Chapter 12, Problem 80E

(a)

To determine

To find: The P-value to conclude about the ANOVA model.

Answer to Problem 80E

Solution: The P-value is obtained as 0.000, which indicates that the mean age of the five different stores are different.

Explanation of Solution

Given: The data on the sample size, mean, and standard deviation of customers of five different stores are provided in the Exercise 12.6, which is shown in the below table.

Store	Sample size $\left(n\right)$	Mean $\left(\overline{x}\right)$	Sample Standard Deviation $\left(s\right)$
A	$n_1=50$	${\overline{x}}_1=38$	$s_1=8$
B	$n_2=50$	${\overline{x}}_2=48$	$s_2=13$
C	$n_3=50$	${\overline{x}}_3=42$	$s_3=11$
D	$n_4=50$	${\overline{x}}_4=28$	$s_4=7$
E	$n_5=50$	${\overline{x}}_5=35$	$s_5=10$

Calculation: The null and alternative hypothesis of the provided is set as below:

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

$H_a$: Not all ${\mu }_i$ ’s are equal

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	$I-1$	${{\displaystyle \sum n_i\left({\overline{x}}_i-\overline{x}\right)}}^2$	$MSG=\frac{SSG}{D{F}_G}$	$\frac{MSG}{MSE}$
Error	$N-I$	${\displaystyle \sum \left(n_i-1\right)s_i{}^2}$	$MSE=\frac{SSE}{D{F}_E}$
Total	$N-1$	${{\displaystyle \sum \left(x_{ij}-\overline{x}\right)}}^2$

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

By the data provided, the number of groups, total number of groups is:

$I=5$

And,

$\begin{array}{c}N=n_1+n_2+n_3+n_4+n_5\\ =50+50+50+50+50\\ =250\end{array}$

The degrees of freedom for groups are calculated as:

$\begin{array}{c}D{F}_G=I-1\\ =5-1\\ =4\end{array}$

The degrees of freedom of error are calculated as:

$\begin{array}{c}D{F}_E=N-I\\ =250-5\\ =245\end{array}$

Total degrees of freedom are:

$\begin{array}{c}\text{Degrees of freedom}=N-1\\ =250-1\\ =249\end{array}$

The pooled mean is calculated as:

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum n_i\times {\overline{x}}_i}}{{\displaystyle \sum n_i}}\\ =\frac{\left(50\times 38\right)+\left(50\times 48\right)+\left(50\times 42\right)+\left(50\times 28\right)+\left(50\times 35\right)}{50+50+50+50+50}\\ =38.2\end{array}$

The sum of squares for groups is calculated as:

$\begin{array}{c}SSG={{\displaystyle \sum n_i\left({\overline{x}}_i-\overline{x}\right)}}^2\\ =50\times \left({\left(38-38.2\right)}^2+{\left(48-38.2\right)}^2+{\left(42-38.2\right)}^2+{\left(28-38.2\right)}^2+{\left(35-38.2\right)}^2\right)\\ =11240\end{array}$

The sum of squares of error is calculated:

$\begin{array}{c}SSE={\displaystyle \sum \left(n_i-1\right)s_i{}^2}\\ =\left(50-1\right)\times \left(8^2+{13}^2+{11}^2+7^2+{10}^2\right)\\ =24647\end{array}$

The Mean Sum of Square of Groups is calculated as:

$\begin{array}{c}MSSG=\frac{SSG}{D{F}_G}\\ =\frac{11240}{4}\\ =2810\end{array}$

The Mean Sum of Square of Error is calculated as:

$\begin{array}{c}MSSE=\frac{SSE}{D{F}_E}\\ =\frac{24647}{245}\\ =100.6\end{array}$

The F-ratio is calculated as:

$\begin{array}{c}F=\frac{MSSG}{MSSE}\\ =\frac{2810}{100.6}\\ =27.93\end{array}$

Now the ANOVA table is:

Sources	Degree of freedom	Sum of square	Mean sum of square	F- ratio
Groups	4	11240	2810	27.93
Error	245	24647	100.6
Total	249

The $P-\text{value}$ for F-statistic is calculated by Excel formula =FDIST(x, deg_freedom1, deg_freedom2). The screenshot given below:

The obtained P-value is approximately 0.000, which is less than the significance level 0.05. So, the null hypothesis is rejected in the provided scenario.

Interpretation: The mean age of the customers at different coffee houses are different.

(b)

To determine

To test: The pairwise comparison using LSD method.

Answer to Problem 80E

Solution: The significant difference between the mean ages of the different coffee houses except the mean age of the store A and store E.

Explanation of Solution

Given: The data on the sample size, mean, and standard deviation of customers of five different stores are provided in the Exercise 12.6, which is shown in the below table.

Store	Sample size $\left(n\right)$	Mean $\left(\overline{x}\right)$	Sample Standard Deviation $\left(s\right)$
A	$n_1=50$	${\overline{x}}_1=38$	$s_1=8$
B	$n_2=50$	${\overline{x}}_2=48$	$s_2=13$
C	$n_3=50$	${\overline{x}}_3=42$	$s_3=11$
D	$n_4=50$	${\overline{x}}_4=28$	$s_4=7$
E	$n_5=50$	${\overline{x}}_5=35$	$s_5=10$

Calculation: The value of the test statistic for LSD method is obtained through the below mentioned formula:

$t_{ij}=\frac{{\overline{x}}_i-{\overline{x}}_j}{\sqrt{s_p^2\times \left(\frac{1}{n_i}+\frac{1}{n_j}\right)}}$

Where $s_p^2$ is the mean sum of square of error, ${\overline{x}}_i$ and ${\overline{x}}_j$ are the mean of i^th and j^th treatment and $n_i$ and $n_j$ are the sample size of the i^th and j^th groups, respectively.

The required calculation is shown in the below table.

ith Condition	jth Condition	Test statistic
Store A	Store B	$-4.985$
	Store C	$-1.994$
	Store D	4.985
	Store E	1.496
Store B	Store A	4.985
	Store C	2.991
	Store D	9.970
	Store E	6.481
Store C	Store A	1.994
	Store B	$-2.991$
	Store C	6.979
	Store D	3.490
Store D	Store A	$-4.985$
	Store B	$-9.970$
	Store C	$-6.979$
	Store E	$-3.490$
Store E	Store A	$-1.496$
	Store B	$-6.481$
	Store C	$-3.490$
	Store D	3.490

The critical value for the test is obtained from the Table D in the textbook. In the table, there is no value available for 245 degrees of freedom. So, the value corresponding to the 1000 degrees of freedom is considered corresponding to $\alpha =0.05$ for two-tailed test. The value is obtained as 1.962. If the value of the test statistic is greater than the critical value, then the difference between the pair of means is considered as significant.

From the obtained result, it can be determined that a significant difference is present between some of the pairs of means. The only pair of means where the difference is not significant is the means of store A and store E. There is significant difference present between the other pairs of means.

Conclusion: The significant difference between the mean ages of the different coffee houses except the mean age of the store A and store E.