Chapter 11.CR, Problem 4CR

To determine

Hypothesis Testing using Analysis of Variance i.e. ANOVA.

Answer to Problem 4CR

Solution:

The null hypothesis is rejected and it is concluded that at least one of the tables is not generating the same mean revenue, at the 0.05 level of significance and the required ANOVA table is,

	SS	df	MS	F
Treatments (T)	$\text{166534735}$	3	555111578.33	67.05
Error (E)	$13245702$	16	827856.38
Total	179780437	19

Explanation of Solution

Given:

The amount of revenue (in whole dollars) received during one particular three-hour period is recorded at each table each night for five nights.

Revenue (in Dollars)
Table 1	Table 2	Table 3	Table 4
8534	9821	10, 542	7367
7845	8997	9982	8021
8901	7905	8934	9034
9371	8923	9344	6703
7782	6675	8745	8432

Formula Used:

Grand Mean is the weighted mean of the $k$ sample means, one from each of the $k$ populations, equivalent to the mean of all the sample data combined, given by

$\overline{\overline{x}}=\frac{{\displaystyle \sum _{i=1}^k{n}_i{\overline{x}}_i}}{{\displaystyle \sum _{i=1}^k{n}_i}}$

Sum of Squares among Treatments (SST) is the measures the variation between the sample means and the grand mean, given by,

$\text{SST=}{\displaystyle \sum _{i=1}^k{n}_i{\left({\overline{x}}_i-\overline{\overline{x}}\right)}^2}$

Sum of Squares for Error (SSE) is the measures the variation in the sample data resulting from the variability within each sample,

$\text{SSE=}{\displaystyle \sum _{j=1}^{n_1}{\left(x_{1j}-{\overline{x}}_1\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(x_{2j}-{\overline{x}}_2\right)}^2}+\mathrm{...}+{\displaystyle \sum _{j=1}^{n_k}{\left(x_{kj}-{\overline{x}}_k\right)}^2}$

Total Variation, it is the sum of the squared deviations from the grand mean for all of the data values in each sample, given by

$\begin{array}{rllllll}\text{Total Variation=}{\displaystyle \sum _j^{}}& =& 1n_1{\left(x_{1j}-\overline{\overline{x}}\right)}^2+{\displaystyle \sum _j^{}}& =& 1n_2{\left(x_{2j}-\overline{\overline{x}}\right)}^2+\mathrm{...}+{\displaystyle \sum _j^{}}& =& 1n_k{\left(x_{kj}-\overline{\overline{x}}\right)}^2\\ & =& \text{SST + SSE}\end{array}$

Mean Square for Treatments (MST) found by dividing the sum of squares among treatments by its degrees of freedom, given by

$\text{MST=}\frac{\text{SST}}{\text{DFT}}\text{with DFT=}k-1$

Mean Square for Error (MSE) found by dividing the sum of squares for error by its degrees of freedom, given by

$\text{MSE=}\frac{\text{SSE}}{\text{DFE}}\text{with DFE=}{n}_T-k$

Test Statistic for an ANOVA Test: Used when independent, simple random samples are taken from populations with variances that are unknown and assumed to be equal, where all of the $k$ population distributions are approximately normal, given by,

$F=\frac{\text{MST}}{\text{MSE}}\text{with }d{f}_1=\text{DFT=}k-1\text{ and }d{f}_2=\text{DFE=}{n}_T-k$

Rejection Region for ANOVA Tests: Reject the null hypothesis, $H_0$ when $F\ge F_{\alpha }$.

ANOVA Table is formed as,

	SS	df	MS	F
Treatments (T)	SST	DFT	$\frac{\text{SST}}{\text{DFT}}$	$\frac{\text{MST}}{\text{MSE}}$
Error (E)	SSE	DFE	$\frac{\text{SSE}}{\text{DFE}}$
Total	SST+SSE	DFT+DFE

Calculation:

The grand Mean is,

$\overline{\overline{x}}=\frac{{\displaystyle \sum _{i=1}^k{n}_i{\overline{x}}_i}}{{\displaystyle \sum _{i=1}^k{n}_i}}$

From the table, substitute the following values as,

$\begin{array}{rll}\overline{\overline{x}}& =& \frac{\left(\left(5\times 42433\right)+\left(5\times 42321\right)+\left(5\times 47547\right)+\left(5\times 39557\right)\right)}{\left(5+5+5+5\right)}\\ & =& 42964.5\end{array}$

Sum of Squares among Treatments (SST) is given as,

$\text{SST=}{\displaystyle \sum _{i=1}^k{n}_i{\left({\overline{x}}_i-\overline{\overline{x}}\right)}^2}$

From the table, substitute the following values as,

$\begin{array}{rll}\text{SST}& =& \left[5\times {\left(42433-42964.5\right)}^2+\mathrm{...}+5\times {\left(39557-42964.5\right)}^2\right]\\ & =& \text{166534735}\end{array}$

Sum of Squares for Error (SSE) is given as,

$\text{SSE=}{\displaystyle \sum _{j=1}^{n_1}{\left(x_{1j}-{\overline{x}}_1\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(x_{2j}-{\overline{x}}_2\right)}^2}+\mathrm{...}+{\displaystyle \sum _{j=1}^{n_k}{\left(x_{kj}-{\overline{x}}_k\right)}^2}$

Substitute the following values as,

$\begin{array}{rll}\text{SSE}& =& \left[{\left(8534-8486.6\right)}^2+\mathrm{...}+{\left(7782-8486.6\right)}^2\right]+\mathrm{...}+\left[{\left(7367-7911.4\right)}^2+\mathrm{...}+{\left(8432-7911.4\right)}^2\right]\\ & =& 13245702\end{array}$

The total Variation is given as,

$\text{Total Variation}=\text{SST + SSE}$

Substitute the following values as,

$\begin{array}{rll}\text{Total Variation}& =& \text{SST + SSE}\\ & =& 166534735+13245702\\ & =& 179780437\end{array}$

The mean Square for Treatments (MST) is,

$\begin{array}{rll}\text{MST }& =& \frac{166534735}{3}\\ & =& \text{55511578}\text{.33}\end{array}$

with degree of freedom as $4-1$ or 3.

The mean Square for Error (MSE) is,

$\begin{array}{rll}\text{MSE }& =& \frac{13245702}{16}\\ & =& \text{827856}\text{.38}\end{array}$

with degree of freedom as $20-4$ or 16.

The test Statistic for an ANOVA Test:

$\begin{array}{rll}F& =& \frac{55511578.33}{827856.38}\\ & =& \text{67}\text{.05 with }d{f}_1& =& \text{DFT=3 and }d{f}_2& =& \text{DFE=16 }\end{array}$

with degree of freedoms as 3 and 16.

Let Table 1, Table 2, and Table 3 and Table 4 be the population 1, 2, 3 and 4 respectively and let ${\text{μ}}_{\text{1}}{\text{, μ}}_{\text{2}}{\text{, μ}}_{\text{3}}{\text{ and μ}}_4$ be the mean revenue for tables 1, 2, 3 and 4 respectively.

Let the null and alternative hypothesis for this test is

${\text{H}}_{\text{0}}{\text{: μ}}_{\text{1}}{\text{= μ}}_{\text{2}}{\text{= μ}}_{\text{3}}={\text{μ}}_{{}_4}$

H₁: At least one mean revenue is different from others

Level of significance is 0.05, thus $\text{α = 0}\text{.05}$. The tabulated F-value for $\left(3,16\right)$ degrees of freedom is $3.24$. Since ${\text{F>F}}_{\alpha }$ i.e. $67.05>3.24$, null hypothesis is rejected. Thus, it is concluded that at least one of the tables is not generating the same mean revenue.