A study involves testing whether or not the amount of caffeine consumed affected memory. Fifteen volunteers took part in this study. They were given three types of drink (type A,B and C) containing different levels of caffeine \((50 \mathrm{mg}, 100 \mathrm{mg}\), and \(150 \mathrm{mg}\),respectively). Volunteers were divided into three groups of five each and were assigned the drink groupwise. They were then given a memory test (In terms of number of words remembered from a list). The results are given in the following table:

\begin{tabular}{|c|c|c|}

\hline Group A (50 mg) & Group B (100 mg) & Group C (150 mg) \\

\hline 7 & 11 & 14 \\

\hline 8 & 14 & 12 \\

\hline 10 & 14 & 10 \\

\hline 12 & 12 & 16 \\

\hline 7 & 10 & 13 \\

\hline

\end{tabular}

At significance level of \(5 \%\), check whether the mean number of words remembered from the list by the participants belonging to the three groups are significantly different.

To determine whether the mean number of words remembered by participants in the three groups (A, B, and C) are significantly different, we will conduct an ANOVA (Analysis of Variance) test. This test is appropriate when comparing the means of three or more groups.

## Hypotheses Formulation

[

\begin{aligned}

&\begin{array}{|c|c|c|}

\hline \boldsymbol{A} & \boldsymbol{B} & \boldsymbol{C} \

\hline 7 & 11 & 14 \

\hline 8 & 14 & 12 \

\hline 10 & 14 & 10 \

\hline 12 & 12 & 16 \

\hline 7 & 10 & 13 \

\hline \sum \boldsymbol{A}=\mathbf{4 4} & \sum \boldsymbol{B}=\mathbf{6 1} & \sum \boldsymbol{C}=\mathbf{6 5} \

\hline

\end{array}\

&\begin{array}{|c|c|c|}

\hline \boldsymbol{A}^{\mathbf{2}} & \boldsymbol{B}^{\mathbf{2}} & \boldsymbol{C}^{\mathbf{2}} \

\hline 49 & 121 & 196 \

\hline 64 & 196 & 144 \

\hline 100 & 196 & 100 \

\hline 144 & 144 & 256 \

\hline 49 & 100 & 169 \

\hline \sum \boldsymbol{A}^{\mathbf{2}}=\mathbf{4 0 6} & \sum \boldsymbol{B}^{\mathbf{2}}=\mathbf{7 5 7} & \sum \boldsymbol{C}^{\mathbf{2}}= \mathbf{8 6 5} \

\hline

\end{array}

\end{aligned}

]

Data table

\begin{tabular}{|c|c|c|c|c|}

\hline Group & (A) & (B) & (C) & Total \

\hline (\mathrm{N}) & (n_1=5) & (n_2=5) & (n_3=5) & (n=15) \

\hline(\sum x_i) & (T_1=\sum x_1=44) & (T_2=\sum x_2=61) & (T_3=\sum x_3=65) & (\sum x=170) \

\hline(\sum x_i^2) & (\sum x_1^2=406) & (\sum x_2^2=757) & (\sum x_3^2=865) & (\sum x^2=2028) \

\hline Mean (\bar{x}_i) & (\bar{x}_1=8.8) & (\bar{x}_2=12.2) & (\bar{x}_3=13) & Overall (\bar{x}=11.3333) \

\hline Std Dev (S_i) & (S_1=2.1679) & (S_2=1.7889) & (S_3=2.2361) & \

\hline

\end{tabular}

Let (\mathrm{k}=) the number of different samples (=3)

[

n=n_1+n_2+n_3=5+5+5=15

]

Overall (\bar{x}=\frac{170}{15}=11.3333)

[

\begin{aligned}

& \sum x=T_1+T_2+T_3=44+61+65=170 \rightarrow(1) \

& \frac{\left(\sum x\right)^2}{n}=\frac{170^2}{15}=1926.6667 \rightarrow(2)

\end{aligned}

]

(\begin{aligned} & \sum \frac{T_i^2}{n_i}=\left(\frac{44^2}{5}+\frac{61^2}{5}+\frac{65^2}{5}\right)=1976.4 \rightarrow(3) \ & \sum x^2=\sum x_1^2+\sum x_2^2+\sum x_3^2=406+757+865=2028 \rightarrow(4)\end{aligned})

ANOVA:

Step-1: sum of squares between samples

[

\begin{aligned}

& \operatorname{SSB}=\left(\sum \frac{T_i^2}{n_i}\right)-\frac{\left(\sum x\right)^2}{n}=(3)-(2) \

& =1976.4-1926.6667 \

& =49.7333

\end{aligned}

]

Or

[

\begin{aligned}

& \operatorname{SSB}=\sum n_j \cdot\left(\bar{x}_j-\bar{x}\right)^2 \

& =5 \times(8.8-11.3333)^2+5 \times(12.2-11.3333)^2+5 \times(13-11.3333)^2 \

& =49.7333

\end{aligned}

]

Step-2 : sum of squares within samples

[

\begin{aligned}

& \operatorname{SSW}=\sum x^2-\left(\sum \frac{T_i^2}{n_i}\right)=(4)-(3) \

& =2028-1976.4 \

& =51.6

\end{aligned}

]

Step-3 : Total sum of squares

[

\begin{aligned}

& \text { SST }=\text { SSB }+ \text { SSW } \

& =49.7333+51.6 \

& =101.3333

\end{aligned}

]

Step-4 : variance between samples

[

\begin{aligned}

& \text { MSB }=\frac{\text { SSB }}{k-1} \

& =\frac{49.7333}{2} \

& =24.8667

\end{aligned}

]

Step-5 : variance within samples

[

\begin{aligned}

& \text { MSW }=\frac{\text { SSW }}{n-k} \

& =\frac{51.6}{15-3} \

& =\frac{51.6}{12} \

& =4.3

\end{aligned}

]

Step-6 : test statistic F for one way ANOVA test

[

\begin{aligned}

F & =\frac{\text { MSB }}{\text { MSW }} \

& =\frac{24.8667}{4.3} \

& =5.7829

\end{aligned}

]

the degree of freedom between samples

[

k-1=2

]

Now, degree of freedom within samples

[

n-k=15-3=12

]

p-value :

(p=F \operatorname{Dist}(F, d f 1, d f 2)=F \operatorname{Dist}(5.7829,2,12)=0.0174) (Using F Distribution calculator)

ANOVA table

\begin{tabular}{|c|c|c|c|c|c|}

\hline \begin{tabular}{c}

Source of \

Variation

\end{tabular} & \begin{tabular}{c}

Sum of \

Squares \

SS

\end{tabular} & df & \begin{tabular}{c}

Mean Squares \

MS

\end{tabular} & (\mathbf{F}) & (\boldsymbol{p})-value \

\hline \begin{tabular}{c}

Between \

samples

\end{tabular} & (S S_B=49.7333) & (k-1=2) & (\frac{49.7333}{2}=24.8667) & 5.7829 & 0.0174 \

\hline \begin{tabular}{c}

Within \

samples

\end{tabular} & (S S_W=51.6) & (n-k=12) & (\frac{51.6}{12}=4.3) & & \

\hline Total & (S S_T=101.3333) & (n-1=14) & & & \

\hline

\end{tabular}

(H_0) : There is no significant differentiating between samples

(H_1) : There is significant differentiating between samples

(F(2,12)) at 0.05 level of significance

[

=3.8853

]

As calculated (F=5.7829>3.8853)

So, (H_0) is rejected, Hence there is significant differentiating between samples

The calculated F-value (5.7829) is greater than the critical F-value (3.8853) at the 0.05 level of significance for the degrees of freedom ((2,12)). The (p)-value ((0.0174)) is also less than the significance level of 0.05 .

Interpretation