Chapter 30, Problem 30.28E

a.

To determine

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

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;

$\begin{array}{l}6.752\text{to}9.021\text{for}{\mu }_{behai}-{\mu }_{red}\\ 10.165\text{to}\text{12}\text{.670}\text{for}{\mu }_{behai}-{\mu }_{yellow}\\ 2.375\text{to}4.688\text{for}{\mu }_{red}-{\mu }_{yellow}\end{array}$

Calculation:

Consider the first confidence interval that is given as;

$6.752\text{to}9.021\text{for}{\mu }_{behai}-{\mu }_{red}$

The null hypothesis is,

${\text{H}}_0:{\mu }_{behai}-{\mu }_{red}=0$

The alternative hypothesis is,

${\text{H}}_a:{\mu }_{behai}-{\mu }_{red}\ne 0$

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 ${\mu }_{bihai}$ and red flowers ${\mu }_{red}$ and 0 does not lie inside this confidence interval. So, the null hypothesis is rejected in this situation and accepts the alternative hypothesis ${\text{H}}_0:{\mu }_{behai}-{\mu }_{red}\ne 0$ in favor of two sided at $5\%$ level of significance level.

Consider the second confidence interval that is given as;

$10.165\text{to}\text{12}\text{.670}\text{for}{\mu }_{behai}-{\mu }_{yellow}$

The null hypothesis is,

${\text{H}}_0:{\mu }_{behai}-{\mu }_{yellow}=0$

The alternative hypothesis is,

${\text{H}}_a:{\mu }_{behai}-{\mu }_{yellow}\ne 0$

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 ${\mu }_{bihai}$ and red flowers ${\mu }_{yellow}$ and 0 does not lie inside this confidence interval. So, the null hypothesis is rejected in this case and accepts the alternative hypothesis ${\text{H}}_0:{\mu }_{behai}-{\mu }_{yellow}\ne 0$ in favor of two sided at $5\%$ level of significance level.

Consider the third confidence interval that is given as;

$2.375\text{to}4.688\text{for}{\mu }_{red}-{\mu }_{yellow}$

The null hypothesis is,

${\text{H}}_0:{\mu }_{red}-{\mu }_{yellow}=0$

The alternative hypothesis is,

${\text{H}}_a:{\mu }_{red}-{\mu }_{yellow}\ne 0$

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 ${\mu }_{red}$ and red flowers ${\mu }_{yellow}$ and 0 does not lie inside this confidence interval. So, the null hypothesis is rejected in this case and accepts the alternative hypothesis ${\text{H}}_0:{\mu }_{red}-{\mu }_{yellow}\ne 0$ 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.

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 ${\mu }_{behai}-{\mu }_{red}$, 10.165 to 12.670 for ${\mu }_{behai}-{\mu }_{yellow}$ and 2.375 to 4.688 for ${\mu }_{red}-{\mu }_{yellow}$.

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,

Fig (1)

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,

${\text{H}}_0:{\mu }_{bihai}={\mu }_{red}={\mu }_{yellow}$

And alternative hypothesis is,

${\text{H}}_a$: 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 ${\mu }_{red}$ and ${\mu }_{bihai}$ is significant because P-value is 0.000 which is very less 5% level of significance.

Similarly for ${\mu }_{yellow}$ and ${\mu }_{bihai}$, the difference is significance and also for ${\mu }_{red}$ and ${\mu }_{yellow}$, 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 $\left(1.799\right)$ is no more than twice the smallest group $\left(0.975\right)$. The one-way ANOVA is used by comparing the mean length of the population.

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.