Loose-leaf Version for The Basic Practice of Statistics 7e & LaunchPad (Twelve Month Access)
Loose-leaf Version for The Basic Practice of Statistics 7e & LaunchPad (Twelve Month Access)
7th Edition
ISBN: 9781319019334
Author: David S. Moore, William I. Notz, Michael A. Fligner
Publisher: W. H. Freeman
Question
Book Icon
Chapter 30, Problem 30.28E

a.

To determine

The confidence interval that capture the true differences between the pairs of population means.

a.

Expert Solution
Check Mark

Answer to Problem 30.28E

All the confidence interval captures the true difference in between the length of pairs of population means of three varieties of the tropical flower Heliconia.

Explanation of Solution

Given info:

The confidence intervals are given below;

6.752to9.021forμbehaiμred10.165to12.670forμbehaiμyellow2.375to4.688forμredμyellow

Calculation:

Consider the first confidence interval that is given as;

6.752to9.021forμbehaiμred

The null hypothesis is,

H0:μbehaiμred=0

The alternative hypothesis is,

Ha:μbehaiμred0

A 95% confidence interval whose lower limit is 6.752 and the upper limit is 9.021 for the mean length of the bihai flowers μbihai and red flowers μred and 0 does not lie inside this confidence interval. So, the null hypothesis is rejected in this situation and accepts the alternative hypothesis H0:μbehaiμred0 in favor of two sided at 5% level of significance level.

Consider the second confidence interval that is given as;

10.165to12.670forμbehaiμyellow

The null hypothesis is,

H0:μbehaiμyellow=0

The alternative hypothesis is,

Ha:μbehaiμyellow0

A 95% confidence interval whose lower limit is 10.165 and the upper limit is 12.670 for the mean length of the bihai flowers μbihai and red flowers μyellow and 0 does not lie inside this confidence interval. So, the null hypothesis is rejected in this case and accepts the alternative hypothesis H0:μbehaiμyellow0 in favor of two sided at 5% level of significance level.

Consider the third confidence interval that is given as;

2.375to4.688forμredμyellow

The null hypothesis is,

H0:μredμyellow=0

The alternative hypothesis is,

Ha:μredμyellow0

A 95% confidence interval whose lower limit is 2.375 and the upper limit is 4.688 for the mean length of the bihai flowers μred and red flowers μyellow and 0 does not lie inside this confidence interval. So, the null hypothesis is rejected in this case and accepts the alternative hypothesis H0:μredμyellow0 in favor of two sided at 5% level of significance level.

b.

To determine

The short summary of the result of the ANOVA which include the multiple comparisons.

b.

Expert Solution
Check Mark

Answer to Problem 30.28E

There are highly significance difference in between the mean length of the flowers varieties fertilized by different Hummingbird species should have different distribution of length. The one-way ANOVA is used by comparing the mean length of the population. But the characteristic of the data that makes ANOVA is risky is lack of normality condition or groups do not follow normal distribution.

Explanation of Solution

Given info:

The confidence intervals are given below;

6.752 to 9.021 for μbehaiμred , 10.165 to 12.670 for μbehaiμyellow and 2.375 to 4.688 for μredμyellow .

Software procedure:

Step by step procedure to obtain one way ANOVA using the MINITAB software:

  • Choose Stat > ANOVA > One way.
  • In Response, select the column of Length.
  • In Factor, select the column of Variety.
  • Choose > Comparisons > Tukey in comparison procedures assuming equal variances and tick on Test in Results and then OK.
  • Click OK.

The Minitab One-way ANOVA output is,

Loose-leaf Version for The Basic Practice of Statistics 7e & LaunchPad (Twelve Month Access), Chapter 30, Problem 30.28E , additional homework tip  1

Fig (1)

Loose-leaf Version for The Basic Practice of Statistics 7e & LaunchPad (Twelve Month Access), Chapter 30, Problem 30.28E , additional homework tip  2

Fig (2)

The means of varieties of the tropical flower Heliconia are obtained as:

The mean for bihai is 47.598, for red is 39.711 and yellow is 36.180 and hypothesis is stated as:

The null hypothesis is,

H0:μbihai=μred=μyellow

And alternative hypothesis is,

Ha : At least one mean is different.

From the analysis of variance table, F-value is 259.12 and their P-value is 0.000. This shows that null hypothesis is rejected because P-value is 0.000 which is very less 5% level of significance.

So, there is highly difference in the mean length of the flowers varieties fertilized by different Hummingbird species should have different distribution of length.

For the pair wise comparisons,

From the Tukey' simultaneous tests for difference of means shows that the difference in between μred and μbihai is significant because P-value is 0.000 which is very less 5% level of significance.

Similarly for μyellow and μbihai , the difference is significance and also for μred and μyellow , the difference is significance due to very small of P-value as compare to the 5% level of significance.

The rule of thumb is satisfied because the standard deviation of largest group (1.799) is no more than twice the smallest group (0.975) . The one-way ANOVA is used by comparing the mean length of the population.

Loose-leaf Version for The Basic Practice of Statistics 7e & LaunchPad (Twelve Month Access), Chapter 30, Problem 30.28E , additional homework tip  3

Fig (3)

The one-way ANOVA is used by comparing the mean length of the population. But the characteristic of the data that makes ANOVA is risky is lack of normality condition or groups do not follow normal distribution.

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!
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman