Elementary Statistics: A Step By Step Approach
Elementary Statistics: A Step By Step Approach
10th Edition
ISBN: 9781259755330
Author: Allan G. Bluman
Publisher: McGraw-Hill Education
bartleby

Videos

Question
Book Icon
Chapter 12, Problem 14CQ
To determine

To check: Whether there is sufficient evidence to conclude a difference in mean prices.

To perform: The appropriate test to find out where the difference exists if the there is a difference in mean prices.

Expert Solution & Answer
Check Mark

Answer to Problem 14CQ

Yes, there is sufficient evidence to conclude a difference in means.

There is a significant difference between the means X¯1 and X¯3 , X¯2 and X¯3 .

Explanation of Solution

Given info:

The table shows the prices of four different bottles of nationwide brands. The level of significance is 0.05.

Calculation:

The hypotheses are given below:

Null hypothesis:

H0:μ1=μ2=μ3

Alternative hypothesis:

H1: At least one mean is different from the others

Here, at least one mean is different from the others is tested. Hence, the claim is that, at least one mean is different from the others.

The level of significance is 0.05. The number of samples k is 3, the sample sizes n1 , n2 and n3 are 4, 4 and 4.

The degrees of freedom are d.f.N=k1 and d.f.D=Nk .

Where

N=n1+n2+n3=4+4+4=12

Substitute 3 for k in d.f.N

d.f.N=k1 =3-1=2

Substitute 12 for N and 3 for k in d.f.D

d.f.D=Nk=123=9

Critical value:

The critical F-value is obtained using the Table H: The F-Distribution with the level of significance α=0.05 .

Procedure:

  • Locate 9 in the degrees of freedom, denominator row of the Table H.
  • Obtain the value in the corresponding degrees of freedom, numerator column below 2.

That is, the critical value is 4.26.

Rejection region:

The null hypothesis would be rejected if F>4.26 .

Software procedure:

Step-by-step procedure to obtain thetest statistic using the MINITAB software:

  • Choose Stat > ANOVA > One-Way.
  • In Response, enter the Prices.
  • In Factor, enter the Factor.
  • Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 12, Problem 14CQ , additional homework tip  1

From the MINITAB output, the test value F is 10.03.

Conclusion:

From the results, the test value is 10.03.

Here, the F-statistic value is greater than the critical value.

That is, 10.03>4.26 .

Thus, it can be concluding that, the null hypothesis is rejected.

Hence, the result concludes that, there is sufficient evidence to conclude a difference in means.

Consider, X¯1 , X¯2 and X¯3 represents the means of Brand X, Brand Y and Store brand, s12,s22 and s32 represents the variances of samples of Brand X, Brand Y and Store brand.

Step-by-step procedure to obtain the test mean and standard deviation using the MINITAB software:

  • Choose Stat > Basic Statistics > Display Descriptive Statistics.
  • In Variables enter the columns Brand X, Brand Y and Store brand.
  • Choose option statistics, and select Mean, Variance and N total.
  • Click OK.

Output using the MINITAB software is given below:

Elementary Statistics: A Step By Step Approach, Chapter 12, Problem 14CQ , additional homework tip  2

The sample sizes n1,n2 and n3 are 4, 4 and 4.

The means are X¯1,X¯2 and X¯3 are 7.015, 7.640 and 4.690.

The sample variances are s12,s22 and s32 are 1.269, 1.297 and 0.327.

Here, the samples of sizes of three states are equal. So, the test used here is Tukey test.

Tukey test:

Critical value:

Here, k is 3 and degrees of freedom v=Nk

Substitute 12 for N and 3 for k in v

v=Nk=123=9

The critical F-value is obtained using the Table N: Critical Values for the Tukey test with the level of significance α=0.05 .

Procedure:

  • Locate 9 in the column of v of the Table H.
  • Obtain the value in the corresponding row below 3.

That is, the critical value is 3.95.

Comparison of the means:

The formula for finding sW2 is,

sW2=(ni1)si2(ni1)

That is,

sW2=(41)1.269+(41)1.297+(41)0.327(41)+(41)+(41)=3.807+3.891+0.9819=8.6799=0.964

Comparison between the means X¯1 and X¯2 :

The hypotheses are given below:

Null hypothesis:

H0: There is no significant difference between X¯1 and X¯2 .

Alternative hypothesis:

H1: There is significant difference between X¯1 and X¯2 .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

Absolute value:

The formula for comparing the means X¯1 and X¯2 is,

q1=X¯1X¯2sW2n

Substitute 7.015 and 7.640 for X¯1 and X¯2 and 0.964 for sW2

q1=7.0157.6400.9644=0.6250.241=0.6250.491=1.27

Thus, the value of q1 is –1.27.

Hence, the absolute value of q1 is 1.27.

Conclusion:

The absolute value is 1.27.

Here, the absolute value is lesser than the critical value.

That is, 1.27<3.95 .

Thus, the null hypothesis is not rejected.

Hence, there is significant difference between the means X¯1 and X¯2 .

Comparison between the means X¯1 and X¯3 :

The hypotheses are given below:

Null hypothesis:

H0: There is no significant difference between X¯1 and X¯3 .

Alternative hypothesis:

H1: There is significant difference between X¯1 and X¯3 .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

Absolute value:

The formula for comparing the means X¯1 and X¯3 is,

q2=X¯1X¯3sW2n

Substitute 7.015 and 4.690 for X¯1 and X¯3 and 0.964 for sW2

q2=7.0154.6900.9644=2.3250.241=0.6250.491=4.74

Thus, the value of q2 is 4.74.

Hence, the absolute value of q2 is 4.74.

Conclusion:

The absolute value is 4.74.

Here, the absolute value is greater than the critical value.

That is, 4.74>3.95 .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means X¯1 and X¯3 .

Comparison between the means X¯2 and X¯3 :

The hypotheses are given below:

Null hypothesis:

H0: There is no significant difference between X¯2 and X¯3 .

Alternative hypothesis:

H1: There is significant difference between X¯2 and X¯3 .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

Absolute value:

The formula for comparing the means X¯2 and X¯3 is,

q3=X¯2X¯3sW2n

Substitute 7.640 and 4.690 for X¯2 and X¯3 and 0.964 for sW2

q3=7.6404.6900.9644=2.950.241=2.950.491=6.01

Thus, the value of q3 is 6.01.

Hence, the absolute value of q3 is 6.01.

Conclusion:

The absolute value is 6.01.

Here, the absolute value is greater than the critical value.

That is, 6.01>3.95 .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means X¯2 and X¯3 .

Justification:

From the results, it can be observed that there is a significant difference between the means X¯1 and X¯3 , X¯2 and X¯3 .

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!

Chapter 12 Solutions

Elementary Statistics: A Step By Step Approach

Ch. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.1 - Prob. 11ECh. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.1 - Prob. 15ECh. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.2 - Colors That Make You Smarter The following set of...Ch. 12.2 - What two tests can be used to compare two means...Ch. 12.2 - Explain the difference between the two tests used...Ch. 12.2 - For Exercises 3 through 8, the null hypothesis was...Ch. 12.2 - For Exercises 3 through 8, the null hypothesis was...Ch. 12.2 - For Exercises 3 through 8, the null hypothesis was...Ch. 12.2 - For Exercises 3 through 8, the null hypothesis was...Ch. 12.2 - For Exercises 3 through 8, the null hypothesis was...Ch. 12.2 - For Exercises 3 through 8, the null hypothesis was...Ch. 12.2 - For Exercises 9 through 13, do a complete one-way...Ch. 12.2 - Prob. 10ECh. 12.2 - For Exercises 9 through 13, do a complete one-way...Ch. 12.2 - For Exercises 9 through 13, do a complete one-way...Ch. 12.2 - Prob. 13ECh. 12.2 - Prob. 14ECh. 12.3 - Automobile Sales Techniques The following outputs...Ch. 12.3 - Prob. 1ECh. 12.3 - Explain what is meant by main effects and...Ch. 12.3 - Prob. 3ECh. 12.3 - How are the F test values computed?Ch. 12.3 - Prob. 5ECh. 12.3 - In a two-way ANOVA, variable A has six levels and...Ch. 12.3 - Prob. 7ECh. 12.3 - When can the main effects for the two-way ANOVA be...Ch. 12.3 - Prob. 9ECh. 12.3 - For Exercises 9 through 15, perform these steps....Ch. 12.3 - For Exercises 9 through 15, perform these steps....Ch. 12.3 - For Exercises 9 through 15, perform these steps....Ch. 12.3 - Prob. 13ECh. 12.3 - For Exercises 9 through 15, perform these steps....Ch. 12.3 - Prob. 15ECh. 12 - If the null hypothesis is rejected in Exercises 1...Ch. 12 - Prob. 12.1.2RECh. 12 - Prob. 12.1.3RECh. 12 - Prob. 12.1.4RECh. 12 - Prob. 12.1.5RECh. 12 - If the null hypothesis is rejected in Exercises 1...Ch. 12 - Prob. 12.1.7RECh. 12 - Prob. 12.1.8RECh. 12 - Review Preparation for Statistics A statistics...Ch. 12 - Effects of Different Types of Diets A medical...Ch. 12 - Prob. 1DACh. 12 - Prob. 2DACh. 12 - Prob. 3DACh. 12 - Prob. 1CQCh. 12 - Prob. 2CQCh. 12 - Prob. 3CQCh. 12 - Determine whether each statement is true or false....Ch. 12 - Prob. 5CQCh. 12 - Prob. 6CQCh. 12 - Prob. 7CQCh. 12 - Prob. 8CQCh. 12 - Complete the following statements with the best...Ch. 12 - Prob. 10CQCh. 12 - For Exercises 11 through 17, use the traditional...Ch. 12 - Prob. 12CQCh. 12 - Prob. 13CQCh. 12 - Prob. 14CQCh. 12 - Prob. 15CQCh. 12 - Prob. 16CQCh. 12 - For Exercises 11 through 17, use the traditional...Ch. 12 - Shown here are the abstract and two tables from a...Ch. 12 - Shown here are the abstract and two tables from a...Ch. 12 - Prob. 3CTCCh. 12 - Prob. 4CTCCh. 12 - Prob. 5CTCCh. 12 - Prob. 6CTCCh. 12 - Prob. 7CTCCh. 12 - Adult Children of Alcoholics Shown here are the...Ch. 12 - Prob. 9CTCCh. 12 - Prob. 10CTCCh. 12 - Prob. 11CTC
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.
Recommended textbooks for you
Text book image
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Hypothesis Testing using Confidence Interval Approach; Author: BUM2413 Applied Statistics UMP;https://www.youtube.com/watch?v=Hq1l3e9pLyY;License: Standard YouTube License, CC-BY
Hypothesis Testing - Difference of Two Means - Student's -Distribution & Normal Distribution; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=UcZwyzwWU7o;License: Standard Youtube License