Chapter 30, Problem 30.29E

a.

To determine

The contrast expression for the comparison of tropical flowers.

Answer to Problem 30.29E

The First hypothesis becomes:

${\text{H}}_0:L_1=0\text{vs}{\text{H}}_a:L_1\ne 0$

The second hypothesis becomes:

${\text{H}}_0:L_2=0\text{vs}{\text{H}}_a:L_2\ne 0$

The third Hypothesis becomes:

${\text{H}}_0:L_3=0\text{vs}{\text{H}}_a:L_3\ne 0$

Explanation of Solution

Given info:

The data represents the comparisons of the tropical flowers. The variables are the lenghts aof bihai variety of Heliconia that is represented in two forms based on the color of caribaea variety.

Calculation:

In the ANOVA construction, the comparison of means ${\mu }_{red},{\mu }_{bihai},\text{and}{\mu }_{yellow}$ of the population, a population contrast is a combination of means.

$L=c_{red}{\mu }_{red}+c_{bihai}{\mu }_{bihai}+c_{yellow}{\mu }_{yellow}$

Group ${\mu }_{bihai}$ indicates the bihai variety of Heliconia, group ${\mu }_{red}$ indicates the red variety of Caribaea and group ${\mu }_{yellow}$ denotes the yellow variety of Caribaea.

The three hypotheses are terms of three contrasts are as follows:

The first hypothesis in terms of three contrasts is,

$\begin{array}{c}{L}_1=c_{bihai}{\mu }_{bihai}-c_{red}{\mu }_{red}\\ =\left(1\right){\mu }_{bihai}+\left(-1\right){\mu }_{red}+\left(0\right){\mu }_{yellow}\end{array}$

Thus, The First hypothesis becomes:

${\text{H}}_0:L_1=0\text{vs}{\text{H}}_a:L_1\ne 0$

The second hypothesis in terms of three contrasts is,

$\begin{array}{c}{L}_2=c_{bihai}{\mu }_{bihai}-c_{yellow}{\mu }_{yellow}\\ =\left(0\right){\mu }_{red}+\left(1\right){\mu }_{bihai}+\left(-1\right){\mu }_{yellow}\end{array}$

Thus, The second hypothesis becomes:

${\text{H}}_0:L_2=0\text{vs}{\text{H}}_a:L_2\ne 0$

The third hypothesis in terms of three contrasts is,

$\begin{array}{c}{L}_3=c_{bihai}{\mu }_{bihai}+c_{red}{\mu }_{red}+c_{yellow}{\mu }_{yellow}\\ =\left(1\right){\mu }_{bihai}-\left(\frac{{\mu }_{red}}{2}+\frac{{\mu }_{yellow}}{2}\right)\\ =\left(1\right){\mu }_{bihai}+\left(-\frac{1}{2}\right){\mu }_{red}+\left(-\frac{1}{2}\right){\mu }_{yellow}\end{array}$

Thus, The third Hypothesis becomes:

${\text{H}}_0:L_3=0\text{vs}{\text{H}}_a:L_3\ne 0$

b.

To determine

The null and alternative hypothesis and the sample contrast and also assess its statistical significance.

Answer to Problem 30.29E

The First hypothesis becomes,

${\text{H}}_0:L_1=0\text{vs}{\text{H}}_a:L_1\ne 0$

The second hypothesis becomes,

${\text{H}}_0:L_2=0\text{vs}{\text{H}}_a:L_2\ne 0$

The third Hypothesis becomes,

${\text{H}}_0:L_3=0\text{vs}{\text{H}}_a:L_3\ne 0$

The first sample contrast is,

${\widehat{L}}_1=7.887,S{E}_1=0.466$

The second sample contrast is,

${\widehat{L}}_2=11.418,S{E}_2=0.517$

The third sample contrast is,

${\widehat{L}}_3=29.495,S{E}_3=0.4319$

The interpretation is that , there is a evidence that the mean bihai variety of Heliconia differs from the average of the two forms of Caribaea.

Explanation of Solution

Calculation:

In the ANOVA construction, the comparison of means ${\mu }_{red},{\mu }_{bihai},\text{and}{\mu }_{yellow}$ of the population, a population contrast is a combination of means.

$L=c_{red}{\mu }_{red}+c_{bihai}{\mu }_{bihai}+c_{yellow}{\mu }_{yellow}$ (1)

The sample contrast $\widehat{L}$ has a standard error is calculated as:

$S{E}_{\widehat{L}}=s_p\sqrt{\frac{c_{red}^2}{n_1}+\frac{c_{bihai}^2}{n_2}+\frac{c_{yellow}^2}{n_3}}$

Software procedure:

Step by step procedure to obtain two 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
• Click on Comparison> Tick on Tukey under Comparison procedures assuming equal variances and then click OK.
• Click OK.

The Minitab output is shown below.

Fig (1)

From the Minitab result, the pooled standard deviation $s_p=1.44$ and the degree of freedom is 51.

The standard errors for the first sample contrast are,

$\begin{array}{c}S{E}_{{\widehat{L}}_1}=s_p\sqrt{\frac{c_{red}^2}{n_1}+\frac{c_{bihai}^2}{n_2}+\frac{c_{yellow}^2}{n_3}}\\ =1.44\sqrt{\frac{{\left(-1\right)}^2}{23}+\frac{1}{16}+\frac{0}{15}}\\ =1.44\sqrt{0.1046}\\ =0.466\end{array}$

Thus, The standard error for first sample contrast is 0.466.

Substitute $-1$ for $c_{red}$, 39.711 for ${\mu }_{red}$, 1 for $c{bihai}_{}$, 47.598 for ${\mu }_{bihai}$, 0 for $c_{yellow}$, and 36.180 in the above equation (1).

$\begin{array}{c}{\widehat{L}}_1=c_{red}{\mu }_{red}+c_{bihai}{\mu }_{bihai}+c_{yellow}{\mu }_{yellow}\\ =\left(-1\right)\times 39.711+\left(1\right)\times 47.598+\left(0\right)\times 36.180\\ =7.887\end{array}$

The standard error for second sample contrast is,

$\begin{array}{c}S{E}_{{\widehat{L}}_2}=s_p\sqrt{\frac{c_{red}^2}{n_1}+\frac{c_{bihai}^2}{n_2}+\frac{c_{yellow}^2}{n_3}}\\ =1.44\sqrt{\frac{{\left(0\right)}^2}{23}+\frac{1}{16}+\frac{{\left(-1\right)}^2}{15}}\\ =1.44\sqrt{0.1291}\\ =0.517\end{array}$

Thus, The standard error for the second sample contrast is 0.5617.

Substitute 0 for $c_{red}$, 39.711 for ${\mu }_{red}$, 1 for $c{bihai}_{}$, 47.598 for ${\mu }_{bihai}$, $\left(-1\right)$ for $c_{yellow}$, and 36.180 for ${\mu }_{yellow}$ in the above equation (1).

$\begin{array}{c}{\widehat{L}}_2=c_{red}{\mu }_{red}+c_{bihai}{\mu }_{bihai}+c_{yellow}{\mu }_{yellow}\\ =\left(0\right)\times 39.711+\left(1\right)\times 47.598+\left(-1\right)\times 36.180\\ =11.418\end{array}$

The standard error for third sample contrast is,

$\begin{array}{c}S{E}_{{\widehat{L}}_3}=s_p\sqrt{\frac{c_{red}^2}{n_1}+\frac{c_{bihai}^2}{n_2}+\frac{c_{yellow}^2}{n_3}}\\ =1.44\sqrt{\frac{{\left(-0.5\right)}^2}{23}+\frac{{\left(1\right)}^2}{16}+\frac{{\left(-0.5\right)}^2}{15}}\\ =1.44\sqrt{\frac{0.25}{23}+\frac{1}{16}+\frac{0.25}{15}}\\ =1.44\sqrt{0.0899}\end{array}$

Further, solve the above expression.

$\begin{array}{c}S{E}_{{\widehat{L}}_3}=1.44\times 0.2999\\ =0.4319\end{array}$

Thus, the standard error for third sample contrast is 0.4319.

Substitute $\left(-0.5\right)$ for $c_{red}$, 39.711 for ${\mu }_{red}$, 1 for $c{bihai}_{}$, 47.598 for ${\mu }_{bihai}$, $\left(-0.5\right)$ for $c_{yellow}$, and 36.180 for ${\mu }_{yellow}$ in the above equation (1).

$\begin{array}{c}{\widehat{L}}_1=c_{red}{\mu }_{red}+c_{bihai}{\mu }_{bihai}+c_{yellow}{\mu }_{yellow}\\ =\left(-0.5\right)\times 39.711+\left(1\right)\times 47.598+\left(-0.5\right)\times 36.180\\ =-0.01259+47.598-18.09\\ =29.495\end{array}$

The hypotheses are given as:

The First hypothesis becomes:

${\text{H}}_0:L_1=0vs{\text{H}}_a:L_1\ne 0$

The second hypothesis becomes:

${\text{H}}_0:L_2=0vs{\text{H}}_a:L_2\ne 0$

The third Hypothesis becomes:

${\text{H}}_0:L_3=0vs{\text{H}}_a:L_3\ne 0$

The t- statistics is,

$\begin{array}{c}t=\frac{\widehat{L_3}}{S{E}_3}\\ =\frac{29.495}{0.4319}\\ =68.2912\end{array}$

The tabulated value with 51 degree of freedom is 1.6753 which is very much lesser than 68.2912. The null hypothesis is rejected for population contrast $L_3$. There is very strong evidence that the population contrast $L_3$ is not equal to zero for the t-statistics with 51 degree of freedom. Therefore, ${\mu }_{bihai}$ and $\frac{\left({\mu }_{red}+{\mu }_{yellow}\right)}{2}$ significantly differ.

There is evidence that the mean bihai variety of Heliconia differs from the average of the two forms of Caribaea.

c.

To determine

The 90% confidence interval for the population contrast.

Answer to Problem 30.29E

The 90% confidence interval for $L_3$ population contrast is $\left(28.7716,30.218\right)$.

Explanation of Solution

Calculation:

The 90% confidence interval from the population contrast for the population contrast is calculated as:

$L_3={\mu }_{bihai}-\left(\frac{{\mu }_{red}+{\mu }_{yellow}}{2}\right)$

The 90% confidence interval for $L_3=29.495$ uses, $t^{\ast }=1.675\left(\text{from}\text{t-table}\text{with}\text{51}\text{df}\right)$.

$\widehat{L_3}\pm t^{\ast }S{E}_3=29.495\pm 1.675\left(0.4319\right)$

The lower limit is,

$\begin{array}{c}\text{Lower}\text{limit}=29.495-1.675\left(0.4319\right)\\ =28.7716\end{array}$

The upper limit is,

$\begin{array}{c}\text{Upper}\text{Limit}=29.495+1.675\left(0.4319\right)\\ =30.218\end{array}$

The 90% confidence interval is $\left(28.7716,30.218\right)$ for $L_3$ population contrast.