Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo

Menu
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 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
1.

two proportions

3.

matched or paired samples

5.

single mean

7.

independent group means, population standard deviations and/or variances unknown

9.

two proportions

11.

independent group means, population standard deviations and/or variances unknown

13.

independent group means, population standard deviations and/or variances unknown

15.

two proportions

17.

The random variable is the difference between the mean amounts of sugar in the two soft drinks.

19.

means

21.

two-tailed

23.

the difference between the mean life spans of White and non-White people

25.

This is a comparison of two population means with unknown population standard deviations.

27.

Check student’s solution.

28.
  1. Cannot accept the null hypothesis
  2. p-value < 0.05
  3. There is not enough evidence at the 5% level of significance to support the claim that life expectancy in the 1900s is different between White and non-White people.
31.

P′OS1 – P′OS2 = difference in the proportions of phones that had system failures within the first eight hours of operation with OS1 and OS2.

34.

proportions

36.

right-tailed

38.

The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota.

40.

Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test.

42.
  1. Cannot accept the null hypothesis.
  2. p-value < alpha
  3. At the 5% significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota.
44.

The difference in mean speeds of the fastball pitches of the two pitchers

46.

–2.46

47.

At the 1% significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s.

49.

Subscripts: 1 = Food, 2 = No Food
H0:μ1≤μ2H0:μ1≤μ2
Ha:μ1>μ2Ha:μ1>μ2

51.

Subscripts: 1 = Gamma, 2 = Zeta
H0:μ1=μ2H0:μ1=μ2
Ha:μ1≠μ2Ha:μ1≠μ2

53.

There is sufficient evidence so we cannot accept the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma.

54.

the mean difference of the system failures

56.

With a p-value 0.0067, we can cannot accept the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures.

60.

H0: μd ≥ 0

Ha: μd < 0

63.

We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective.

65.

Subscripts: 1: two-year colleges; 2: four-year colleges

  1. H0:μ1≥μ2H0:μ1≥μ2
  2. Ha:μ1<μ2Ha:μ1<μ2
  3. X ¯ 1 – X ¯ 2 X ¯ 1 – X ¯ 2 is the difference between the mean enrollments of the two-year colleges and the four-year colleges.
  4. Student’s-t
  5. test statistic: -0.2480
  6. p-value: 0.4019
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Cannot reject
    3. Reason for Decision: p-value > alpha
    4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges.
67.

Subscripts: 1: mechanical engineering; 2: electrical engineering

  1. H0:μ1≥μ2H0:μ1≥μ2
  2. Ha:μ1<μ2Ha:μ1<μ2
  3. X ¯ 1 − X ¯ 2 X ¯ 1 − X ¯ 2 is the difference between the mean entry level salaries of mechanical engineers and electrical engineers.
  4. t108
  5. test statistic: t = –0.82
  6. p-value: 0.2061
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Cannot 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 mean entry-level salaries of mechanical engineers is lower than that of electrical engineers.
69.
  1. H0:μ1=μ2H0:μ1=μ2
  2. Ha:μ1≠μ2Ha:μ1≠μ2
  3. X ¯ 1 − X ¯ 2 X ¯ 1 − X ¯ 2 is the difference between the mean times for completing a lap in races and in practices.
  4. t20.32
  5. test statistic: –4.70
  6. p-value: 0.0001
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Cannot accept the null hypothesis.
    3. Reason for Decision: p-value < alpha
    4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.
71.
  1. H0:μ1=μ2H0:μ1=μ2
  2. Ha:μ1≠μ2Ha:μ1≠μ2
  3. is the difference between the mean times for completing a lap in races and in practices.
  4. t40.94
  5. test statistic: –5.08
  6. p-value: zero
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Cannot accept the null hypothesis.
    3. Reason for Decision: p-value < alpha
    4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices.
74.

c

76.

Test: two independent sample means, population standard deviations unknown.

μ1μ1 = the mean price of a sociology text on the selected site.

μ2μ2 = the mean price of a math/science text on the selected site.

Random variable: X1¯-X1¯X1¯-X1¯ = the difference in the sample mean textbook price between sociology texts and math/science texts.

Hypotheses: H0 : Î¼1-μ2 = 0, Ha : Î¼1 - Î¼2 < Î¼2H0 : Î¼1-μ2 = 0, Ha : Î¼1 - Î¼2 < Î¼2 which can be expressed as H0s: Î¼1-μ2, Ha Î¼1 < Î¼2H0s: Î¼1-μ2, Ha Î¼1 < Î¼2.

Distribution for the test: Use tdftdf; because each sample has more than 30 observations, df=n1+n2-2=33+33-2=64df=n1+n2-2=33+33-2=64.

Estimate the critical value on the tt-table using the nearest available degrees of freedom, 60. The critical value, 2.660, is found in the .0005 column.

Calculate the test statistic: tc=(X¯1-X¯2)-0s12n2+s22n2=(74.64-111.56)-049.36233+66.90233=-2.55tc=(X¯1-X¯2)-0s12n2+s22n2=(74.64-111.56)-049.36233+66.90233=-2.55.

Using a calculator with tc=-2.55tc=-2.55 and df=64df=64, the left-tailed pp-value: Decision: Reject H0H0. Conclusion: At the 1% level of significance, from the sample data, there is sufficient evidence to conclude that the mean price of sociology textbooks is less than the mean price of textbooks for math/science.

78.

d

80.
  1. H0: PW = PB
  2. Ha: PW ≠ PB
  3. The random variable is the difference in the proportions of White and Black suicide victims, aged 15 to 24.
  4. normal for two proportions
  5. test statistic: –0.1944
  6. p-value: 0.8458
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Cannot accept 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 proportions of White and Black female suicide victims, aged 15 to 24, are different.
82.

Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College

  1. H0:p1=p2H0:p1=p2
  2. Ha:p1≠p2Ha:p1≠p2
  3. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College.
  4. normal for two proportions
  5. test statistic: 4.29
  6. p-value: 0.00002
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Cannot accept the null hypothesis.
    3. Reason for decision: p-value < alpha
    4. Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different.
84.

a

85.

Test: two independent sample proportions.

Random variable: p′1 - p′2

Distribution:
H0:p1=p2H0:p1=p2
Ha:p1≠p2Ha:p1≠p2

The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users.

Graph: two-tailed

87.

Test: two independent sample proportions

Random variable: p′1 − p′2

Distribution:

H 0 : p 1 = p 2 H 0 : p 1 = p 2
H a : p 1 > p 2 H a : p 1 > p 2

A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.

Graph: right-tailed

Do not reject the H0.

Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older.

89.

Subscripts: 1: men; 2: women

  1. H0:p1≤p2H0:p1≤p2
  2. Ha:p1>p2Ha:p1>p2
  3. P ′ 1 − P ′ 2 P ′ 1 − P ′ 2 is the difference between the proportions of men and women who enjoy shopping for electronic equipment.
  4. normal for two proportions
  5. test statistic: 0.22
  6. p-value: 0.4133
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Cannot 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 proportion of men who enjoy shopping for electronic equipment is more than the proportion of women.
91.
  1. H0:p1=p2H0:p1=p2
  2. Ha:p1≠p2Ha:p1≠p2
  3. P ′ 1 − P ′ 2 P ′ 1 − P ′ 2 is the difference between the proportions of men and women that have at least one pierced ear.
  4. normal for two proportions
  5. test statistic: –4.82
  6. p-value: zero
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Cannot accept the null hypothesis.
    3. Reason for Decision: p-value < alpha
    4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different.
92.
  1. H0: µd = 0
  2. Ha: µd > 0
  3. The random variable Xd is the mean difference in work times on days when eating breakfast and on days when not eating breakfast.
  4. t9
  5. test statistic: 4.8963
  6. p-value: 0.0004
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Cannot accept the null hypothesis.
    3. Reason for Decision: p-value < alpha
    4. Conclusion: At the 5% level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased.
94.

Subscripts: 1 = boys, 2 = girls

  1. H0:μ1≤μ2H0:μ1≤μ2
  2. Ha:μ1>μ2Ha:μ1>μ2
  3. The random variable is the difference in the mean auto insurance costs for boys and girls.
  4. normal
  5. test statistic: z = 2.50
  6. p-value: 0.0062
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Cannot accept the null hypothesis.
    3. Reason for Decision: p-value < alpha
    4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls.
96.

Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans

  1. H0:μ1≥μ2H0:μ1≥μ2
  2. Ha:μ1<μ2Ha:μ1<μ2
  3. The random variable is the difference in the mean miles per gallon of non-hybrid sedans and hybrid sedans.
  4. normal
  5. test statistic: 6.36
  6. p-value: 0
  7. Check student’s solution.
    1. Alpha: 0.05
    2. Decision: Cannot accept the null hypothesis.
    3. Reason for decision: p-value < alpha
    4. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans.
98.
  1. H0: µd = 0
  2. Ha: µd < 0
  3. The random variable Xd is the average difference between husband’s and wife’s satisfaction level.
  4. t9
  5. test statistic: t = –1.86
  6. p-value: 0.0479
  7. Check student’s solution
    1. Alpha: 0.05
    2. Decision: Cannot accept the null hypothesis, but run another test.
    3. Reason for Decision: p-value < alpha
    4. Conclusion: This is a weak test because alpha and the p-value are close. However, there is insufficient evidence to conclude that the mean difference is negative.
99.

p-value = 0.1494

At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks.

103.

Test: two matched pairs or paired samples (t-test)

Random variable: X – d X – d

Distribution: t12

H0: μd = 0 Ha: μd > 0

The mean of the differences of new female breast cancer cases in the south between 2013 and 2012 is greater than zero. The estimate for new female breast cancer cases in the south is higher in 2013 than in 2012.

Graph: right-tailed

p-value: 0.0004

Decision: Cannot accept H0

Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in 2013 than in 2012.

105.

Test: matched or paired samples (t-test)

Difference data: {–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, 1.8}

Random Variable: X – d X – d

Distribution: H0: μd = 0 Ha: μd < 0

The mean of the differences of the rate of underemployment in the northeastern states between 2012 and 2011 is less than zero. The underemployment rate went down from 2011 to 2012.

Graph: left-tailed.

Decision: Cannot reject H0.

Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from 2011 to 2012.

107.

e

109.

d

111.

f

113.

e

115.

f

117.

a

Order a print copy

As an Amazon Associate we earn from qualifying purchases.

Citation/Attribution

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/introductory-business-statistics/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/introductory-business-statistics/pages/1-introduction
Citation information

© Jun 23, 2022 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.