The objective is to check whether the mean test scores of the two groups significantly differ at the 0.05 level.

Let \(\mu_{1}\) denote the mean of the participants participated in the survey after studying for one hour continuously.

Let \(\mu_{2}\) denote the mean of the participants participated in the survey after studying for three twenty-minute sessions.

The null hypothesis is that there is no difference between the mean test of scores between the two groups. That is,

\(H_{0}\ :\ \mu_{1} = mu_{2}\)

The alternative hypothesis is that there is a difference between the mean test of the score between the two groups. That is,

\(H_{1}\ :\ \mu_{1}\ \neq\ \mu_{2}\)

The level of significance is,

\(\alpha = 0.05\)

Group 1:

Sample mean, \(\bar{x}_{1}= 75\)

Sample variance, \(S^{\frac{2}{1}} = 120\)

Sample Size, \(n1 = 12\)

Group 2:

Sample mean, \(\bar{x}_{2} = 86\)

Sample variance, \(S^{\frac{2}{2}} = 100\)

Sample Size, \(n2 = 12\)

\(t=\frac{(\bar{x}_{1}\ -\ \bar{x}_{2})\ -\ (\mu_{1}\ -\ \mu_{2})}{\sqrt{S^{\frac{2}{p}}\left(\frac{1}{n}\ 1\ +\ \frac{1}{n}\ 2\right)}}\)

The formula for pooled variance is,

\(S^{\frac{2}{p}}=\frac{(n1\ -\ 1)S^{\frac{2}{1}}\ +\ (n2\ -\ 1)S^{\frac{2}{2}}}{n1\ +\ n2\ -\ 2}\)

\(=\frac{(12\ -\ 1)120\ +\ (12\ -\ 1)100}{12\ +\ 12\ -\ 2}\)

\(=\frac{11(120)\ +\ 11(100)}{24\ -\ 2}\)

\(=\frac{1320\ +\ 1100}{22}\)

\(= \frac{2420}{22}\)

\(= 110\)

The test statistic is computed below.

\(t=\frac{(\bar{x}_{1}\ -\ \bar{x}_{2})\ -\ (\mu_{1}\ -\ \mu_{2})}{\sqrt{S^{\frac{2}{p}}\left(\frac{1}{n}\ 1\ +\ \frac{1}{n}\ 2\right)}}\)

\(=\frac{75\ -\ 86\ -\ 0}{\sqrt{110\left(\frac{1}{12}\ +\ \frac{1}{12}\right)}}\)

\(=\frac{-11}{\sqrt{110\left(0.08333\ +\ 0.08333\right)}}\)

\(=\frac{-11}{\sqrt{110\left(166667\right)}}\)

\(=\frac{-11}{\sqrt{18.33333}}\)

\(=\frac{-11}{4.281744}\)

\(=\ -2.569047\)

The degrees of freedom is,

\(df = n1\ +\ n2\ -\ 2 = 12\ +\ 12\ -\ 2 = 24\ -\ 2 = 22\)

Computation of p value:

from the alternative hypothesis,it is clear that the test is a two- tailed test.

t represents the test statistic and df is the degrees of freedom p value

\(= t\ dist (t,\ df\ tails)\)

\(= t\ dist(2.569047,\ 22,\ 2)\)

\(- 0.017503\)

Decision rule:

Reject the null hypothesis if p value is less than the level of significance.

That is, p value \(<\ 0.05\)</span>

Conclusion:

Here the p value is 0.017503 is less than 0.05. So, reject the null hypothesis and conclude that there is a significant difference between the means of the two groups.

Let \(\mu_{1}\) denote the mean of the participants participated in the survey after studying for one hour continuously.

Let \(\mu_{2}\) denote the mean of the participants participated in the survey after studying for three twenty-minute sessions.

The null hypothesis is that there is no difference between the mean test of scores between the two groups. That is,

\(H_{0}\ :\ \mu_{1} = mu_{2}\)

The alternative hypothesis is that there is a difference between the mean test of the score between the two groups. That is,

\(H_{1}\ :\ \mu_{1}\ \neq\ \mu_{2}\)

The level of significance is,

\(\alpha = 0.05\)

Group 1:

Sample mean, \(\bar{x}_{1}= 75\)

Sample variance, \(S^{\frac{2}{1}} = 120\)

Sample Size, \(n1 = 12\)

Group 2:

Sample mean, \(\bar{x}_{2} = 86\)

Sample variance, \(S^{\frac{2}{2}} = 100\)

Sample Size, \(n2 = 12\)

\(t=\frac{(\bar{x}_{1}\ -\ \bar{x}_{2})\ -\ (\mu_{1}\ -\ \mu_{2})}{\sqrt{S^{\frac{2}{p}}\left(\frac{1}{n}\ 1\ +\ \frac{1}{n}\ 2\right)}}\)

The formula for pooled variance is,

\(S^{\frac{2}{p}}=\frac{(n1\ -\ 1)S^{\frac{2}{1}}\ +\ (n2\ -\ 1)S^{\frac{2}{2}}}{n1\ +\ n2\ -\ 2}\)

\(=\frac{(12\ -\ 1)120\ +\ (12\ -\ 1)100}{12\ +\ 12\ -\ 2}\)

\(=\frac{11(120)\ +\ 11(100)}{24\ -\ 2}\)

\(=\frac{1320\ +\ 1100}{22}\)

\(= \frac{2420}{22}\)

\(= 110\)

The test statistic is computed below.

\(t=\frac{(\bar{x}_{1}\ -\ \bar{x}_{2})\ -\ (\mu_{1}\ -\ \mu_{2})}{\sqrt{S^{\frac{2}{p}}\left(\frac{1}{n}\ 1\ +\ \frac{1}{n}\ 2\right)}}\)

\(=\frac{75\ -\ 86\ -\ 0}{\sqrt{110\left(\frac{1}{12}\ +\ \frac{1}{12}\right)}}\)

\(=\frac{-11}{\sqrt{110\left(0.08333\ +\ 0.08333\right)}}\)

\(=\frac{-11}{\sqrt{110\left(166667\right)}}\)

\(=\frac{-11}{\sqrt{18.33333}}\)

\(=\frac{-11}{4.281744}\)

\(=\ -2.569047\)

The degrees of freedom is,

\(df = n1\ +\ n2\ -\ 2 = 12\ +\ 12\ -\ 2 = 24\ -\ 2 = 22\)

Computation of p value:

from the alternative hypothesis,it is clear that the test is a two- tailed test.

t represents the test statistic and df is the degrees of freedom p value

\(= t\ dist (t,\ df\ tails)\)

\(= t\ dist(2.569047,\ 22,\ 2)\)

\(- 0.017503\)

Decision rule:

Reject the null hypothesis if p value is less than the level of significance.

That is, p value \(<\ 0.05\)</span>

Conclusion:

Here the p value is 0.017503 is less than 0.05. So, reject the null hypothesis and conclude that there is a significant difference between the means of the two groups.