Chapter 15, Problem 44CR

a.

To determine

Check whether there is evidence that the true average iron content is the same for all four storage periods at 0.05 level of significance.

Answer to Problem 44CR

There is convincing evidence that the mean average iron content is not same for all four storage periods at 0.05 level of significance.

Explanation of Solution

Calculation:

The given table information is that there are four storage periods and twenty-four containers of maize. These twenty-four containers are divide into four groups with six of each.

The given summary statistics is as follows:

Storage Period	$\overline{x}$	$s^2$
0	4.923	0.000107
1	4.923	0.000067
2	4.917	0.000147
4	4.902	0.000057

It is assumed that ${\mu }_1,{\mu }_2,{\mu }_3\text{, and }{\mu }_4$ represents the mean average iron content for the storage period 0, 1, 2, and 4, respectively.

Hypotheses:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4$.

That is, the mean average iron content does not depend on the storage period.

Alternative hypothesis:

$H_1:$ At least two among ${\mu }_1,{\mu }_2,{\mu }_3\text{, and }{\mu }_4$ are different.

That is, the mean average iron content depends on the storage period.

Significance level: $\alpha =0.05$

Test statistic:

$F=\frac{\text{MSTr}}{\text{MSE}}$

Where, MSTr be the mean sum of squares for treatments and MSE be the mean sum of squares for errors.

The grand mean is calculated as follows:

$\begin{array}{c}\overline{\overline{x}}=\frac{\text{Grand Total}}{N}\\ =\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2+n_3{\overline{x}}_3+n_4{\overline{x}}_4}{N}\\ =\frac{6\left(4.923\right)+6\left(4.923\right)+6\left(4.917\right)+6\left(4.902\right)}{24}\\ =\frac{29.538+29.538+29.502+29.412}{24}\\ =\frac{117.99}{24}\\ =4.916\end{array}$

The value of sum of squares for treatments is calculated as follows:

$\begin{array}{c}\text{SSTr}=n_1{\left({\overline{x}}_1-\overline{\overline{x}}\right)}^2+n_2{\left({\overline{x}}_2-\overline{\overline{x}}\right)}^2+n_3{\left({\overline{x}}_3-\overline{\overline{x}}\right)}^2+n_4{\left({\overline{x}}_4-\overline{\overline{x}}\right)}^2\\ =6{\left(4.923-4.916\right)}^2+6{\left(4.923-4.916\right)}^2+6{\left(4.917-4.916\right)}^2\\ +6{\left(4.902-4.916\right)}^2\\ =6\left(0.000049\right)+6\left(0.000049\right)+6\left(0.000001\right)+6\left(0.000196\right)\\ =0.000294+0.000294+0.000006+0.001176\\ =0.00177\end{array}$

The value of sum of squares for error is calculated as follows:

$\begin{array}{c}\text{SSE}=\left(n_1-1\right)s_1{}^2+\left(n_2-1\right)s_2{}^2+\left(n_3-1\right)s_3{}^2+\left(n_4-1\right)s_4{}^2\\ =\left(6-1\right)\left(0.000107\right)+\left(6-1\right)\left(0.000067\right)+\left(6-1\right)\left(0.000147\right)\\ +\left(6-1\right)\left(0.000057\right)\\ =5\left(0.000107\right)+5\left(0.000067\right)+5\left(0.000147\right)+5\left(0.000057\right)\\ =0.000535+0.000335+0.000735+0.000285\\ =0.00189\end{array}$

The degrees of freedom for treatment is calculated as follows:

$\begin{array}{c}df=k-1\\ =4-1\\ =3\end{array}$

The degrees of freedom for error is calculated as follows:

$\begin{array}{c}df=N-k\\ =24-4\\ =20\end{array}$

The value of mean sum of squares for treatments is calculated as follows:

$\begin{array}{c}\text{MSTr}=\frac{\text{SSTr}}{df\text{for treatments}}\\ =\frac{0.00177}{3}\\ =0.00059\end{array}$

The value of mean sum of squares for error is calculated as follows:

$\begin{array}{c}\text{MSE}=\frac{\text{SSE}}{df\text{for error}}\\ =\frac{0.00189}{20}\\ =0.0000945\end{array}$

The value of the F test statistic is calculated as follows:

$\begin{array}{c}F=\frac{\text{MSTr}}{\text{MSE}}\\ =\frac{0.00059}{0.0000945}\\ =6.24\end{array}$

P-value:

Software procedure:

A step-by-step procedure to find P-value using MINITAB software is as follows:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘F’ distribution.
• Enter the Numerator df as 3 and Denominator df as 20.
• Click the Shaded Area tab.
• Choose X Value and Right Tail for the region of the curve to shade.
• Enter the X value as 6.24.
• Click OK.

Output obtained using MINITAB software is as follows:

From MINITAB output, the P-value is approximately 0.00364.

Conclusion:

If $P-\text{value}\le \alpha$, then reject the null hypothesis.

Therefore, the P-value of 0.00364 is less than 0.05 (level of significance).

Hence, reject the null hypothesis.

Thus, it can be concluded that there is convincing evidence that the mean average iron content is not the same for all four storage periods at 0.05 level of significance.

b.

To determine

Perform a multiple comparison analysis.

Explanation of Solution

Calculation:

The formula for the Tukey–Kramer multiple comparison procedure is as follows:

$\text{For }{\mu }_1-{\mu }_2:\left({\overline{x}}_1-{\overline{x}}_2\right)\pm q\sqrt{\frac{\text{MSE}}{l}}$ .

Where , ${\overline{x}}_1\text{ and }{\overline{x}}_2$ be the sample means for Treatment 1 and 2, respectively and l be the sample size. Here, q represents the relevant Studentized range critical value and MSE be the mean sum of squares for error.

From APPENDIX: Table 7 Critical Values of q for the Studentized Range Distribution, the critical value for q is noted as follows:

• Locate the value 20 or the nearest value in the column of Error df.
• Locate 95% in the column of confidence interval corresponding to the above error df.
• Locate the value 4 in the row of number of populations, treatments, or levels being compared.
• The intersecting value that corresponds to the error df 20 at 95% confidence level with 4 treatments is 3.96.

From Part a, the value of MSE is 0.0000945. The sample mean values for storage periods 0,1, 2, and 3 are 4.923, 4.923, 4.917, and 4.902, respectively.

$\text{For }{\mu }_1-{\mu }_2:$

95% T–K intervals for ${\mu }_1-{\mu }_2$ are calculated as follows:

$\begin{array}{c}\left({\overline{x}}_1-{\overline{x}}_2\right)\pm q\sqrt{\frac{\text{MSE}}{l}}=\left(4.923-4.923\right)\pm 3.96\sqrt{\frac{0.0000945}{6}}\\ =0\pm 3.96\left(\sqrt{0.00001575}\right)\\ =0\pm 3.96\left(0.0040\right)\\ =0\pm 0.01584\\ =\left(-0.01584,0.01584\right)\end{array}$

From the MINITAB output, the 95% Tukey–Kramer confidence intervals for storage group 0 and 1 are $\left(-0.01584,0.01584\right)$. It includes the value 0 and it can be concluded that storage groups 0 and 1 do not differ from each other.

$\text{For }{\mu }_1-{\mu }_3:$

The 95% T–K interval for ${\mu }_1-{\mu }_3$ is calculated as follows:           

$\begin{array}{c}\left({\overline{x}}_1-{\overline{x}}_3\right)\pm q\sqrt{\frac{\text{MSE}}{l}}=\left(4.923-4.917\right)\pm 3.96\sqrt{\frac{0.0000945}{6}}\\ =0.006\pm 3.96\left(\sqrt{0.00001575}\right)\\ =0.006\pm 3.96\left(0.0040\right)\\ =0.006\pm 0.01584\\ =\left(0.006-0.01584,0.006+0.01584\right)\\ =\left(-0.00954,0.02184\right)\end{array}$

From the MINITAB output, the 95% Tukey–Kramer confidence intervals for the storage group 0 and 2 are $\left(-0.00954,0.02184\right)$. It includes the value 0 and it can be concluded that the storage group 0 and 2 do not differ from each other.

$\text{For }{\mu }_1-{\mu }_4:$

The 95% T–K interval for ${\mu }_1-{\mu }_4$ is calculated as follows:

$\begin{array}{c}\left({\overline{x}}_1-{\overline{x}}_4\right)\pm q\sqrt{\frac{\text{MSE}}{l}}=\left(4.923-4.902\right)\pm 3.96\sqrt{\frac{0.0000945}{6}}\\ =0.021\pm 3.96\left(\sqrt{0.00001575}\right)\\ =0.021\pm 3.96\left(0.0040\right)\\ =0.021\pm 0.01584\\ =\left(0.021-0.01584,0.021+0.01584\right)\\ =\left(0.00516,0.03684\right)\end{array}$

$\text{For }{\mu }_2-{\mu }_3:$

The 95% T–K interval for ${\mu }_2-{\mu }_3$ is calculated as follows:

$\begin{array}{c}\left({\overline{x}}_2-{\overline{x}}_3\right)\pm q\sqrt{\frac{\text{MSE}}{l}}=\left(4.923-4.917\right)\pm 3.96\sqrt{\frac{0.0000945}{6}}\\ =0.006\pm 3.96\left(\sqrt{0.00001575}\right)\\ =0.006\pm 3.96\left(0.0040\right)\\ =0.006\pm 0.01584\\ =\left(0.006-0.01584,0.006+0.01584\right)\\ =\left(-0.00984,0.02184\right)\end{array}$

From the MINITAB output, the 95% Tukey–Kramer confidence intervals for the storage group 1 and 2 is $\left(-0.00984,0.02184\right)$. It includes the value 0 and it can be concluded that the storage groups 2 and 4 do not differ from each other.

$\text{For }{\mu }_2-{\mu }_4:$

The 95% T–K interval for ${\mu }_2-{\mu }_4$ is calculated as follows:

$\begin{array}{c}\left({\overline{x}}_2-{\overline{x}}_4\right)\pm q\sqrt{\frac{\text{MSE}}{l}}=\left(4.923-4.902\right)\pm 3.96\sqrt{\frac{0.0000945}{6}}\\ =0.021\pm 3.96\left(\sqrt{0.00001575}\right)\\ =0.021\pm 3.96\left(0.0040\right)\\ =0.021\pm 0.01584\\ =\left(0.021-0.01584,0.021+0.01584\right)\\ =\left(0.00516,0.03684\right)\end{array}$

From the MINITAB output, the 95% Tukey–Kramer confidence intervals for the storage group 1 and 2 are $\left(-0.00984,0.02184\right)$. It does not include the value 0 and it can be concluded that storage group 2 and storage group 4 differ from each other.

$\text{For }{\mu }_3-{\mu }_4:$

The 95% T–K intervals for ${\mu }_3-{\mu }_4$ is calculated as follows:

$\begin{array}{c}\left({\overline{x}}_3-{\overline{x}}_4\right)\pm q\sqrt{\frac{\text{MSE}}{l}}=\left(4.917-4.902\right)\pm 3.96\sqrt{\frac{0.0000945}{6}}\\ =0.015\pm 3.96\left(\sqrt{0.00001575}\right)\\ =0.015\pm 3.96\left(0.0040\right)\\ =0.015\pm 0.01584\\ =\left(0.015-0.01584,0.015+0.01584\right)\\ =\left(-0.00084,0.03084\right)\end{array}$

From the MINITAB output, the 95% Tukey–Kramer confidence intervals for the storage groups 1 and 2 are $\left(-0.00984,0.02184\right)$. It includes the value 0 and it can be concluded that the storage group 2 and storage group 4 do not differ from each other.

Summarizing the results of the Tukey–Kramer Procedure:

1. 1. List the sample means in an increasing order and identify the corresponding population just above the value of each $\overline{x}$.

1. 2. Here, the population mean of storage group 0 does not differ from storage group 1 and storage group 2 but storage group 0 differs from storage group 4. Storage group 1 does not differ from storage group 2. Draw a horizontal line as follows:

1. 3. Here, the population mean of storage group 4 does not differ from storage group 2 but storage group 4 differs from storage group 0 and storage group 1. Draw a horizontal line as follows: