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  1. Preface
  2. 1 Sampling and Data
    1. Introduction
    2. 1.1 Definitions of Statistics, Probability, and Key Terms
    3. 1.2 Data, Sampling, and Variation in Data and Sampling
    4. 1.3 Levels of Measurement
    5. 1.4 Experimental Design and Ethics
    6. Key Terms
    7. Chapter Review
    8. Homework
    9. References
    10. Solutions
  3. 2 Descriptive Statistics
    1. Introduction
    2. 2.1 Display Data
    3. 2.2 Measures of the Location of the Data
    4. 2.3 Measures of the Center of the Data
    5. 2.4 Sigma Notation and Calculating the Arithmetic Mean
    6. 2.5 Geometric Mean
    7. 2.6 Skewness and the Mean, Median, and Mode
    8. 2.7 Measures of the Spread of the Data
    9. Key Terms
    10. Chapter Review
    11. Formula Review
    12. Practice
    13. Homework
    14. Bringing It Together: Homework
    15. References
    16. Solutions
  4. 3 Probability Topics
    1. Introduction
    2. 3.1 Terminology
    3. 3.2 Independent and Mutually Exclusive Events
    4. 3.3 Two Basic Rules of Probability
    5. 3.4 Contingency Tables and Probability Trees
    6. 3.5 Venn Diagrams
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Bringing It Together: Practice
    12. Homework
    13. Bringing It Together: Homework
    14. References
    15. Solutions
  5. 4 Discrete Random Variables
    1. Introduction
    2. 4.1 Hypergeometric Distribution
    3. 4.2 Binomial Distribution
    4. 4.3 Geometric Distribution
    5. 4.4 Poisson Distribution
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  6. 5 Continuous Random Variables
    1. Introduction
    2. 5.1 Properties of Continuous Probability Density Functions
    3. 5.2 The Uniform Distribution
    4. 5.3 The Exponential Distribution
    5. Key Terms
    6. Chapter Review
    7. Formula Review
    8. Practice
    9. Homework
    10. References
    11. Solutions
  7. 6 The Normal Distribution
    1. Introduction
    2. 6.1 The Standard Normal Distribution
    3. 6.2 Using the Normal Distribution
    4. 6.3 Estimating the Binomial with the Normal Distribution
    5. Key Terms
    6. Chapter Review
    7. Formula Review
    8. Practice
    9. Homework
    10. References
    11. Solutions
  8. 7 The Central Limit Theorem
    1. Introduction
    2. 7.1 The Central Limit Theorem for Sample Means
    3. 7.2 Using the Central Limit Theorem
    4. 7.3 The Central Limit Theorem for Proportions
    5. 7.4 Finite Population Correction Factor
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  9. 8 Confidence Intervals
    1. Introduction
    2. 8.1 A Confidence Interval for a Population Standard Deviation, Known or Large Sample Size
    3. 8.2 A Confidence Interval for a Population Standard Deviation Unknown, Small Sample Case
    4. 8.3 A Confidence Interval for A Population Proportion
    5. 8.4 Calculating the Sample Size n: Continuous and Binary Random Variables
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  10. 9 Hypothesis Testing with One Sample
    1. Introduction
    2. 9.1 Null and Alternative Hypotheses
    3. 9.2 Outcomes and the Type I and Type II Errors
    4. 9.3 Distribution Needed for Hypothesis Testing
    5. 9.4 Full Hypothesis Test Examples
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  11. 10 Hypothesis Testing with Two Samples
    1. Introduction
    2. 10.1 Comparing Two Independent Population Means
    3. 10.2 Cohen's Standards for Small, Medium, and Large Effect Sizes
    4. 10.3 Test for Differences in Means: Assuming Equal Population Variances
    5. 10.4 Comparing Two Independent Population Proportions
    6. 10.5 Two Population Means with Known Standard Deviations
    7. 10.6 Matched or Paired Samples
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. Bringing It Together: Homework
    14. References
    15. Solutions
  12. 11 The Chi-Square Distribution
    1. Introduction
    2. 11.1 Facts About the Chi-Square Distribution
    3. 11.2 Test of a Single Variance
    4. 11.3 Goodness-of-Fit Test
    5. 11.4 Test of Independence
    6. 11.5 Test for Homogeneity
    7. 11.6 Comparison of the Chi-Square Tests
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. Bringing It Together: Homework
    14. References
    15. Solutions
  13. 12 F Distribution and One-Way ANOVA
    1. Introduction
    2. 12.1 Test of Two Variances
    3. 12.2 One-Way ANOVA
    4. 12.3 The F Distribution and the F-Ratio
    5. 12.4 Facts About the F Distribution
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  14. 13 Linear Regression and Correlation
    1. Introduction
    2. 13.1 The Correlation Coefficient r
    3. 13.2 Testing the Significance of the Correlation Coefficient
    4. 13.3 Linear Equations
    5. 13.4 The Regression Equation
    6. 13.5 Interpretation of Regression Coefficients: Elasticity and Logarithmic Transformation
    7. 13.6 Predicting with a Regression Equation
    8. 13.7 How to Use Microsoft Excel® for Regression Analysis
    9. Key Terms
    10. Chapter Review
    11. Practice
    12. Solutions
  15. A | Statistical Tables
  16. B | Mathematical Phrases, Symbols, and Formulas
  17. Index
2.

X is the number of hours a patient waits in the emergency room before being called back to be examined. X - X - is the mean wait time of 70 patients in the emergency room.

4.

CI: (1.3808, 1.6192)

This is a normal distribution curve. The peak of the curve coincides with the point 1.5 on the horizontal axis.  A central region is shaded between points 1.38 and 1.62.
Figure 8.13

EBM = 0.12

6.
  1. x - x - = 151
  2. s x s x = 32
  3. n = 108
  4. n – 1 = 107
8.

X - X - is the mean number of hours spent watching television per month from a sample of 108 Americans.

10.

CI: (142.92, 159.08)

This is a normal distribution curve. The peak of the curve coincides with the point 151 on the horizontal axis.  A central region is shaded between points 142.92 and 159.08.
Figure 8.14


EBM = 8.08

12.
  1. 3.26
  2. 1.02
  3. 39
14.

μ

16.

t38

18.

0.025

20.

(2.93, 3.59)

22.

We are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors.

23.

The error bound would become EBM = 0.245. This error bound decreases because as sample sizes increase, variability decreases and we need less interval length to capture the true mean.

26.

It would decrease, because the z-score would decrease, which reducing the numerator and lowering the number.

28.

X is the number of “successes” where the woman makes the majority of the purchasing decisions for the household. P′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household.

30.

CI: (0.5321, 0.6679)

This is a normal distribution curve. The peak of the curve coincides with the point 0.6 on the horizontal axis.  A central region is shaded between points 0.5321 and 0.6679.
Figure 8.15

EBM: 0.0679

32.

X is the number of “successes” where an executive prefers a truck. P′ is the percentage of executives sampled who prefer a truck.

34.

CI: (0.19432, 0.33068)

This is a normal distribution curve. The peak of the curve coincides with the point 0.26 on the horizontal axis.  A central region is shaded between points 0.1943 and 0.3307.
Figure 8.16
36.

The sampling error means that the true mean can be 2% above or below the sample mean.

38.

P′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election.

40.

CI: (0.62735, 0.67265)

EBM: 0.02265

42.

The number of girls, ages 8 to 12, in the 5 P.M. Monday night beginning ice-skating class.

44.
  1. x = 64
  2. n = 80
  3. p′ = 0.8
46.

p

48.

P ~N(0.8, (0.8)(0.2) 80 ) P ~N(0.8, (0.8)(0.2) 80 ) . (0.72171, 0.87829).

50.

0.04

52.

(0.72; 0.88)

54.

With 92% confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72% and 88%.

56.

The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error.

58.
  1. 244
  2. 15
  3. 50
60.

N( 244, 15 50 ) N( 244, 15 50 )

62.

As the sample size increases, there will be less variability in the mean, so the interval size decreases.

64.

X is the time in minutes it takes to complete the U.S. Census short form. X - X - is the mean time it took a sample of 200 people to complete the U.S. Census short form.

66.

CI: (7.9441, 8.4559)

This is a normal distribution curve. The peak of the curve coincides with the point 8.2 on the horizontal axis.  A central region is shaded between points 7.94 and 8.46.
Figure 8.17
68.

The level of confidence would decrease because decreasing n makes the confidence interval wider, so at the same error bound, the confidence level decreases.

70.
  1. x - x - = 2.2
  2. σ = 0.2
  3. n = 20
72.

X - X - is the mean weight of a sample of 20 heads of lettuce.

74.

EBM = 0.07
CI: (2.1264, 2.2736)

This is a normal distribution curve. The peak of the curve coincides with the point 2.2 on the horizontal axis. A central region is shaded between points 2.13 and 2.27.
Figure 8.18
76.

The interval is greater because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals.

78.

The confidence level would increase.

80.

30.4

82.

σ

84.

μ

86.

normal

88.

0.025

90.

(24.52,36.28)

92.

We are 95% confident that the true mean age for Winger Foothill College students is between 24.52 and 36.28.

94.

The error bound for the mean would decrease because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean.

96.

2,185

98.

6,765

100.

595

103.
    1. 8629
    2. 6944
    3. 35
    4. 34
  1. t 34 t 34
    1. CI: (6244, 11,014)
    2. Figure 8.19
  2. It will become smaller
105.
    1. x - x - = 2.51
    2. s x s x = 0.318
    3. n = 9
    4. n - 1 = 8
  1. the effective length of time for a tranquilizer
  2. the mean effective length of time of tranquilizers from a sample of nine patients
  3. We need to use a Student’s-t distribution, because we do not know the population standard deviation.
    1. CI: (2.27, 2.76)
    2. Check student's solution.
  4. If we were to sample many groups of nine patients, 95% of the samples would contain the true population mean length of time.
107.

x - =$251,854.23 x - =$251,854.23

s= $521,130.41 s= $521,130.41

Note that we are not given the population standard deviation, only the standard deviation of the sample.

There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29

CL = 0.96, so α = 1 - CL = 1 - 0.96 = 0.04

α 2 =0.02 t α 2 = t 0.02 α 2 =0.02 t α 2 = t 0.02 = 2.150

EBM= t α 2 ( s n )=2.150( 521,130.41 30 ) ~ $204,561.66 EBM= t α 2 ( s n )=2.150( 521,130.41 30 ) ~ $204,561.66

x - x - - EBM = $251,854.23 - $204,561.66 = $47,292.57

x - x - + EBM = $251,854.23+ $204,561.66 = $456,415.89

We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89.

109.
    1. x - x - = 11.6
    2. s x s x = 4.1
    3. n = 225
    4. n - 1 = 224
  1. X is the number of unoccupied seats on a single flight. X - X - is the mean number of unoccupied seats from a sample of 225 flights.
  2. We will use a Student’s-t distribution, because we do not know the population standard deviation.
    1. CI: (11.12 , 12.08)
    2. Check student's solution.
111.
    1. CI: (7.64 , 9.36)
    2. Figure 8.20
  1. The sample should have been increased.
  2. Answers will vary.
  3. Answers will vary.
  4. Answers will vary.
113.

b

114.
  1. 1,068
  2. The sample size would need to be increased since the critical value increases as the confidence level increases.
116.
  1. X = the number of people who feel that the president is doing an acceptable job;

    P′ = the proportion of people in a sample who feel that the president is doing an acceptable job.

  2. N( 0.61, (0.61)(0.39) 1200 ) N( 0.61, (0.61)(0.39) 1200 )
    1. CI: (0.59, 0.63)
    2. Check student’s solution
118.
    1. (0.72, 0.82)
    2. (0.65, 0.76)
    3. (0.60, 0.72)
  1. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.
  2. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families.
  3. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a black person into their families.
120.
  1. X = the number of adult Americans who feel that crime is the main problem; P′ = the proportion of adult Americans who feel that crime is the main problem
  2. Since we are estimating a proportion, given P′ = 0.2 and n = 1000, the distribution we should use is N( 0.2, (0.2)(0.8) 1000 ) N( 0.2, (0.2)(0.8) 1000 ) .
    1. CI: (0.18, 0.22)
    2. Check student’s solution.
  3. One way to lower the sampling error is to increase the sample size.
  4. The stated “± 3%” represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3%. Thus, they estimate the percentage of adult Americans who feel that crime is the main problem to be between 18% and 22%.
122.

c

125.

a

127.
  1. p′ = (0.55 + 0.49) 2 (0.55 + 0.49) 2 = 0.52; EBP = 0.55 - 0.52 = 0.03
  2. No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this.
  3. CL = 0.75, so α = 1 – 0.75 = 0.25 and α 2 =0.125  z α 2 =1.150 α 2 =0.125  z α 2 =1.150 . (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.)
    EBP=(1.150) 0.52(0.48) 1,000 0.018 EBP=(1.150) 0.52(0.48) 1,000 0.018
    (p′ - EBP, p′ + EBP) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538)
  4. Yes – this interval does not fall less than 0.50 so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75% confidence.
130.
    1. 71
    2. 2.8
    3. 48
  1. X is the height of a male Swede, and x_x_ is the mean height from a sample of 48 male Swedes.
  2. Normal. We know the standard deviation for the population, and the sample size is greater than 30.
    1. CI: (70.151, 71.85)
    2. Figure 8.21
  3. The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean.
132.
    1. x-x- = 23.6
    2. σσ = 7
    3. n = 100
  1. X is the time needed to complete an individual tax form. X-X- is the mean time to complete tax forms from a sample of 100 customers.
  2. N( 23.6, 7 100 )N( 23.6, 7 100 ) because we know sigma.
    1. (22.228, 24.972)
    2. Figure 8.22
  3. It will need to change the sample size. The firm needs to determine what the confidence level should be, then apply the error bound formula to determine the necessary sample size.
  4. The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval.
  5. According to the error bound formula, the firm needs to survey 206 people. Since we increase the confidence level, we need to increase either our error bound or the sample size.
134.
    1. 7.9
    2. 2.5
    3. 20
  1. X is the number of letters a single camper will send home. X-X- is the mean number of letters sent home from a sample of 20 campers.
  2. N 7.9(2.5 20 )7.9(2.5 20 )

    1. CI: (6.98, 8.82)
    2. Figure 8.23
  3. The error bound and confidence interval will decrease.
136.
  1. x - x - = $568,873
  2. CL = 0.95 α = 1 – 0.95 = 0.05 z α 2 z α 2 = 1.96
    EBM = z 0.025 σ n z 0.025 σ n = 1.96 909200 40 909200 40 = $281,764
  3. x - x - EBM = 568,873 − 281,764 = 287,109
    x - x - + EBM = 568,873 + 281,764 = 850,637
  4. We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637.
139.

Higher

140.

It would increase to four times the prior value.

141.

No, It could have no affect if it were to change to 1 – p, for example. If it gets closer to 0.5 the minimum sample size would increase.

142.

Yes

143.
  1. No
  2. No
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