Chapter 12, Problem 46E

Section 1:

To determine

To find: The pooled two-sample t-statistic for provided scenario.

Answer to Problem 46E

Solution: The pooled two-sample t- statistic is 4.17.

Explanation of Solution

Given: The data provided in exercise 7.68 is

Group	Sample size $\left(n\right)$	Mean $\left(\overline{x}\right)$	Sample Standard Deviation $\left(s\right)$
LC	$n_1=32$	${\overline{x}}_1=47.3$	${\sigma }_1=28.3$
LF	$n_2=33$	${\overline{x}}_2=19.3$	${\sigma }_2=25.8$

Calculation: The two-sample pooled t-statistic is calculated by the formula is

$t=\frac{\left(\overline{x_1}-\overline{x_2}\right)}{s_p\times \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled sample standard deviation is calculated as

$\begin{array}{c}{s}_p=\sqrt{\frac{\left(n_1-1\right)\times s_1^2+\left(n_2-1\right)\times s_2^2}{n_1+n_2-2}}\\ =\sqrt{\frac{\left(32-1\right)\times {\left(28.3\right)}^2+\left(33-1\right)\times {\left(25.8\right)}^2}{32+33-2}}\\ =\sqrt{\frac{24827.6+21300.48}{63}}\\ =27.06\end{array}$

Therefore, substituting all values in the above mentioned formula, the value of required t- statistic is

$\begin{array}{c}t=\frac{\left(47.3-19.3\right)}{27.06\times \sqrt{\frac{1}{32}+\frac{1}{33}}}\\ =\frac{28}{6.714}\\ =4.17\end{array}$

Section 2:

To determine

To find: An ANOVA table.

Answer to Problem 46E

Solution: The ANOVA table is

Sources	Degree of freedom	Sum of square	Mean sum of square	F- ratio
Groups	1	12737.01	12737.01	17.4
Error	63	46128.07	732.19
Total	64

The analysis show that the means are not equal.

Explanation of Solution

Calculation: The null and alternative hypotheses are set as below:

$H_0:{\mu }_1={\mu }_2$

$H_a:{\mu }_1\ne {\mu }_2$

The formulation of ANOVA table will be by the formula provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	$I-1$	${{\displaystyle \sum n_i\left({\overline{x}}_i-\overline{x}\right)}}^2$	$MSSG=\frac{SSG}{D{F}_G}$	$\frac{MSSG}{MSSE}$
Error	$N-I$	${\displaystyle \sum \left(n_i-1\right)s_i{}^2}$	$MSSE=\frac{SSE}{D{F}_E}$
Total	$N-1$

Here, N is the total number of samples, I is the number of cases, $D{F}_G$ is the degrees of freedom for the groups, and $D{F}_E$ is the degrees of freedom for the error.

By the data provided, the total number of groups is

$I=2$

And,

$\begin{array}{c}N=n_1+n_2\\ =32+33\\ =65\end{array}$

The degree of freedom for groups is calculated as

$\begin{array}{c}D{F}_G=I-1\\ =2-1\\ =1\end{array}$

The degree of freedom of error is calculated as

$\begin{array}{c}D{F}_E=N-I\\ =65-2\\ =63\end{array}$

Total degree of freedom is

$\begin{array}{c}\text{Degrees of freedom}=N-1\\ =65-1\\ =64\end{array}$

The pooled mean is calculated as

$\begin{array}{c}\overline{\overline{x}}=\frac{{\displaystyle \sum n_i\times {\overline{x}}_1}}{{\displaystyle \sum n_i}}\\ =\frac{\left(32\times 47.3\right)+\left(33\times 19.3\right)}{32+33}\\ =33.08\end{array}$

The sum of squares for groups is calculated as

$\begin{array}{c}SSG={{\displaystyle \sum n_i\left({\overline{x}}_i-\overline{\overline{x}}\right)}}^2\\ =\left(32\times {\left(47.3-33.08\right)}^2\right)+\left(33\times {\left(19.3-33.08\right)}^2\right)\\ =6470.69+6266.32\\ =12737.01\end{array}$

The sum of squares of error is calculated as

$\begin{array}{c}SSE={\displaystyle \sum \left(n_i-1\right)s_i{}^2}\\ =\left(\left(32-1\right)\times {\left(28.3\right)}^2\right)+\left(\left(33-1\right)\times {\left(25.8\right)}^2\right)\\ =\text{24827}\text{.59}+\text{21300}\text{.48}\\ =46128.07\end{array}$

The mean sum of square of groups is calculated as

$\begin{array}{c}MSSG=\frac{SSG}{D{F}_G}\\ =\frac{12737.01}{1}\\ =12737.01\end{array}$

The mean sum of square of error is calculated as

$\begin{array}{c}MSSE=\frac{SSE}{D{F}_E}\\ =\frac{46128.07}{63}\\ =732.19\end{array}$

The F-ratio is calculated as

$\begin{array}{c}F=\frac{MSSG}{MSSE}\\ =\frac{12737.01}{732.19}\\ =17.4\end{array}$

Now the ANOVA table is

Sources	Degree of freedom	Sum of square	Mean sum of square	F- ratio
Groups	1	12737.01	12737.01	17.4
Error	63	46128.07	732.19
Total	64

The critical value for the test is calculated using the Excel. The screenshot of the used formula is shown below:

The critical value is obtained as 3.99, which is smaller than the calculated value. So, the null hypothesis will be rejected at 5% significance level.

Section 3:

To determine

To explain: Whether $F=t^2$ or not.

Answer to Problem 46E

Solution: Yes, the F- statistic is equal to square of t- statistic.

Explanation of Solution

The values $F=17.4$ and $t=4.17$ are obtained in previous parts.

The calculation can be done as

$\begin{array}{c}F=t^2\\ ={\left(4.17\right)}^2\\ =17.4\end{array}$

Therefore, it is verified that $F=t^2$ whenever there are only two groups for one way ANOVA test.