Chapter 10.2, Problem 21E

The article “The Effect of Enzyme Inducing Agents on the Survival Times of Rats Exposed to Lethal Levels of Nitrogen Dioxide” (Toxicology and Applied Pharmacology, 1978: 169–174) reports the following data on survival times for rats exposed to nitrogen dioxide (70 ppm) via different injection regimens. There were J = 14 rats in each group.

	Regimen	${\bar{x}}_i$. (min)	si
1.	Control	166	32
2.	3-Methylcholanthrene	303	53
3.	Allylisopropylacetamide	266	54
4.	Phénobarbital	212	35
5.	Chlorpromazine	202	34
6.	p-Aminobenzoic Acid	184	31

a. Test the null hypothesis that true average survival time does not depend on an injection regimen against the alternative that there is some dependence on an injection regimen using α = .01.

b. Suppose that 100(1 – α)% CIs for k different parametric functions are computed from the same ANOVA data set. Then it is easily verified that the simultaneous confidence level is at least 100(1 – ka)%. Compute CIs with a simultaneous confidence level of at least 98% for

µ₁ -1/5(µ₂ + µ₃ + µ₄ + µ₅ + µ₆) and 1/4 (µ₂ + µ₃ + µ₄ + µ₅) - µ₆

To determine

Conduct a test for the null hypothesis that the true average survival time of rats is independent on the injection regimen, versus the alternative that there is dependence of the true average survival time on the injection regime, at $\alpha =0.01$.

Answer to Problem 21E

The test suggests that there is dependence of the true average survival time on the injection regime, at $\alpha =0.01$.

Explanation of Solution

Given info:

The data provides with the summary data on the survival time (minutes) of rats, which are exposed to nitrogen dioxide (70 ppm) via different injection regimes, namely, the Control group, 3-Methycholanthrene, Allylisopropylacetamide, Phenobarbital, Chlorpromazine and p-Aminobenzoic Acid. The injection regimes are numbered 1 to 6, respectively. Here, the sample means are ${\overline{X}}_{1.}=166$, ${\overline{X}}_{2.}=303$, ${\overline{X}}_{3.}=266$, ${\overline{X}}_{4.}=212$, ${\overline{X}}_{5.}=202$ and ${\overline{X}}_{6.}=184$. The sample standard deviations are $s_1=32$, $s_2=53$, $s_3=54$, $s_4=35$, $s_5=34$ and $s_6=31$. The number of rats in each group is $J=14$

Calculation:

Suppose there are I treatments and J observations corresponding to each treatment in a designed experiment, resulting in a total of IJ observations. Then, the following properties hold true:

• Degrees of freedom (df): The treatment df is $I-1$, the total df is $IJ-1$. The error df is the difference between these, that is, $I\left(J-1\right)$.
• Sum of squares: The total sum of squares (SST), treatment sum of squares (SSTr) and error sum of squares (SSE) are related as: $\text{SST}=\text{SSTr}+\text{SSE}$.
• Mean of squares: The mean of squares is the ratio of the sum of squares to the corresponding df. The mean square error (MSE) is, $\text{MSE}=\frac{\text{SSE}}{I\left(J-1\right)}$. The mean square treatment (MSTr) is, $\text{MSTr}=\frac{\text{SSTr}}{I-1}$.
• The F-statistic: The F- statistic is the ratio of MSTr and MSE, that is, $F=\frac{\text{MSTr}}{\text{MSE}}$.

Here, each injection regime is a treatment.

Denote $X_{ij}$ as the j^th observation corresponding to the i^th treatment, for $j=1,2,\mathrm{...},J\left(=14\right)$, corresponding to each $i=1,2,\mathrm{...},I\left(=6\right)$.

State the test hypotheses.

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5={\mu }_6=0$.

That is, the effects of all the treatments are similar, that is, survival time is independent of injection regime.

Alternative hypothesis:

$H_a:{\mu }_i\ne 0$, for at least one $i=1,2,3,4,5,6$.

That is, the effects of all the treatments are not similar, that is, survival time is not independent of injection regime.

Test statistic:

The suitable test statistic is the F- statistic, which is the ratio of the mean square treatment (MSTr) and the mean square error (MSE), that is,

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

Degrees of freedom:

In this case, number of treatments is $I=6$. Thus, the treatment df is:

$\begin{array}{c}I-1=6-1\\ =5.\end{array}$

The number of observations corresponding to each treatment is $J=14$.

The total df is:

$\begin{array}{c}IJ-1=\left(6\times 14\right)-1\\ =84-1\\ =83.\end{array}$

The error df is:

$\begin{array}{c}I\left(J-1\right)=6\times \left(14-1\right)\\ =6\times 13\\ =78.\end{array}$

The numerator degrees of freedom is:

$\begin{array}{c}{\nu }_1=I-1\\ =5.\end{array}$

The denominator degrees of freedom is:

$\begin{array}{c}{\nu }_2=I\left(J-1\right)\\ =78.\end{array}$

Calculation for test statistic:

The grand mean is:

$\begin{array}{c}{\overline{X}}_{\mathrm{..}}=\frac{1}{I}{\displaystyle \sum _{i=1}^I{\overline{X}}_{i.}}\\ =\frac{166+303+266+212+202+184}{6}\\ =\frac{1,333}{6}\\ =\mathrm{222.167.}\end{array}$

The mean square of treatments, MSTr is:

$\begin{array}{c}\text{MSTr}=\frac{\text{SSTr}}{I-1}\\ =\frac{J}{I-1}{\displaystyle \sum _{i=1}^I{\left({\overline{X}}_{i.}-{\overline{X}}_{\mathrm{..}}\right)}^2}\\ =\frac{14}{5}\left[\begin{array}{l}{\left(166-222.167\right)}^2+{\left(303-222.167\right)}^2+{\left(266-222.167\right)}^2\\ +{\left(212-222.167\right)}^2+{\left(202-222.167\right)}^2+{\left(184-222.167\right)}^2\end{array}\right]\\ =\frac{14}{5}\times \left[3,154.732+6,533.974+1,921.332+103.368+406.708+1,456.720\right]\end{array}$

$\begin{array}{l}=\frac{14}{5}\times 13,576.83\\ =38,\mathrm{015.13.}\end{array}$

The sample variance corresponding to the i^th treatment is $s_i^2$ and the corresponding standard deviation is $s_i$.

The mean square of error, MSE is:

$\begin{array}{c}\text{MSE}=\frac{\text{SSE}}{I\left(J-1\right)}\\ =\frac{1}{I}\cdot \frac{\text{SSE}}{\left(J-1\right)}\\ =\frac{1}{I}\cdot \frac{1}{J-1}{\displaystyle \sum _{i=1}^I{\displaystyle \sum _{j=1}^J\left(X_{ij}-{\overline{X}}_{i\cdot }\right)}}\\ =\frac{1}{I}\cdot {\displaystyle \sum _{i=1}^I\left[\frac{1}{J-1}{\displaystyle \sum _{j=1}^J\left(X_{ij}-{\overline{X}}_{i\cdot }\right)}\right]}\end{array}$

$\begin{array}{c}=\frac{1}{I}\cdot {\displaystyle \sum _{i=1}^I{s}_i^2}\\ =\frac{1}{6}\left[{32}^2+{53}^2+{54}^2+{35}^2+{34}^2+{31}^2\right]\\ =\frac{1}{6}\left[1,024+2,809+2,916+1,225+1,156+961\right]\\ =\frac{10,091}{6}\end{array}$

$=1,681.83$.

Thus, the F-statistic is:

$\begin{array}{c}F=\frac{\text{MSTr}}{\text{MSE}}\\ =\frac{38,015.13}{1,681.83}\\ =\mathrm{22.60.}\end{array}$

Level of significance:

The level of significance here is $\alpha =0.01$.

Critical value:

The critical value for the $F_{{\nu }_1,{\nu }_2}$ distribution at level of significance $\alpha$ is $F_{\alpha ;{\nu }_1,{\nu }_2}$, which is the value of the $F_{{\nu }_1,{\nu }_2}$-distribution, the probability above which is $\alpha$.

Here, the critical value would be $F_{0.01;5,78}$. In the Table A.9, “Critical Values for F Distributions”, the value of $F_{0.01;5,78}$ is not available. Thus, consider the values $F_{0.01;5,60}=3.34$ and $F_{0.01;5,100}=3.21$.

The critical value $F_{0.01;5,78}$ must lie between that of $F_{0.01;5,60}$ and $F_{0.01;5,100}$. As a result, the required critical value can neither be less than 3.21, nor greater than 3.34. Again, the denominator df 78 lies between denominator df’s 60 and 100.  Thus, it is logical to assume that the value of $F_{0.01;5,78}$ must lie between $F_{0.01;5,60}=3.34$ and $F_{0.01;5,100}=3.21$.

Assume that the reduction in the critical value for increase in the df $\left(n-I\right)$, between the corresponding df values 60 and 100 is uniform.

Now, for increase in 40 denominator df’s, the critical value reduces by $0.13\left(=3.34-3.21\right)$. Thus, by using unitary method, the reduction in critical value is 0.00325 for unit increase in df from df 60. Thus, the approximate critical value for 18 units increase in df from 60 is $F_{0.01;5,78}\approx 3.28$.

Again, if $\alpha =0.001$, $F_{0.001;5,60}=4.76$ and $F_{0.001;5,100}=4.48$. Using the same logic as before, the approximate critical value would be $F_{0.001;5,78}\approx 4.63$.

P-value:

The P-value is the area to the right of the F-statistic value f, under the $F_{{\nu }_1,{\nu }_2}$ distribution curve, that is, $P\text{-value}=P\left(F_{{\nu }_1,{\nu }_2}>f\right)$

Hence, the P-value is $P\text{-value}=P\left(F_{5,78}>22.60\right)$.

Here, the test statistic value, $f=22.60$, which is higher than the approximate critical value $F_{0.01;5,78}\approx 3.28$. In fact, the value $f=22.60$ is even higher than the approximate critical value $F_{0.001;5,78}\approx 4.63$.

That is, $f\left(=22.60\right)>F_{0.001;5,78}\left(\approx 4.63\right)>F_{0.01;5,78}\left(\approx 3.28\right)$.

Rejection rule:

If the P-value is less than the level of significance $\alpha$, such that $P\text{-value}<\alpha$, then reject the null hypothesis at level $\alpha$.

Conclusion:

Here, the P-value is less than 0.001 and thus, evidently, the significance level 0.01.

That is, $P\text{-value}<0.001<0.01$.

Thus, the decision is “reject the null hypothesis”.

Therefore, the data provides sufficient evidence to conclude that the effects of all the treatments are not similar.

Thus, the test suggests that there is dependence of the true average survival time on the injection regime, at $\alpha =0.01$.

To determine

Find the simultaneous CI for ${\mu }_1-\frac{1}{5}\left({\mu }_2+{\mu }_3+{\mu }_4+{\mu }_5+{\mu }_6\right)$, with a simultaneous confidence level of at least 98%.

Find the simultaneous CI for $\frac{1}{4}\left({\mu }_2+{\mu }_3+{\mu }_4+{\mu }_5\right)-{\mu }_6$, with a simultaneous confidence level of at least 98%.

Answer to Problem 21E

The simultaneous CI for ${\mu }_1-\frac{1}{5}\left({\mu }_2+{\mu }_3+{\mu }_4+{\mu }_5+{\mu }_6\right)$, with a simultaneous confidence level of at least 98% is $\underset{\_}{\left(-95.916,-38.884\right)}$.

The simultaneous CI for $\frac{1}{4}\left({\mu }_2+{\mu }_3+{\mu }_4+{\mu }_5\right)-{\mu }_6$, with a simultaneous confidence level of at least 98% is $\underset{\_}{\left(32.647,90.853\right)}$.

Explanation of Solution

Given info:

The $100\left(1-\alpha \right)\%$ simultaneous confidence intervals (CI) for k different parametric functions, using the same ANOVA data set, is at least $100\left(1-k\alpha \right)\%$.

Calculation:

Confidence interval for parametric functions:

The $100\left(1-\alpha \right)\%$ confidence interval for a function of ${\mu }_i$’s, $\theta ={\displaystyle \sum _i{c}_i{\mu }_i}$, where $c_i$’s are constants for $i=1,2,\mathrm{...},I$, is given as:

${\displaystyle \sum _i{c}_i{\overline{x}}_i}\pm t_{\frac{\alpha }{2},I\left(J-1\right)}\sqrt{\frac{\text{MSE}{\displaystyle \sum _i{c}_i^2}}{J}}$.

Here, MSE is the mean square of error.

Here, two parametric functions are considered, for which, simultaneous CI’s have to be calculated. Thus, $k=2$

Now, from part a, the level of significance $\alpha =0.01$.

Hence, the least simultaneous confidence level for each the two parametric functions would be:

$\begin{array}{c}100\left(1-k\alpha \right)\%=100\left(1-\left(2\times 0.01\right)\right)\%\\ =100\left(1-0.02\right)\%\\ =98\%.\end{array}$

Simultaneous CI for ${\mu }_1-\frac{1}{5}\left({\mu }_2+{\mu }_3+{\mu }_4+{\mu }_5+{\mu }_6\right)$:

The first parametric function is:

$\begin{array}{c}{\theta }_1={\mu }_1-\frac{1}{5}\left({\mu }_2+{\mu }_3+{\mu }_4+{\mu }_5+{\mu }_6\right)\\ ={\mu }_1-\frac{1}{5}{\mu }_2-\frac{1}{5}{\mu }_3-\frac{1}{5}{\mu }_4-\frac{1}{5}{\mu }_5-\frac{1}{5}{\mu }_6\\ ={\mu }_1-0.2{\mu }_2-0.2{\mu }_3-0.2{\mu }_4-0.2{\mu }_5-0.2{\mu }_6.\end{array}$

Compare this for the general expression $\theta ={\displaystyle \sum _i{c}_i{\mu }_i}$. Thus, here, $c_1=1$ and $c_2=c_3=c_4=c_5=c_6=-0.2$.

As a result,

$\begin{array}{c}{\displaystyle \sum _i{c}_i^2}={\left(1\right)}^2+{\left(-0.2\right)}^2+{\left(-0.2\right)}^2+{\left(-0.2\right)}^2+{\left(-0.2\right)}^2+{\left(-0.2\right)}^2\\ =1+0.04+0.04+0.04+0.04+0.04\\ =\mathrm{1.2.}\end{array}$

Use the available values ${\overline{X}}_{1.}=166$, ${\overline{X}}_{2.}=303$, ${\overline{X}}_{3.}=266$, ${\overline{X}}_{4.}=212$, ${\overline{X}}_{5.}=202$ and ${\overline{X}}_{6.}=184$ to compute ${\displaystyle \sum _i{c}_i{\overline{x}}_i}$:

$\begin{array}{c}{\displaystyle \sum _i{c}_i{\overline{x}}_i}=166-\left(0.2\times 303\right)-\left(0.2\times 266\right)-\left(0.2\times 212\right)-\left(0.2\times 202\right)-\left(0.2\times 184\right)\\ =166-60.6-53.2-42.4-40.4-36.8\\ =-\mathrm{67.4.}\end{array}$

The level of confidence is at least 98%. Thus,

$\begin{array}{c}100\left(1-\alpha \right)\%=98\%\\ 1-\alpha =0.98\\ \alpha =1-0.98\\ =\mathrm{0.02.}\end{array}$

Thus, the level of significance for the test is:

$\begin{array}{c}\frac{\alpha }{2}=\frac{0.02}{2}\\ =\mathrm{0.01.}\end{array}$

Thus, $t_{\frac{\alpha }{2};I\left(J-1\right)}=t_{0.01,78}$.

Critical value:

Software procedure:

Step by step procedure to the critical value using MINITAB software is given as,

• Choose Graph > Probability Distribution Plot > View Probability.
• In Distribution, select t and in Degrees of freedom enter 78.
• Go to Shaded Area, select Probability and Right Tail, enter probability value as 0.01.
• Click OK.

Output using MINITAB software is given as,

From the output, $t_{0.01,78}=2.375$.

Substitute the values in the expression for the confidence interval (CI):

$\begin{array}{c}{\displaystyle \sum _i{c}_i{\overline{x}}_i}\pm t_{\frac{\alpha }{2},I\left(J-1\right)}\sqrt{\frac{\text{MSE}{\displaystyle \sum _i{c}_i^2}}{J}}=-67.4\pm \left[2.375\times \sqrt{\frac{\left(1,681.83\right)\times \left(1.2\right)}{14}}\right]\\ =-67.4\pm \left[2.375\times \sqrt{\frac{2,018.196}{14}}\right]\\ =-67.4\pm \left[2.375\times \sqrt{144.1569}\right]\\ =-67.4\pm \left[2.375\times 12.007\right]\end{array}$

$\begin{array}{c}=-67.4\pm 28.516\\ =\left(-67.4-28.516,-67.4+28.516\right)\\ =\underset{\_}{\left(-95.916,-38.884\right)}.\end{array}$

Thus, the simultaneous CI for ${\mu }_1-\frac{1}{5}\left({\mu }_2+{\mu }_3+{\mu }_4+{\mu }_5+{\mu }_6\right)$, with a simultaneous confidence level of at least 98% is $\underset{\_}{\left(-95.916,-38.884\right)}$.

Simultaneous CI for $\frac{1}{4}\left({\mu }_2+{\mu }_3+{\mu }_4+{\mu }_5\right)-{\mu }_6$:

The second parametric function is:

$\begin{array}{c}{\theta }_2=\frac{1}{4}\left({\mu }_2+{\mu }_3+{\mu }_4+{\mu }_5\right)-{\mu }_6\\ =\frac{1}{4}{\mu }_2+\frac{1}{4}{\mu }_3+\frac{1}{4}{\mu }_4+\frac{1}{4}{\mu }_5-{\mu }_6\\ =0.25{\mu }_2+0.25{\mu }_3+0.25{\mu }_4+0.25{\mu }_5-{\mu }_6.\end{array}$

Compare this for the general expression $\theta ={\displaystyle \sum _i{c}_i{\mu }_i}$. Thus, here, $c_1=0$, $c_2=c_3=c_4=c_5=0.25$ and $c_6=-1$.

As a result,

$\begin{array}{c}{\displaystyle \sum _i{c}_i^2}=0+{\left(0.25\right)}^2+{\left(0.25\right)}^2+{\left(0.25\right)}^2+{\left(0.25\right)}^2+{\left(-1\right)}^2\\ =0.0625+0.0625+0.0625+0.0625+1\\ =\mathrm{1.25.}\end{array}$

Thus,

$\begin{array}{c}{\displaystyle \sum _i{c}_i{\overline{x}}_i}=\left(0\times 166\right)+\left(0.25\times 303\right)+\left(0.25\times 266\right)+\left(0.25\times 212\right)+\left(0.25\times 202\right)-\left(1\times 184\right)\\ =0+75.75+66.5+53+50.5-184\\ =\mathrm{61.75.}\end{array}$

Here, $t_{\frac{\alpha }{2};I\left(J-1\right)}=t_{0.01,78}=2.375$.

Substitute the values in the expression for the confidence interval (CI):

$\begin{array}{c}{\displaystyle \sum _i{c}_i{\overline{x}}_i}\pm t_{\frac{\alpha }{2},I\left(J-1\right)}\sqrt{\frac{\text{MSE}{\displaystyle \sum _i{c}_i^2}}{J}}=61.75\pm \left[2.375\times \sqrt{\frac{\left(1,681.83\right)\times \left(1.25\right)}{14}}\right]\\ =61.75\pm \left[2.375\times \sqrt{\frac{2,102.289}{14}}\right]\\ =61.75\pm \left[2.375\times \sqrt{150.1634}\right]\\ =61.75\pm \left[2.375\times 12.254\right]\end{array}$

$\begin{array}{c}=61.75\pm 29.103\\ =\left(61.75-29.103,61.75+29.103\right)\\ =\underset{\_}{\left(32.647,90.853\right)}.\end{array}$

Thus, the simultaneous CI for $\frac{1}{4}\left({\mu }_2+{\mu }_3+{\mu }_4+{\mu }_5\right)-{\mu }_6$, with a simultaneous confidence level of at least 98% is $\underset{\_}{\left(32.647,90.853\right)}$.