = 0.025 2 2 While either technology or a table of critical values of t can be used to find the critical value, for this problem, use a table. Use a table of critical values of t with df = n – 1 = 63 and  2 = 0.025 to find the critical value for a 95% confidence interval. t 2 = 1.9983.

So, X − t 2

S n

   X  t 2

S n

= 66  (1.9983)

16 64

= 66  3.9966.

Therefore, the confidence interval is 62.00    70.00.

8.12

8.13

(a)

df = 9,  = 0.05, t /2 = 2.2622

(b)

df = 9,  = 0.01, t /2 = 3.2498

(c)

df = 31,  = 0.05, t /2 = 2.0395

(d)

df = 64,  = 0.05, t /2 = 1.9977

(e)

df = 15,  = 0.1, t /2 = 1.7531

3.7417 8 2.4495 Set 2: 4.5  2.3646 8 Set 1: 4.5  2.3646

1.3719    7.6281 2.4522    6.5478

The data sets have different confidence interval widths because they have different values for the standard deviation.

8.14

6.4660 7 2.1602 Altered data: 4.00  2.4469  7 Original data: 5.8571  2.4469 

– 0.1229    11.8371 2.0022    5.9978

The presence of an outlier in the original data increases the value of the sample mean and greatly inflates the sample standard deviation.

8.15

(a)

PHStat output: Confidence Interval Estimate for the Mean

Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

200 1500 100 95%

Intermediate Calculations Standard Error of the Mean 20 Degrees of Freedom 99 t Value 1.9842 Interval Half Width 39.6843 Confidence Interval Interval Lower Limit Interval Upper Limit

X t (b)

S 200 = 1500  1.9842  n 100

1460.32 1539.68 $ 1460.32    $ 1539.68

You can be 95% confident that the population mean spending for all Amazon Prime member shoppers is somewhere between $1460.32 and $1539.68.

8.16

(a)

S 9 = 82  1.9890  n 84 PHStat output:

$80.05    $83.95

X t

Confidence Interval Estimate for the Mean Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

9 82 84 95%

Intermediate Calculations Standard Error of the Mean 0.981980506 Degrees of Freedom 83 t Value 1.9890 Interval Half Width 1.9531

(b)

8.17

(a)

Confidence Interval Interval Lower Limit 80.05 Interval Upper Limit 83.95 You can be 95% confident that the population mean one-time gift donation is somewhere between $80.05 and $83.95.

The equation below defines the (1 −  )  100% confidence interval estimate for the mean with  unknown, where t 2 is the critical value corresponding to an upper-tail probability  2 from the t distribution with n − 1 degrees of freedom.

S X  t 2 n  = 0.10 n = 24 X = 141.3 S = 19.1 t 2 = 1.7139

(b) (c)

141.3  6.6821 Therefore, the confidence interval is 134.62    147.98.

Yes, because a grade of 150 is not in the interval.

It is not unusual because it is only 0.56 standard deviations above the sample mean.

8.18

PHStat output: Confidence Interval Estimate for the Mean

Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

1.406031226 7.09 15 95%

Intermediate Calculations Standard Error of the Mean 0.363035702 Degrees of Freedom 14 t Value 2.1448 Interval Half Width 0.7786 Confidence Interval Interval Lower Limit Interval Upper Limit

8.19

6.31 7.87

S 1.4060 = 7.09  2.1448 n 15

6.31    7.87

(a)

X t

(b)

You can be 95% confident that the population mean amount spent for lunch ($) at a fastfood restaurant is somewhere between $6.31 and $7.87.

PHStat output: Confidence Interval Estimate for the Mean

Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

5.320012804 28.70833333 24 95%

Intermediate Calculations Standard Error of the Mean 1.085943066 Degrees of Freedom 23 t Value 2.0687 Interval Half Width 2.2464 Confidence Interval Interval Lower Limit Interval Upper Limit

26.46 30.95

S  5.3200  = 28.7083  2.0687   n  24 

26.46    30.95

8.19

(a)

X t

cont.

(b)

You can be 95% confident that the population mean MPG of 2015 midsized sedans is somewhere between 26.46 and 30.95.

Because the upper limit of the 95% confidence interval for population mean miles per gallon of 2015 small SUVs is lower than the lower limit of the 95% confidence interval for population mean miles per gallon of 2015 midsized sedans, you are able to conclude that the population mean miles per gallon of 2015 small SUVs is lower than that of 2015 midsized sedans.

(c)

8.20

PHStat output:

Confidence Interval Estimate for the Mean Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

1.954078057 23.15 20 95%

Intermediate Calculations Standard Error of the Mean 0.436945137 Degrees of Freedom 19 t Value 2.0930 Interval Half Width 0.9145 Confidence Interval Interval Lower Limit Interval Upper Limit

22.24 24.06

S  1.9541  = 23.15  2.0930   n  20 

22.24    24.06

(a)

X t

(b)

You can be 95% confident that the population mean MPG of 2015 small SUVs is somewhere between 22.24 and 24.06.

Because the upper limit of the 95% confidence interval for population mean miles per gallon of 2015 small SUVs is lower than the lower limit of the 95% confidence interval for population mean miles per gallon of 2015 midsized sedans, you are able to conclude that the population mean miles per gallon of 2015 small SUVs is lower than that of 2015 midsized sedans.

(c)

8.21

(a)

PHStat output: One-Year CD

Confidence Interval Estimate for the Mean Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

0.289327957 0.86 25 95%

Intermediate Calculations Standard Error of the Mean 0.057865591 Degrees of Freedom 24 t Value 2.0639 Interval Half Width 0.1194 Confidence Interval Interval Lower Limit Interval Upper Limit

0.74 0.98

S  0.2893  = 0.86  2.0639   n  25 

(a)

X t

(b)

PHStat output: Five-Year CD

Confidence Interval Estimate for the Mean Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

0.424447484 1.64 25 95%

Intermediate Calculations Standard Error of the Mean 0.084889497 Degrees of Freedom 24 t Value 2.0639 Interval Half Width 0.1752 Confidence Interval Interval Lower Limit Interval Upper Limit

1.46 1.81

0.74    0.98

8.21

(b)

cont. (c)

1.46    1.81 Because the 95% confidence interval for population mean yield of one-year certificates of deposit does not overlap with that for the population mean yield of fiveyear certificates of deposit, you are able to conclude that the population mean yield of one-year certificates of deposit is lower than that of five-year certificates of deposit.

S 41.9261 = 43.04  2.0096  n 50

31.12    54.96

(a)

X t

(b) (c)

The population distribution needs to be normally distribution.

Normal Probability Plot 180 160 140 120

Days

8.22

S  0.4244  = 1.64  2.0639   n  25 

X t

100 80 60 40 20 0

-3

-2

-1

0

1

2

3

Z Value

(c) Box-and-whisker Plot

Days

0

(d)

50

100

150

Both the normal probability plot and the boxplot suggest that the distribution is skewed to the right.

Even though the population distribution is not normally distributed, with a sample of 50, the t distribution can still be used due to the Central Limit Theorem.

(a)

X t

S 89.5508 = 1723.4  2.0452  n 30

1689.96    1756.84

Confidence Interval Estimate for the Mean Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

89.55083319 1723.4 30 95%

Intermediate Calculations Standard Error of the Mean 16.34967046 Degrees of Freedom 29 t Value 2.0452 Interval Half Width 33.4388

(b) (c)

Confidence Interval Interval Lower Limit 1689.96 Interval Upper Limit 1756.84 The population distribution needs to be normally distributed.

Normal Probability Plot 2000 1800 1600 1400 1200 1000 800 600 400 200 0 -3

Force

8.23

-2

-1

0 Z Value

1

2

3

The normal probability plot indicates that the population distribution is normally distributed.

(a)

PHStat output: Confidence Interval Estimate for the Mean

Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

6.651868497 16.39285714 28 95%

Intermediate Calculations Standard Error of the Mean 1.257084985 Degrees of Freedom 27 t Value 2.0518 Interval Half Width 2.5793 Confidence Interval Interval Lower Limit Interval Upper Limit

X t (b) (c)

13.81 18.97

S  6.6519  = 16.3929  2.0518   n  28 

13.81    18.97

The population distribution needs to be normally distributed.

Normal Probability Plot 40 35

Mobile Commerce Penetration (%)

8.24

30

25 20

Mobile Commerce Penetration (%)

15 10

5 0

-3

-2

-1

0 Z Value

1

2

3

The normal probability plot suggests a right skewed distribution due to the outlier from the Republic of Korea.

8.25

S 0.0017 = −0.00023  1.9842  n 100

–0.000566    0.000106

(a)

X t

(b)

The population distribution needs to be normally distributed. However, with a sample of 100, the t distribution can still be used as a result of the Central Limit Theorem even if the population distribution is not normal.

(c) Normal Probability Plot 0.006 0.005 0.004

Error

0.003 0.002 0.001 0 -0.001 -3

-2

-1

0

1

2

0.004

0.006

3

-0.002 -0.003 -0.004

Z Value Box-and-whisker Plot

Error

-0.006

(d)

8.26

8.27

p=

p=

-0.004

-0.002

0

0.002

Both the normal probability plot and the boxplot suggest that the distribution is skewed to the right.

We are 95% confident that the mean difference between the actual length of the steel part and the specified length of the steel part is between -0.000566 and 0.000106 inch , which is narrower than the plus or minus 0.005 inch requirement. The steel mill is doing a good job at meeting the requirement. This is consistent with the finding in Problem 2.39.

X 50 = 0.25 = n 200

X 25 = 0.0625 = n 400

pZ

p(1– p) 0.25(0.75) = 0.25  1.96 n 200

pZ

0.19    0.31

p(1 − p) 0.0625(0.9375) = 0.0625  2.58 n 400 0.0313    0.0937

8.28

(a)

p=

(b)

8.29

(a) (b) (c)

X 135 = 0.27 = n 500

pZ

p(1 – p) 0.27(1 − 0.27) = 0.27  2.5758 n 500

0.22    0.32 The manager in charge of promotional programs concerning residential customers can infer that the proportion of households that would purchase a new cellphone if it were made available at a substantially reduced installation cost is between 0.22 and 0.32 with a 99% level of confidence.

The confidence interval is 0.3709    0.3968.

The confidence interval is 0.2971    0.3217.

One can be 90% confident that the population proportion of all travelers who said that location was very important for choosing a hotel lies within the interval in (a). One can be 90% confident that the population proportion of all travelers who said that reputation was very important in choosing an airline lies within the interval in (b).

8.30

(a)

PHStat output: Confidence Interval Estimate for the Proportion Data Sample Size Number of Successes Confidence Level

1000 260 95%

Intermediate Calculations Sample Proportion Z Value Standard Error of the Proportion Interval Half Width

0.26 -1.9600 0.0139 0.0272

Confidence Interval Interval Lower Limit Interval Upper Limit

0.2328 0.2872

p = = 0.26

pZ

p(1– p) 0.26(1 − 0.26) = 0.26  1.96 n 1000

(b)

0.2328    0.2872 No, you cannot because the interval estimate includes 0.25 (25%).

(c)

(a)

(d)

p = 0.26

pZ

p(1– p) 0.26(1 − 0.26) = 0.26  1.96 n 10000

0.2514    0.2686 (b) Yes, you can, because the interval is above 0.25 (25%).

The larger the sample size, the narrow is the confidence interval holding everything else constant.

8.31

PHStat output: Confidence Interval Estimate for the Proportion

Data Sample Size Number of Successes Confidence Level Intermediate Calculations Sample Proportion Z Value Standard Error of the Proportion Interval Half Width

0.44 -1.9600 0.0375 0.0735

Confidence Interval Interval Lower Limit Interval Upper Limit

0.3665 0.5135

p=

8.32

(a)

175 77 95%

X 77 = 0.44 = n 175

pZ

p(1– p) 0.44(1 − 0.44) = 0.44  1.96 n 175 0.3665    0.5135

PHStat output: Confidence Interval Estimate for the Proportion

Data Sample Size Number of Successes Confidence Level

1074 935 95%

Intermediate Calculations Sample Proportion Z Value Standard Error of the Proportion Interval Half Width

0.8706 -1.9600 0.0102 0.0201

Confidence Interval Interval Lower Limit Interval Upper Limit

0.8505 0.8907

8.32

X 935 = 0.8706 = n 1074 p(1– p) 0.8706(1 − 0.8706) pZ = 0.8706  1.96 n 1074

(a)

p=

(b)

PHStat output: Confidence Interval Estimate for the Proportion

cont.

0.8505    0.8907

Data Sample Size Number of Successes Confidence Level

1074 419 95%

Intermediate Calculations Sample Proportion Z Value Standard Error of the Proportion Interval Half Width

0.3901 -1.9600 0.0149 0.0292

Confidence Interval Interval Lower Limit Interval Upper Limit

0.3610 0.4193

X 419 = 0.3901 = n 1074 p(1– p) 0.3901(1 − 0.3901) pZ = 0.3901  1.96 n 1074

p=

(c)

0.3610    0.4193 There is a significantly higher proportion of Facebook users who are connected to friends from the past than connected to people they have never met in person because the two confidence intervals do not overlap.

8.33

(a)

p=

157 = 0.8920 176

pZ

p (1 − p ) n

= 0.8920  1.96

0.8920 (1 − 0.8920 ) 176

0.8462    0.9379

Confidence Interval Estimate for the Proportion Data Sample Size Number of Successes Confidence Level

(b)

176 157 95%

Intermediate Calculations Sample Proportion Z Value Standard Error of the Proportion Interval Half Width

0.8920 -1.9600 0.0234 0.0458

Confidence Interval Interval Lower Limit Interval Upper Limit

0.8462 0.9379

p=

79 = 0.4489 176

pZ

p (1 − p ) n

= 0.4489  1.96

176

0.3754    0.5223

Confidence Interval Estimate for the Proportion Data Sample Size Number of Successes Confidence Level

0.4489 (1 − 0.4489 )

176 79 95%

Intermediate Calculations Sample Proportion Z Value Standard Error of the Proportion Interval Half Width

0.4489 -1.9600 0.0375 0.0735

Confidence Interval Interval Lower Limit Interval Upper Limit

0.3754 0.5223

8.33 cont.

You can be 95% confident that the population proportion of tech CEOs who indicate cybersecurity is strategically important for their organization is somewhere between 0.8462 and 0.9379, and the population proportion of tech CEOs who indicate robotics is strategically important for their organization is somewhere between 0.3754 and 0.5223.

(c)

8.34

n=

Z 2  2 1.962 152 = 34.57 = e2 52

Use n = 35

8.35

n=

Z 2  2 2.582 100 2 = 166.41 = e2 202

Use n = 167

8.36

n=

Z 2 (1–  ) 2.582 (0.5)(0.5) = 1,040.06 = e2 (0.04) 2

Use n = 1,041

8.37

n=

Z 2 (1–  ) 1.962 (0.4)(0.6) = 2,304.96 = e2 (0.02)2

Use n = 2,305

8.38

(a) (b)

Z 2  2 1.962  4002 = 245.86 = e2 502 Z 2 2 1.962  4002 = 983.41 n= 2 = e 252

n=

Use n = 246 Use n = 984

8.39

n=

Z 2 2 1.962  (0.02) 2 = 96.04 = e2 (0.004) 2

Use n = 97

8.40

n=

Z 2 2 1.962  (1000)2 = 96.04 = e2 (200)2

Use n = 97

8.41

n=

Z 2  2 1.962 (0.05) 2 = = 96.04 e2 (0.01) 2

Use n = 97

8.42

(a) (b)

8.43

(a) (b)

Z 2 2 2.57582  202 = 106.1583 = e2 52 Z 2 2 1.962  202 = 61.4633 n= 2 = e 52

n=

Z 2  2 1.6452 45 2 = = 219.19 e2 52 Z 2 2 2.57582  452 = 537.4266 n= 2 = e 52

n=

Use n = 107 Use n = 62

Use n = 220 Use n = 538

8.44

Note: All the answers are computed using PHStat. Answers computed otherwise may be slightly different due to rounding. (a) (b) (c) (d)

8.45

Use n = 246 Use n = 385 Use n = 554

When there is more variability in the population, a larger sample is needed to accurately estimate the mean.

Note: All the answers are computed using PHStat. Answers computed otherwise may be slightly different due to rounding. (a) (b) (c)

8.46

Z 2 2 1.96 2  2 2 = 245.85 = e2 0.25 2 Z 2 2 1.96 2  2.5 2 = 384.15 n= = e2 0.25 2 Z 2 2 1.96 2  3.0 2 = 553.17 n= = e2 0.25 2 n=

Z 2 (1 –  ) 1.962 (0.31)(1 − 0.310.31) = 513.5550 n= = e2 (0.04) 2 Z 2 (1–  ) 2.57582 (0.31)(1 − 0.31) = 887.0027 n= = e2 (0.04)2 Z 2 (1–  ) 1.962 (0.31)(1 − 0.31) = 2,054.2201 n= = e2 (0.02) 2

Use n = 514 Use n = 888 Use n = 2,055

(d) n =

Z 2 (1–  ) 2.57582 (0.31)(1 − 0.31) = 3548.0110 = e2 (0.02)2

(e)

The higher the level of confidence desired, the larger is the sample size required. The smaller the sampling error desired, the larger is the sample size required.

Use n = 3,549

Note: All the answers are computed using PHStat. Answers computed otherwise may be slightly different due to rounding. (a)

(b)

(c)

p = 0.47

p = 0.33

p = 0.42

pZ

pZ

pZ

p (1 − p ) n p (1 − p ) n p (1 − p ) n

= 0.47  1.96

= 0.33  1.96

= 0.42  1.96

0.47 (1 − 0.47 ) 300

0.4135    0.5265

0.33 (1 − 0.33) 300

0.2768    0.3832

0.42 (1 − 0.42 ) 300

0.3641    0.4759

8.46

(d)

(a)

n=

cont.

8.47

(b)

n=

(c)

n=

Z 2 (1 −  ) e2 Z 2 (1 −  ) e2 Z 2 (1 −  ) e2

=

= =

1.962 ( 0.47 )(1 − 0.47 ) 0.022 1.962 ( 0.33)(1 − 0.33) 0.022 1.962 ( 0.42 )(1 − 0.42 ) 0.022

= 2,392.2685

Use n = 2,393

= 2,123.3664

Use n = 2,124

= 2,339.4484

Use n = 2,340

Note: All the answers are computed using PHStat. Answers computed otherwise may be slightly different due to rounding. (a)

p = 98/362 = 0.2707

pZ

p (1 − p ) n

= 0.2707  1.96

0.2707 (1 − 0.2707 ) 362

0.2249    0.3165

Confidence Interval Estimate for the Proportion Data Sample Size Number of Successes Confidence Level

362 98 95%

Intermediate Calculations Sample Proportion Z Value Standard Error of the Proportion Interval Half Width

(b)

(c) 8.48

0.2707 -1.9600 0.0234 0.0458

Confidence Interval Interval Lower Limit 0.2249 Interval Upper Limit 0.3165 You are 95% confident that the population proportion of nonprofits that indicate inability to pay competitively as the greatest employee retention challenge at their organization. is somewhere between 0.2249 and 0.3165.

n=

Z 2 (1 −  ) e2

=

1.962 ( 0.2707 )(1 − 0.2707 ) 0.012

= 7,584.1872

Use n = 7,585

Note: All the answers are computed using PHStat. Answers computed otherwise may be slightly different due to rounding. (a) If you conducted a follow-up study, you would use p = 0.33 in the sample size formula because it is based on past information on the proportion. (b)

n=

Z 2 (1 −  ) e2

=

1.962 ( 0.33)(1 − 0.33) 0.032

=943.7184

Use n = 944

8.49

Note: All the answers are computed using PHStat. Answers computed otherwise may be slightly different due to rounding.

Z 2 (1 −  )

n=

(b)

n=

(c)

The higher the precision you require of your confidence interval, the larger is the sample size required.

e2 Z 2 (1 −  ) e2

=

2.57582 ( 0.67 )(1 − 0.67 )

(a)

=

0.032 2.57582 ( 0.67 )(1 − 0.67 ) 0.052

= 1,629.9729

Use n = 1,630

= 586.7903

Use n = 578

8.50

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.

8.51

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

8.52

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.

8.53

The term  (1 −  ) reaches its largest value when the population proportion is at 0.5. Hence, the sample size n =

Z 2 (1 −  ) e2

needed to determine the proportion is smaller when the population

proportion is 0.20 than when the population proportion is 0.50.

8.54

(a)

Computer:

pZ

p (1 − p ) n

= 0.91  1.96

0.91(1 − 0.91) 1000

0.8923    0.9277

Smartphone:

pZ

p (1 − p ) n

= 0.80  1.96

0.80 (1 − 0.80 ) 1000

0.7752    0.8248

Tablet:

pZ

p (1 − p ) n

= 0.47  1.96

0.47 (1 − 0.47 ) 1000

0.4391    0.5009

Smart watch:

pZ

p (1 − p ) n

= 0.09  1.96

0.09 (1 − 0.09 ) 1000

0.0723    0.1077

8.54

(b)

We are 95% confident that the population proportion of the adult Internet users that own a computer is between 0.8923 and 0.9277.

We are 95% confident that the population proportion of the adult Internet users that own a smartphone is between 0.7752 and 0.8248.

We are 95% confident that the population proportion of the adult Internet users that own a tablet is between 0.4391 and 0.5009.

We are 95% confident that the population proportion of the adult Internet users that own a smart watch is between 0.0723 and 0.1077.

8.55

Note: All the answers are computed using PHStat. Answers computed otherwise may be slightly different due to rounding. (a) Buy store brands:

p = 0.9296

pZ

p (1 − p ) n

= 0.9296  1.96

0.9296 (1 − 0.9296 ) 1008

0.9138    0.9454

Use print coupons:

p = 0.8304

pZ

p (1 − p ) n

= 0.8304  1.96

0.8304 (1 − 0.8304 ) 1008

0.8072    0.8535

Plan meals around sale items:

p = 0.7202

pZ

p (1 − p ) n

= 0.7202  1.96

0.7202 (1 − 0.7202 ) 1008

0.6925    0.7479

Buy groceries online:

p = 0.3304

(b)

pZ

p (1 − p ) n

= 0.3304  1.96

0.3304 (1 − 0.3304 ) 1008

0.3013    0.3594

We are 95% confident that the population proportion of adult women in the U. S. try to save when shopping for groceries through buying store brand is somewhere between 0.9138 and 0.9454.

We are 95% confident that the population proportion of adult women in the U. S. try to save when shopping for groceries through using print coupon is somewhere between 0.8072 and 0.8535.

We are 95% confident that the population proportion of adult women in the U. S. try to save when shopping for groceries through planning meals around sale items is somewhere between 0.6925 and 0.7479.

We are 95% confident that the population proportion of adult women in the U. S. try to save when shopping for groceries through buying groceries online is somewhere between 0.3013 and 0.3594.

8.56

Note: All the answers are computed using PHStat. Answers computed otherwise may be slightly different due to rounding. (a) PHStat output: Confidence Interval Estimate for the Mean Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

3.5 51 40 95%

Intermediate Calculations Standard Error of the Mean 0.553398591 Degrees of Freedom 39 t Value 2.0227 Interval Half Width 1.1194 Confidence Interval Interval Lower Limit Interval Upper Limit

(b)

49.88 52.12

49.88    52.12

PHStat output: Confidence Interval Estimate for the Proportion Data Sample Size Number of Successes Confidence Level

(c) (d) (e)

40 32 95%

Intermediate Calculations Sample Proportion Z Value Standard Error of the Proportion Interval Half Width

0.8000 -1.9600 0.0632 0.1240

Confidence Interval Interval Lower Limit Interval Upper Limit

0.6760 0.9240

Z 2  2 1.96 2 5 2 n= = = 24.01 e2 22 Z 2    (1–  ) 1.962  (0.5)  (0.5) = 266.7680 n= = e2 (0.06) 2

0.6760    0.9240 Use n = 25 Use n = 267

If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 267) should be used.

8.57

(a)

PHStat output: Confidence Interval Estimate for the Mean

Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

10 45 70 99%

Intermediate Calculations Standard Error of the Mean 1.195228609 Degrees of Freedom 69 t Value 2.6490 Interval Half Width 3.1661 Confidence Interval Interval Lower Limit Interval Upper Limit (b)

41.83 48.17

41.83    48.17

PHStat output: Confidence Interval Estimate for the Proportion

Data Sample Size Number of Successes Confidence Level

70 36 95%

Intermediate Calculations Sample Proportion 0.514285714 Z Value -1.9600 Standard Error of the Proportion 0.0597 Interval Half Width 0.1171 Confidence Interval Interval Lower Limit Interval Upper Limit

0.3972 0.6314

0.3972    0.6314

8.58

(a)

PHStat output: Confidence Interval Estimate for the Mean

Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

7.3 6.2 25 95%

Intermediate Calculations Standard Error of the Mean 1.46 Degrees of Freedom 24 t Value 2.0639 Interval Half Width 3.0133 Confidence Interval Interval Lower Limit Interval Upper Limit (b)

3.19 9.21

3.19    9.21

PHStat output: Confidence Interval Estimate for the Proportion

Data Sample Size Number of Successes Confidence Level

(c) (d) (e)

25 13 95%

Intermediate Calculations Sample Proportion Z Value Standard Error of the Proportion Interval Half Width

0.52 -1.9600 0.0999 0.1958

Confidence Interval Interval Lower Limit Interval Upper Limit

0.3242 0.7158

Z  1.96  8 = 109.2682 = 2 e 1.52 Z 2    (1–  ) 1.6452  (0.5)  (0.5) = 120.268 n= = e2 (0.075)2 n=

2

2

2

0.3241    0.7158

2

Use n = 110 Use n = 121

If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 121) should be used.

8.59

Note: All the answers are computed using PHStat. Answers computed otherwise may be slightly different due to rounding. (a) PHStat output: Confidence Interval Estimate for the Mean Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

1000 12500 100 95%

Intermediate Calculations Standard Error of the Mean 100 Degrees of Freedom 99 t Value 1.9842 Interval Half Width 198.4217 Confidence Interval Interval Lower Limit Interval Upper Limit

(b)

12301.58 12698.42

$12,301.58    $12,698.42

PHStat output: Confidence Interval Estimate for the Proportion Data Sample Size Number of Successes Confidence Level

(c) (d)

100 30 95%

Intermediate Calculations Sample Proportion Z Value Standard Error of the Proportion Interval Half Width

0.3000 -1.9600 0.0458 0.0898

Confidence Interval Interval Lower Limit Interval Upper Limit

0.2102 0.3898

Z  2.58 1000 = 106.1583 = 2 e 2502 Z 2    (1–  ) 1.6452  (0.3)  (0.3) = 280.5749 n= = e2 (0.045)2

n=

2

2

2

0.2102    0.3898

2

Use n = 107 Use n = 281

If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 281) should be used.

8.60

(a) (b)

8.61

3.22    3.78

X t

S 3000 = 18000  1.9720  n 200

$17,581.68    $18,418.32

(a)

X t

S $9.22 = $21.34  1.9949  n 70

$19.14    $23.54

(b)

pZ

p(1– p) 0.3714(0.6286) = 0.3714  1.645  n 70

(d)

Z 2  2 1.96 2 10 2 n= = = 170.74 e2 1.5 2 Z 2    (1–  ) 1.6452  (0.5)  (0.5) = 334.08 n= = e2 (0.045)2

0.2764    0.4664 Use n = 171 Use n = 335

(e)

If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 335) should be used.

(a)

X t

(b) (c) (d)

8.63

0.2562    0.3638

(c)

(c)

8.62

p(1– p) 0.31(0.69) = 0.31  1.645  n 200 S 2 X t = 3.5  1.9720  n 200 pZ

S $7.26 = $38.54  2.0010  n 60

p(1– p) 0.30(0.70) = 0.30  1.645  n 60 2 2 2 2 Z  1.96 8 n= = = 109.27 2 e 1.5 2 Z 2    (1–  ) 1.6452  (0.5)  (0.5) = 422.82 n= = e2 (0.04)2 pZ

$36.66    $40.42 0.2027    0.3973 Use n = 110 Use n = 423

(e)

If a single sample were to be selected for both purposes, the larger of the two sample sizes (n = 423) should be used.

(a)

n=

Z 2    (1 −  ) 1.962  (0.5)  (0.5) = 384.16 = e2 (0.05)2

Use n = 385

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.

(b) (c) 8.64

(a)

pZ

p(1 − p) 0.84(0.16) = 0.84  1.96  n 50

0.7384    0.9416

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 accordingly perform a cost-benefit analysis to decide if he/she wants to sell the Ice Melt product.

Confidence Interval Estimate for the Proportion Data Sample Size Number of Successes Confidence Level

90 51 95%

Intermediate Calculations Sample Proportion 0.566666667 Z Value -1.9600 Standard Error of the Proportion 0.0522 Interval Half Width 0.1024 Confidence Interval Interval Lower Limit Interval Upper Limit

pZ

0.4643 0.6690

p(1 − p) 0.5667(1 − 0.5667) = 0.5667  1.96  n 90 0.4643    0.6690

(b)

Confidence Interval Estimate for the Mean Data Sample Standard Deviation Sample Mean Sample Size Confidence Level

1103.6491 563.38 51 95%

Intermediate Calculations Standard Error of the Mean 154.5417855 Degrees of Freedom 50 t Value 2.0086 Interval Half Width 310.4063 Confidence Interval Interval Lower Limit Interval Upper Limit

X t

252.97 873.79

S  1103.6491  = 563.38  2.0086   n 51  

$252.97    $873.79

8.67

5.46    5.54

X t

(b) (c)

Since 5.5 grams is within the 99% confidence interval, the company can claim that the mean weight of tea in a bag is 5.5 grams with a 99% level of confidence.

The assumption is valid as the weight of the tea bags is approximately normally distributed.

(a)

X t

(b)

(c)

With 95% confidence, the population mean 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.

The assumption is valid as the width of the troughs is approximately normally distributed.

(a)

X t

(b)

S 0.0461 = 8.4209  2.0106  n 49

8.41    8.43

S 34.713 = 3124.2147  1.9665  n 368 S 46.7443 X t = 3704.0424  1.9672  n 330

3120.66    3127.77 3698.98<  3709.10

(c) Normal Probability Plot 3300 3250

Boston

8.66

S 0.1058 = 5.5014  2.6800  n 50

(a)

3200 3150 3100 3050 3000 -4

-3

-2

-1

0

1

2

3

4

Z Value

Normal Probability Plot 3900 3850 3800

Vermont

8.65

3750 3700 3650 3600 3550 -4

-3

-2

-1

0

1

2

3

4

Z Value

(d)

The weight for Boston shingles is slightly skewed to the right while the weight for Vermont shingles appears to be slightly skewed to the left.

Since the two confidence intervals do not overlap, the mean weight of Vermont shingles is greater than the mean weight of Boston shingles.

(a) (b)

S 0.1424 = 0.2641  1.9741 n 170 S 0.1227 X t = 0.218  1.9772  n 140

0.2425    0.2856

X t

0.1975    0.2385

(c) Normal Probability Plot 0.9 0.8

Vermont

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -3

-2

-1

0

1

2

3

Z Value

Normal Probability Plot 1.2 1

Boston

8.68

0.8 0.6 0.4 0.2 0 -3

-2

-1

0

1

2

3

Z Value

(d)

The amount of granule loss for both brands are skewed to the right but the sample sizes are large enough so the violation of the normality assumption is not critical.

Because the two confidence intervals do not overlap, you can conclude that the mean granule loss of Boston shingles is higher than that of Vermont shingles