Chapter 13.7, Problem 30E

a.

To determine

Check whether there is sufficient evidence to indicate a difference in the mean wear among the four treatments using $\alpha =0.05$ significance.

Answer to Problem 30E

There is sufficient evidence that at least one mean is different.

Explanation of Solution

The null and alternative hypotheses are as follows:

Null hypothesis: There is no significant difference in mean wear among the four treatments.

Alternative hypothesis: There is significant difference in mean wear among the four treatments.

The sample means are as follows:

$\begin{array}{c}{\overline{Y}}_{1.}=\frac{{\displaystyle \sum _{j=1}^{n_1}{y}_{1j}}}{n_1}\\ =\frac{9.16+13.29+12.07+11.97+13.31+12.32+11.78}{7}\\ =11.98571\end{array}$

$\begin{array}{c}{\overline{Y}}_{2.}=\frac{{\displaystyle \sum _{j=1}^{n_2}{y}_{2j}}}{n_2}\\ =\frac{11.95+15.15+14.75+14.79+15.48+13.47+13.06}{7}\\ =14.09286\end{array}$

$\begin{array}{c}{\overline{Y}}_{3.}=\frac{{\displaystyle \sum _{j=1}^{n_3}{y}_{3j}}}{n_1}\\ =\frac{11.47+9.54+11.26+13.66+11.18+15.03+14.86}{7}\\ =12.42857\end{array}$

$\begin{array}{c}{\overline{Y}}_{4.}=\frac{{\displaystyle \sum _{j=1}^{n_4}{y}_{4j}}}{n_2}\\ =\frac{11.35+8.73+10+9.75+11.71+12.45+12.38}{7}\\ =10.91\end{array}$

$\begin{array}{c}\overline{Y}=\frac{{\overline{Y}}_{1.}+{\overline{Y}}_{2.}+{\overline{Y}}_{3.}+{\overline{Y}}_{4.}}{4}\\ =12.35429\end{array}$

Total sum of square is calculated as follows:

$\begin{array}{c}\text{SSE}\text{=}{\displaystyle \sum _{i=1}^4{\displaystyle \sum _{j=1}^7{\left(Y_{ij}-{\overline{Y}}_{i.}\right)}^2}}\\ =\left[\begin{array}{l}{\left(9.16-11.98571\right)}^2+{\left(13.29-11.98571\right)}^2+{\left(12.07-11.98571\right)}^2+\\ {\left(11.97-11.98571\right)}^2+{\left(13.31-11.98571\right)}^2+{\left(12.32-11.98571\right)}^2+\\ {\left(11.78-11.98571\right)}^2+{\left(11.95-14.09286\right)}^2+{\left(15.15-14.09286\right)}^2+\\ {\left(14.75-14.09286\right)}^2+{\left(14.79-14.09286\right)}^2+{\left(15.48-14.09286\right)}^2+\\ {\left(13.47-14.09286\right)}^2+{\left(13.06-14.09286\right)}^2+{\left(11.47-12.42857\right)}^2+\\ {\left(9.54-12.42857\right)}^2+{\left(11.26-12.42857\right)}^2+{\left(13.66-12.42857\right)}^2+\\ {\left(11.18-12.42857\right)}^2+{\left(15.03-12.42857\right)}^2+{\left(14.86-12.42857\right)}^2+\\ {\left(11.35-10.91\right)}^2+{\left(8.73-10.91\right)}^2+{\left(10-10.91\right)}^2+{\left(9.75-10.91\right)}^2+\\ {\left(11.71-10.91\right)}^2+{\left(12.45-10.91\right)}^2+{\left(12.38-10.91\right)}^2+\end{array}\right]\\ =60.282\end{array}$

Sum of square for treatments is calculated as follows:

$\begin{array}{c}\text{SST=}{\displaystyle \sum _{i=1}^4{n}_i{\left({\overline{Y}}_{i.}-\overline{Y}\right)}^2}\\ =0.951+21.158+0.039+14.602\\ =36.750\end{array}$

The mean squares for the treatments are as follows:

$\begin{array}{c}MST=\frac{SST}{k-1}\\ =\frac{36.75}{4-1}\\ =12.25\end{array}$

The mean square for the error is as follows:

$\begin{array}{c}MSE=\frac{SSE}{n-4}\\ =\frac{60.282}{28-4}\\ =2.5118\end{array}$

The F-value is as follows:

$\begin{array}{c}F=\frac{MST}{MSE}\\ =\frac{12.25}{2.5118}\\ =4.877\end{array}$

Critical value:

The critical value is obtained from F table with 3 degrees of freedom for the numerator and 24 for the denominator.

Step-by-step procedure to obtain F-value using Table 7 of Appendix 3:

• Locate the value 3 and $\alpha =0.05$ in the right column of Table 7 of Appendix 3.
• Go through the row corresponding to the value 3 and column corresponding to the value 24 of the Table 7.
• Locate the value corresponding to (3, 24).

The required critical value for F-distribution with the level of significance 0.05, $d{f}_1=3$ and $d{f}_2=24$ is 3.01.

Rejection rule:

If $F\ge F\text{-critical}$, reject $H_0$. Otherwise fail to reject $H_0$.

Conclusion:

Here, the F-test statistic is 4.877.

The test statistic $\left(=4.877\right)$ is greater than the critical value $\left(=3.01\right)$.

Therefore, the null hypothesis is rejected.

Thus, there is sufficient evidence to conclude that at least one mean is different from others.

b.

To determine

Construct a 99% confidence interval for the difference in haze increase between treatments B and C.

Answer to Problem 30E

The 99% confidence interval for the difference in haze increase between treatments B and C is $\left(-0.70521,4.03379\right)$.

Explanation of Solution

${\overline{Y}}_B=14.09286,{\overline{Y}}_C=12.42857,\text{ MSE }=2.5118=S^2,n-k=24,n_A=7\text{ and }{n}_B=7$.

The confidence interval for the difference in the means for treatments i and i’ is as follows:

$\text{CI}=\left({\overline{Y}}_{i.}-{\overline{Y}}_{i^{\text{'}}.}\pm t_{\frac{\alpha }{2}}S\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\right)$

Here, $t_{\frac{\alpha }{2}}$ is based on the $\left(n-k\right)$ degrees of freedom.

From “Appendix 3, Table 5 Percentage Points of the t Distributions”, the critical value under the level of significance $0.01\left(=\alpha \right)$ and 24 degrees of freedom is 2.797.

Substitute the values in the confidence interval and the 99% confidence interval for the difference in haze increase between treatments B and C:

$\begin{array}{c}\text{CI}=\left({\overline{Y}}_{i.}-{\overline{Y}}_{i^{\text{'}}.}\pm t_{\frac{\alpha }{2}}S\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\right)\\ =\left(14.09286-12.42857\pm \left(2.797\right)\sqrt{2.5118}\sqrt{\frac{1}{7}+\frac{1}{7}}\right)\\ =\left(1.66429\pm 2.3695\right)\\ =\left(-0.70521,4.03379\right)\end{array}$

Thus, the 99% confidence interval for the difference in haze increase between treatments B and C is $\left(-0.70521,4.03379\right)$.

c.

To determine

Construct a 90% confidence interval for the mean wear for lenses receiving treatment A.

Answer to Problem 30E

The 90% confidence interval for the mean wear for lenses receiving treatment A is $\left(10.961,13.011\right)$.

Explanation of Solution

For mean treatment A, ${\overline{Y}}_A=11.98571,\text{MSE }=2.5118=S^2,n-k=24,n_A=7$.

The confidence interval for the mean treatment i is as follows:

$\text{CI}=\left({\overline{Y}}_{i.}\pm t_{\frac{\alpha }{2}}\frac{S}{\sqrt{n_i}}\right)$

Here, $t_{\frac{\alpha }{2}}$ is based on the $\left(n-k\right)$ degrees of freedom.

From “Appendix 3, Table 5 Percentage Points of the t Distributions”, the critical value under the level of significance $0.05\left(=\alpha \right)$ and 24 degrees of freedom is 1.711.

Substitute the values in the confidence interval and the 90% confidence interval for the mean wear for lenses receiving treatment A:

$\begin{array}{c}\text{CI}=\left({\overline{Y}}_{i.}\pm t_{\frac{\alpha }{2}}\frac{S}{\sqrt{n_i}}\right)\\ =\left(11.98571\pm \left(1.711\right)\frac{\sqrt{2.5118}}{\sqrt{7}}\right)\\ =\left(10.961,13.011\right)\end{array}$

Thus, the 90% confidence interval for the mean wear for lenses receiving treatment A is $\left(10.961,13.011\right)$.