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Introductory Statistics

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Introductory StatisticsSolutions
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Table of contents
  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 Frequency, Frequency Tables, and Levels of Measurement
    5. 1.4 Experimental Design and Ethics
    6. 1.5 Data Collection Experiment
    7. 1.6 Sampling Experiment
    8. Key Terms
    9. Chapter Review
    10. Practice
    11. Homework
    12. Bringing It Together: Homework
    13. References
    14. Solutions
  3. 2 Descriptive Statistics
    1. Introduction
    2. 2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
    3. 2.2 Histograms, Frequency Polygons, and Time Series Graphs
    4. 2.3 Measures of the Location of the Data
    5. 2.4 Box Plots
    6. 2.5 Measures of the Center of the Data
    7. 2.6 Skewness and the Mean, Median, and Mode
    8. 2.7 Measures of the Spread of the Data
    9. 2.8 Descriptive Statistics
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. Bringing It Together: Homework
    16. References
    17. 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
    6. 3.5 Tree and Venn Diagrams
    7. 3.6 Probability Topics
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Bringing It Together: Practice
    13. Homework
    14. Bringing It Together: Homework
    15. References
    16. Solutions
  5. 4 Discrete Random Variables
    1. Introduction
    2. 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable
    3. 4.2 Mean or Expected Value and Standard Deviation
    4. 4.3 Binomial Distribution
    5. 4.4 Geometric Distribution
    6. 4.5 Hypergeometric Distribution
    7. 4.6 Poisson Distribution
    8. 4.7 Discrete Distribution (Playing Card Experiment)
    9. 4.8 Discrete Distribution (Lucky Dice Experiment)
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. References
    16. Solutions
  6. 5 Continuous Random Variables
    1. Introduction
    2. 5.1 Continuous Probability Functions
    3. 5.2 The Uniform Distribution
    4. 5.3 The Exponential Distribution
    5. 5.4 Continuous Distribution
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. 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 Normal Distribution (Lap Times)
    5. 6.4 Normal Distribution (Pinkie Length)
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  8. 7 The Central Limit Theorem
    1. Introduction
    2. 7.1 The Central Limit Theorem for Sample Means (Averages)
    3. 7.2 The Central Limit Theorem for Sums
    4. 7.3 Using the Central Limit Theorem
    5. 7.4 Central Limit Theorem (Pocket Change)
    6. 7.5 Central Limit Theorem (Cookie Recipes)
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. References
    13. Solutions
  9. 8 Confidence Intervals
    1. Introduction
    2. 8.1 A Single Population Mean using the Normal Distribution
    3. 8.2 A Single Population Mean using the Student t Distribution
    4. 8.3 A Population Proportion
    5. 8.4 Confidence Interval (Home Costs)
    6. 8.5 Confidence Interval (Place of Birth)
    7. 8.6 Confidence Interval (Women's Heights)
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. References
    14. 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 Rare Events, the Sample, Decision and Conclusion
    6. 9.5 Additional Information and Full Hypothesis Test Examples
    7. 9.6 Hypothesis Testing of a Single Mean and Single Proportion
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. References
    14. Solutions
  11. 10 Hypothesis Testing with Two Samples
    1. Introduction
    2. 10.1 Two Population Means with Unknown Standard Deviations
    3. 10.2 Two Population Means with Known Standard Deviations
    4. 10.3 Comparing Two Independent Population Proportions
    5. 10.4 Matched or Paired Samples
    6. 10.5 Hypothesis Testing for Two Means and Two Proportions
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. Bringing It Together: Homework
    13. References
    14. Solutions
  12. 11 The Chi-Square Distribution
    1. Introduction
    2. 11.1 Facts About the Chi-Square Distribution
    3. 11.2 Goodness-of-Fit Test
    4. 11.3 Test of Independence
    5. 11.4 Test for Homogeneity
    6. 11.5 Comparison of the Chi-Square Tests
    7. 11.6 Test of a Single Variance
    8. 11.7 Lab 1: Chi-Square Goodness-of-Fit
    9. 11.8 Lab 2: Chi-Square Test of Independence
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. Bringing It Together: Homework
    16. References
    17. Solutions
  13. 12 Linear Regression and Correlation
    1. Introduction
    2. 12.1 Linear Equations
    3. 12.2 Scatter Plots
    4. 12.3 The Regression Equation
    5. 12.4 Testing the Significance of the Correlation Coefficient
    6. 12.5 Prediction
    7. 12.6 Outliers
    8. 12.7 Regression (Distance from School)
    9. 12.8 Regression (Textbook Cost)
    10. 12.9 Regression (Fuel Efficiency)
    11. Key Terms
    12. Chapter Review
    13. Formula Review
    14. Practice
    15. Homework
    16. Bringing It Together: Homework
    17. References
    18. Solutions
  14. 13 F Distribution and One-Way ANOVA
    1. Introduction
    2. 13.1 One-Way ANOVA
    3. 13.2 The F Distribution and the F-Ratio
    4. 13.3 Facts About the F Distribution
    5. 13.4 Test of Two Variances
    6. 13.5 Lab: One-Way ANOVA
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. References
    13. Solutions
  15. A | Review Exercises (Ch 3-13)
  16. B | Practice Tests (1-4) and Final Exams
  17. C | Data Sets
  18. D | Group and Partner Projects
  19. E | Solution Sheets
  20. F | Mathematical Phrases, Symbols, and Formulas
  21. G | Notes for the TI-83, 83+, 84, 84+ Calculators
  22. H | Tables
  23. Index
1.

mean = 25 and standard deviation = 7.0711

3.

when the number of degrees of freedom is greater than 90

5.

df = 2

7.

a goodness-of-fit test

9.

3

11.

2.04

13.

We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.

15.

H0: the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County.

17.

right-tailed

19.

2016.136

21.

Graph: Check student’s solution.

Decision: Reject the null hypothesis.

Reason for the Decision: p-value < alpha

Conclusion (write out in complete sentences): The make-up of AIDS cases does not fit the ethnicities of the general population of Santa Clara County.

23.

a test of independence

25.

a test of independence

27.

8

29.

6.6

31.

0.0435

33.
Smoking Level Per Day African American Native Hawaiian Latino Japanese people White Totals
1-10 9,886 2,745 12,831 8,378 7,650 41,490
11-20 6,514 3,062 4,932 10,680 9,877 35,065
21-30 1,671 1,419 1,406 4,715 6,062 15,273
31+ 759 788 800 2,305 3,970 8,622
Totals 18,830 8,014 19,969 26,078 27,559 10,0450
Table 11.60
35.
Smoking Level Per Day African American Native Hawaiian Latino Japanese people White
1-10 7777.57 3310.11 8248.02 10771.29 11383.01
11-20 6573.16 2797.52 6970.76 9103.29 9620.27
21-30 2863.02 1218.49 3036.20 3965.05 4190.23
31+ 1616.25 687.87 1714.01 2238.37 2365.49
Table 11.61
37.

10,301.8

39.

right

41.
  1. Reject the null hypothesis.
  2. p-value < alpha
  3. There is sufficient evidence to conclude that smoking level is dependent on ethnic group.
43.

test for homogeneity

45.

test for homogeneity

47.

All values in the table must be greater than or equal to five.

49.

3

51.

0.00005

53.

a goodness-of-fit test

55.

a test for independence

57.

Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way ∑ (ij) (O-E) 2 E ∑ (ij) (O-E) 2 E . In addition, all values must be greater than or equal to five.

59.

a test of a single variance

61.

a left-tailed test

63.

H0: σ2 = 0.812;

Ha: σ2 > 0.812

65.

a test of a single variance

67.

0.0542

69.

true

71.

false

73.
Marital Status Percent Expected Frequency
never married 31.3 125.2
married 56.1 224.4
widowed 2.5 10
divorced/separated 10.1 40.4
Table 11.62
  1. The data fits the distribution.
  2. The data does not fit the distribution.
  3. 3
  4. chi-square distribution with df = 3
  5. 19.27
  6. 0.0002
  7. Check student’s solution.
    1. Alpha = 0.05
    2. Decision: Reject null
    3. Reason for decision: p-value < alpha
    4. Conclusion: Data does not fit the distribution.
75.
  1. H0: The local results follow the distribution of the U.S. AP examinee population
  2. Ha: The local results do not follow the distribution of the U.S. AP examinee population
  3. df = 5
  4. chi-square distribution with df = 5
  5. chi-square test statistic = 13.4
  6. p-value = 0.0199
  7. Check student’s solution.
    1. Alpha = 0.05
    2. Decision: Reject null when a = 0.05
    3. Reason for Decision: p-value < alpha
    4. Conclusion: Local data do not fit the AP Examinee Distribution.
    5. Decision: Do not reject null when a = 0.01
    6. Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution.
77.
  1. H0: The actual college majors of graduating females fit the distribution of their expected majors
  2. Ha: The actual college majors of graduating females do not fit the distribution of their expected majors
  3. df = 10
  4. chi-square distribution with df = 10
  5. test statistic = 11.48
  6. p-value = 0.3211
  7. Check student’s solution.
    1. Alpha = 0.05
    2. Decision: Do not reject null when a = 0.05 and a = 0.01
    3. Reason for decision: p-value > alpha
    4. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females do not fit the distribution of their expected majors.
79.

true

81.

true

83.

false

85.

The hypotheses for the goodness-of-fit test are:

  1. H0: Surveyed obese fit the distribution of expected obese
  2. Ha: Surveyed obese do not fit the distribution of expected obese

Use a chi-square distribution with df = 4 to evaluate the data.

The test statistic is X2 = 9.85

The P-value = 0.0431

At 5% significance level, α = 0.05. For this data, P < α. Reject the null hypothesis.

At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese.

87.
  1. H0: Car size is independent of family size.
  2. Ha: Car size is dependent on family size.
  3. df = 9
  4. chi-square distribution with df = 9
  5. test statistic = 15.8284
  6. p-value = 0.0706
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject the null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent.
89.
  1. H0: Honeymoon locations are independent of bride’s age.
  2. Ha: Honeymoon locations are dependent on bride’s age.
  3. df = 9
  4. chi-square distribution with df = 9
  5. test statistic = 15.7027
  6. p-value = 0.0734
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject the null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent.
91.
  1. H0: The types of fries sold are independent of the location.
  2. Ha: The types of fries sold are dependent on the location.
  3. df = 6
  4. chi-square distribution with df = 6
  5. test statistic =18.8369
  6. p-value = 0.0044
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Reject the null hypothesis.
    3. Reason for decision: p-value < alpha
    4. Conclusion: At the 5% significance level, There is sufficient evidence that types of fries and location are dependent.
93.
  1. H0: Salary is independent of level of education.
  2. Ha: Salary is dependent on level of education.
  3. df = 12
  4. chi-square distribution with df = 12
  5. test statistic = 255.7704
  6. p-value = 0
  7. Check student’s solution.

  8. Alpha: 0.05

    Decision: Reject the null hypothesis.

    Reason for decision: p-value < alpha

    Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent.

95.

true

97.

true

99.
  1. H0: Age is independent of the youngest online entrepreneurs’ net worth.
  2. Ha: Age is dependent on the net worth of the youngest online entrepreneurs.
  3. df = 2
  4. chi-square distribution with df = 2
  5. test statistic = 1.76
  6. p-value 0.4144
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject the null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that age and net worth for the youngest online entrepreneurs are dependent.
101.
  1. H0: The distribution for personality types is the same for both majors
  2. Ha: The distribution for personality types is not the same for both majors
  3. df = 4
  4. chi-square with df = 4
  5. test statistic = 3.01
  6. p-value = 0.5568
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject the null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors.
103.
  1. H0: The distribution for fish caught is the same in Green Valley Lake and in Echo Lake.
  2. Ha: The distribution for fish caught is not the same in Green Valley Lake and in Echo Lake.
  3. 3
  4. chi-square with df = 3
  5. 11.75
  6. p-value = 0.0083
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Reject the null hypothesis.
    3. Reason for decision: p-value < alpha
    4. Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake
105.
  1. H0: The distribution of average energy use in the USA is the same as in Europe between 2005 and 2010.
  2. Ha: The distribution of average energy use in the USA is not the same as in Europe between 2005 and 2010.
  3. df = 4
  4. chi-square with df = 4
  5. test statistic = 2.7434
  6. p-value = 0.7395
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Do not reject the null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the average energy use values in the US and EU are not derived from different distributions for the period from 2005 to 2010.
107.
  1. H0: The distribution for technology use is the same for community college students and university students.
  2. Ha: The distribution for technology use is not the same for community college students and university students.
  3. 2
  4. chi-square with df = 2
  5. 7.05
  6. p-value = 0.0294
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Reject the null hypothesis.
    3. Reason for decision: p-value < alpha
    4. Conclusion: There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities.
110.

225

112.

H0: σ2 ≤ 150

114.

36

116.

Check student’s solution.

118.

The claim is that the variance is no more than 150 minutes.

120.

a Student's t- or normal distribution

122.
  1. H0: σ = 15
  2. Ha: σ > 15
  3. df = 42
  4. chi-square with df = 42
  5. test statistic = 26.88
  6. p-value = 0.9663
  7. Check student’s solution.
    1. Alpha = 0.05
    2. Decision: Do not reject null hypothesis.
    3. Reason for decision: p-value > alpha
    4. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15.
124.
  1. H0: σ ≤ 3
  2. Ha: σ > 3
  3. df = 17
  4. chi-square distribution with df = 17
  5. test statistic = 28.73
  6. p-value = 0.0371
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Reject the null hypothesis.
    3. Reason for decision: p-value < alpha
    4. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three.
126.
  1. H0: σ = 2
  2. Ha: σ ≠ 2
  3. df = 14
  4. chi-square distiribution with df = 14
  5. chi-square test statistic = 5.2094
  6. p-value = 0.0346
  7. Check student’s solution.
    1. Alpha = 0.05
    2. Decision: Reject the null hypothesis
    3. Reason for decision: p-value < alpha
    4. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2.
128.

The sample standard deviation is $34.29.

H0 : σ2 = 252
 Ha : σ2 > 252


df = n – 1 = 7.


test statistic: x 2 =  x 7 2 =  (n–1) s 2 25 2 =  (8–1) (34.29) 2 25 2 =13.169 x 2 =  x 7 2 =  (n–1) s 2 25 2 =  (8–1) (34.29) 2 25 2 =13.169 ;


p-value: P( x 7 2 >13.169 )=1–P( x 7 2  ≤13.169 )=0.0681 P( x 7 2 >13.169 )=1–P( x 7 2  ≤13.169 )=0.0681


Alpha: 0.05


Decision: Do not reject the null hypothesis.


Reason for decision: p-value > alpha


Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625.

130.
  1. The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected.
  2. Testing to see if the data fits the distribution “too well” or is too perfect.
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