Step 1

We are authorised to answer three subpart at a time, since you have not mentioned which part you are looking for, so we are answering the first three subpart, please re post your question separately for the remaining subpart.

Given:

\(n_{1} = 40, s_{1} = 0.76\)

\(n_{2} = 25, s_{2} = 0.64\)

Step 2

8.2

Hypothesis

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

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

Test statistics would be

\(F=\frac{S_{1}^{2}}{S_{2}^{2}}\)

\(=\frac{0.76^{2}}{0.64^{2}}\)

\(=1.41\)

Degree of freedom of numerator \(40 - 1 = 39\)

Degree of freedom of denominator \(25 - 1 = 24\)

P-value of the test \(= 0.3759\)

Therefore P-value is greater than 0.05, therefore it is fail to reject the null hypothesis.

Conclusion is that there is no significant difference between the populations.

Step 3

8.3

For this two sample t-test is used the reason is that population variances are equal.

Step 4

8.4

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

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

Degree of freedom

\(= n_{1} + n_{2} - 2\)

\(= 40 + 25 - 2\)

\(= 63\)

For two tailed test critical values \(\pm1.998\)

If t is less than -1.998 or greater than 1.998, fall to reject the null hypothesis.

Pooled standard deviation

\(S_{p}=\sqrt{\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{n_{1}+n_{2}-2}}\)

\(=\sqrt{\frac{(40-1)0.76^{2}+(25-1)0.64^{2}}{40+25-2}}\)

\(= 0.71666\)

And the t-statistics would be

\(t=\frac{\overline{x}_{1}-\overline{x}_{2}}{S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}\)

\(= \frac{6.8-6.56}{0.71666\times \sqrt{\frac{1}{40}+\frac{1}{25}}}\)

\(= 1.3136\)

By seeing the t value, it is not lying in the rejection region, therefore it is fail to reject the null hypothesis.

We are authorised to answer three subpart at a time, since you have not mentioned which part you are looking for, so we are answering the first three subpart, please re post your question separately for the remaining subpart.

Given:

\(n_{1} = 40, s_{1} = 0.76\)

\(n_{2} = 25, s_{2} = 0.64\)

Step 2

8.2

Hypothesis

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

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

Test statistics would be

\(F=\frac{S_{1}^{2}}{S_{2}^{2}}\)

\(=\frac{0.76^{2}}{0.64^{2}}\)

\(=1.41\)

Degree of freedom of numerator \(40 - 1 = 39\)

Degree of freedom of denominator \(25 - 1 = 24\)

P-value of the test \(= 0.3759\)

Therefore P-value is greater than 0.05, therefore it is fail to reject the null hypothesis.

Conclusion is that there is no significant difference between the populations.

Step 3

8.3

For this two sample t-test is used the reason is that population variances are equal.

Step 4

8.4

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

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

Degree of freedom

\(= n_{1} + n_{2} - 2\)

\(= 40 + 25 - 2\)

\(= 63\)

For two tailed test critical values \(\pm1.998\)

If t is less than -1.998 or greater than 1.998, fall to reject the null hypothesis.

Pooled standard deviation

\(S_{p}=\sqrt{\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{n_{1}+n_{2}-2}}\)

\(=\sqrt{\frac{(40-1)0.76^{2}+(25-1)0.64^{2}}{40+25-2}}\)

\(= 0.71666\)

And the t-statistics would be

\(t=\frac{\overline{x}_{1}-\overline{x}_{2}}{S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}\)

\(= \frac{6.8-6.56}{0.71666\times \sqrt{\frac{1}{40}+\frac{1}{25}}}\)

\(= 1.3136\)

By seeing the t value, it is not lying in the rejection region, therefore it is fail to reject the null hypothesis.