Applied Statistics in Business and Economics
Applied Statistics in Business and Economics
5th Edition
ISBN: 9780077837303
Author: David Doane, Lori Seward Senior Instructor of Operations Management
Publisher: McGraw-Hill Education
bartleby

Videos

Question
Book Icon
Chapter 15, Problem 30CE
To determine

State the null and alternative hypothesis.

Find the degrees of freedom for the contingency table.

Find the critical value of chi-square from Appendix E or from Excel’s function.

Calculate the chi-square test statistics at 0.05 level of significance.

Interpret the p-value.

Check whether the conclusion is sensitive to the level of significance chosen, identify the cells that contribute to the chi-square test statistic and check for the small expected frequencies.

Perform a two-tailed, two-sample z test for π1=π2 and verify that z2 is the same as the chi-square statistics.

Expert Solution & Answer
Check Mark

Answer to Problem 30CE

The null hypothesis is:

H0: Correct response and type of cola are independent.

And the alternative hypothesis is:

H1: Correct response and type of cola are not independent.

The degrees of freedom for the contingency table is 1.

The critical-value using EXCEL is 3.841.

The chi-square test statistics at 0.05 level of significance is 1.50.

The p-value for the hypothesis test is 0.221.

There is enough evidence to conclude that the correct response and type of cola are independent.

The conclusion is not sensitive to the level of significance chosen.

The cells (1, 1) and (1, 2) contribute the most to the chi-square test statistic.

There is no expected frequencies that are too small.

It is verified that z2 is the same as the chi-square statistics.

Explanation of Solution

The table summarizes the grade and order of papers handed in.

The claim is to test whether the data provide sufficient evidence to conclude that the grade and order handed in are independent. If the claim is rejected, then the grade and order handed in are not independent.

The test hypotheses are given below:

Null hypothesis:

H0: Grade and order handed in are independent.

Alternative hypothesis:

H1: Grade and order handed in are not independent.

The degrees of freedom can be obtained as follows:

d.f=(r1)(c1)

Substitute 2 for r and 2 for c.

d.f =(21)(21)=(1)(1)=1

Thus, the degrees of freedom for the contingency table is 1.

Procedure for critical-value using EXCEL:

Step-by-step software procedure to obtain critical-value using EXCEL software is as follows:

  • Open an EXCEL file.
  • In cell A1, enter the formula “=CHISQ.INV.RT(0.05,1)”
  • Output using EXCEL software is given below:

Applied Statistics in Business and Economics, Chapter 15, Problem 30CE , additional homework tip  1

Thus, the critical-value using EXCEL is 3.841.

Test statistic:

Software procedure:

Step by step procedure to obtain the chi-square test statistics and p-value using the MINITAB software:

  • Choose Stat > Tables >Cross Tabulation and Chi-Square.
  • Choose Row data (categorical variables).
  • In Rows, choose Grade.
  • In Columns, choose Order handed in.
  • In Frequencies, choose Count.
  • In Display, select Counts.
  • In chi-square, select Chi-square test, Expected cell counts and Each cell’s contribution to chi-square.
  • Click OK.

Output using the MINITAB software is given below:

Applied Statistics in Business and Economics, Chapter 15, Problem 30CE , additional homework tip  2

Thus, the test statistic is 1.50 and the p-value for the hypothesis test is 0.221.

  Rejection rule:

If the p-value is less than or equal to the significance level, then reject the null hypothesis H0. Otherwise, do not reject H0.

Conclusion:

Here, the p-value is greater than the level of significance.

That is, p-value(=0.221)>α(=0.05).

Therefore, the null hypothesis is not rejected.

Thus, the data provide sufficient evidence to conclude that the correct response and type of cola are independent.

Take α=0.01

Here, the p-value is greater than the level of significance.

That is, p-value(=0.221)>α(=0.01).

Therefore, the null hypothesis is not rejected.

Thus, the data provide sufficient evidence to conclude that the correct response and type of cola are independent.

Thus, the conclusion is same for both the significance levels.

Hence, the conclusion is not sensitive to the level of significance chosen.

The cells (1, 1) and (1, 2) contribute the most to the chi-square test statistic.

Since all ej2, there is no expected frequencies that are too small.

Two-tailed, two-sample z test:

The test hypotheses are given below:

Null hypothesis:

H0:π1π2=0

Alternative hypothesis:

H1:π1π20

The proportion of “yes” responses to the regular cola is denoted as p1 and is obtained as follows:

p1=x1n1=624=0.25

Where x1 is the number of “yes” responses to the regular cola, n1 is the number of response for regular cola.

The proportion of number of “yes” responses to the diet cola is denoted as p2 and is obtained as follows:

p2=x2n1=1024=0.4167

Where x2 is the number of “yes” responses to the diet cola and n2 is the number of response for diet cola.

The pooled proportion is denoted as p¯ obtained as follows:

p¯=x1+x2n1+n2=6+1024+24=1648=0.333

Test statistic:

The z-test statistics can be obtained as follows:

zcalc=p1p2p¯(1p¯)1n1+1n2=0.250.4167(0.333)(10.333)124+124=0.1667(0.333)(0.667)(0.083)=0.1667(0.018435213)=0.16670.135776334=1.22

Thus, the z-test statistic is –1.22.

The square of the z-test statistic is,

z2=(1.22)×(1.22)=1.48841.5

Thus the square of the z-test statistic is same as the chi-square statistics.

Procedure for p-value using EXCEL:

Step-by-step software procedure to obtain p-value using EXCEL software is as follows:

  • Open an EXCEL file.
  • In cell A1, enter the formula “=2*(1-NORM.S.DIST(–1.22,1))”
  • Output using EXCEL software is given below:

Applied Statistics in Business and Economics, Chapter 15, Problem 30CE , additional homework tip  3

Thus, the p-value using EXCEL is 1.778, which is not same as the p-value obtained in chi-square test. But the square of the z-test statistic is same as the chi-square statistics.

Thus, it is verified that z2 is the same as the chi-square statistics.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Using the accompanying Home Market Value data and associated regression​ line, Market ValueMarket Valueequals=​$28,416+​$37.066×Square ​Feet, compute the errors associated with each observation using the formula e Subscript ieiequals=Upper Y Subscript iYiminus−ModifyingAbove Upper Y with caret Subscript iYi and construct a frequency distribution and histogram.   LOADING... Click the icon to view the Home Market Value data.       Question content area bottom Part 1 Construct a frequency distribution of the​ errors, e Subscript iei.   ​(Type whole​ numbers.) Error Frequency minus−15 comma 00015,000less than< e Subscript iei less than or equals≤minus−10 comma 00010,000 0   minus−10 comma 00010,000less than< e Subscript iei less than or equals≤minus−50005000 5   minus−50005000less than< e Subscript iei less than or equals≤0 21   0less than< e Subscript iei less than or equals≤50005000 9…
The managing director of a consulting group has the accompanying monthly data on total overhead costs and professional labor hours to bill to clients. Complete parts a through c   Overhead Costs    Billable Hours345000    3000385000    4000410000    5000462000    6000530000    7000545000    8000
Using the accompanying Home Market Value data and associated regression​ line, Market ValueMarket Valueequals=​$28,416plus+​$37.066×Square ​Feet, compute the errors associated with each observation using the formula e Subscript ieiequals=Upper Y Subscript iYiminus−ModifyingAbove Upper Y with caret Subscript iYi and construct a frequency distribution and histogram. Square Feet    Market Value1813    911001916    1043001842    934001814    909001836    1020002030    1085001731    877001852    960001793    893001665    884001852    1009001619    967001690    876002370    1139002373    1131001666    875002122    1161001619    946001729    863001667    871001522    833001484    798001589    814001600    871001484    825001483    787001522    877001703    942001485    820001468    881001519    882001518    885001483    765001522    844001668    909001587    810001782    912001483    812001519    1007001522    872001684    966001581    86200

Chapter 15 Solutions

Applied Statistics in Business and Economics

Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt
Text book image
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Text book image
Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Type I and II Errors, Power, Effect Size, Significance and Power Analysis in Quantitative Research; Author: NurseKillam;https://www.youtube.com/watch?v=OWn3Ko1WYTA;License: Standard YouTube License, CC-BY