Chapter 7.2, Problem 97E

(a)

To determine

To find: The degrees of freedom using the formula as well as using software.

Answer to Problem 97E

Solution: The degrees of freedom is 137.

Explanation of Solution

Given: Consider the provided data $n_1=115$ and $n_2=220$, for drill and blast workers, ${\overline{x}}_1=18.0$ and $s_1=7.8$, also for outdoor concrete workers ${\overline{x}}_2=6.5$ and $s_2=3.4$.

Calculation: The degrees of freedom for the two sample t test using pooled method can be calculated as:

$\begin{array}{c}df=\frac{{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}^2}{\frac{1}{n_1-1}{\left(\frac{s_1^2}{n_1}\right)}^2+\frac{1}{n_2-1}{\left(\frac{s_2^2}{n_2}\right)}^2}\\ =\frac{{\left(\frac{{7.8}^2}{115}+\frac{{3.4}^2}{220}\right)}^2}{\frac{1}{115-1}{\left(\frac{{7.8}^2}{115}\right)}^2+\frac{1}{220-1}{\left(\frac{{3.4}^2}{220}\right)}^2}\\ =137\end{array}$

The degrees of freedom using software can be obtained by using the below mentioned steps.

Step 1: Go to Stat > Basic Statistics > 2-Sample t.

Step 2: Choose the option of ‘Summarized data’ from the dropdown and put the data of sample size, sample mean and sample standard deviation of both the samples.

Step 3: Substitute the value of confidence level, test difference and alternative with 95.0, 0, and not equal respectively.

Step 4: Click OK twice.

The degrees of freedom is obtained as 137 in Minitab output.

Interpretation: The degrees of freedom is obtained as 137 by using the manual method and software.

(b)

To determine

To find: The pooled standard deviation and the comparison of the pooled standard deviation with the standard deviation of each group.

Answer to Problem 97E

Solution: The poled standard deviation is 5.33 which is closer to the sample standard deviation o the second group.

Explanation of Solution

Given: Consider the provided data $n_1=115$ and $n_2=220$, for drill and blast workers, ${\overline{x}}_1=18.0$ and $s_1=7.8$, also for outdoor concrete workers ${\overline{x}}_2=6.5$ and $s_2=3.4$.

Calculation: The below steps are followed in Minitab software to calculate the pooled standard deviation.

Step 1: Go to Stat > Basic Statistics > 2-Sample t.

Step 2: Choose the option of ‘Summarized data’ from the dropdown and put the data of sample size, sample mean and sample standard deviation of both the samples. And select the option “Assume equal variances”.

Step 3: Substitute the value of confidence level, test difference and alternative with 95.0, 0, and not equal respectively.

Step 4: Click OK twice.

Interpretation: The pooled standard deviation is obtained as 5.33. The obtained pooled standard deviation is closer to the sample standard deviation of the second group when the large sample is taken.

(c)

To determine

To find: The standard error of the difference of means without assuming equal variances and with assuming equal variance and compare both values.

Answer to Problem 97E

Solution: The standard error of the difference of means without assuming equal variances is 0.7626 whereas the standard error for the pooled case is 0.6133.

Explanation of Solution

Calculation: The standard error of the difference of the mean, when assumption of the equal variances is not made, can be calculated as:

$\begin{array}{c}S{E}_1=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\sqrt{\frac{{7.8}^2}{115}+\frac{{3.4}^2}{220}}\\ =0.7626\end{array}$

The standard error of the difference of the mean, when assumption of the equal variances is made, can be calculated as:

$\begin{array}{c}S{E}_2=s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\\ =5.33\times \sqrt{\frac{1}{115}+\frac{1}{220}}\\ =0.6133\end{array}$

Interpretation: The standard error for the cases where the assumption of the equal variances is not made and assumption of the equal variances is made are obtained as 0.7626 and 0.6133.

(d)

To determine

To test: The hypothesis test of difference in population mean and determine the value of confidence interval.

Answer to Problem 97E

Solution: There is a sufficient evidence at $\alpha =0.05$ to reject the null hypothesis and the confidence interval is obtained as $\left(10.293,12.707\right)$.

Explanation of Solution

Given: According to the exercise 7.77, there is a significant difference in the mean exposure to dust between the two groups at 5% significance level. Consider the provided data $n_1=115$ and $n_2=220$, for drill and blast workers, ${\overline{x}}_1=18.0$ and $s_1=7.8$, also for outdoor concrete workers ${\overline{x}}_2=6.5$ and $s_2=3.4$.

Calculation: To test hypothesis for the difference of population means proceed as follows:

First of all, define the null and alternate hypotheses for a two-tailed test as:

$\begin{array}{c}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1\ne {\mu }_2\end{array}$

Here, the alternate hypothesis is two-sided as the provided problem does not provide any specific information regarding the group, which is more exposed to dust than the other one.

To obtain the value of the test statistic, Minitab is used. Below mentioned steps are followed to obtain the result.

Step 1: Go to Stat > Basic Statistics > 2-Sample t.

Step 2: Choose the option of ‘Summarized data’ from the dropdown and put the data of sample size, sample mean and sample standard deviation of both the samples. And select the option “Assume equal variances”.

Step 3: Substitute the value of confidence level, test difference and alternative with 95.0, 0, and not equal respectively.

Step 4: Click OK twice.

The value of the test statistic, degrees of freedom and the P-value is obtained as 18.74, 333 and 0.000 respectively. Moreover, the 95% confidence interval is obtained as $\left(10.293,12.707\right)$.

Conclusion: There is adequate evidence to reject the null hypothesis because P-value of 0.000 is less than the level of significance, that is, $\alpha =0.05$. Therefore, it can be concluded that ${\mu }_1\ne {\mu }_2$, and it can be claimed there is a significant difference in the mean exposure to dust between the two groups. The result is same with the result obtained in the exercise 7.77.

(e)

To determine

To find: The degrees of freedom using the formula as well as using software, the pooled standard deviation and the comparison of the pooled standard deviation with the standard deviation of each group, the standard error of the difference of means without assuming equal variances and with assuming equal variance and compare both values, and the hypothesis test of difference in population mean and determine the value of confidence interval.

Answer to Problem 97E

Solution: The degrees of freedom is 121.

The poled standard deviation is 1.734 which is closer to the sample standard deviation o the second group.

The standard error of the difference of means without assuming equal variances is 0.2653 whereas the standard error for the pooled case is 0.1995.

There is a sufficient evidence at $\alpha =0.05$ to reject the null hypothesis and the confidence interval is obtained as $\left(4.508,5.292\right)$.

Explanation of Solution

Given: According to the exercise 7.78, there is a significant difference in the mean exposure to dust between the two groups at 5% significance level. Consider the provided data $n_1=115$ and $n_2=220$, for drill and blast workers, ${\overline{x}}_1=6.3$ and $s_1=2.8$, also for outdoor concrete workers ${\overline{x}}_2=1.4$ and $s_2=0.7$.

Calculation: The degrees of freedom for the two sample t test using pooled method can be calculated as:

$\begin{array}{c}df=\frac{{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}^2}{\frac{1}{n_1-1}{\left(\frac{s_1^2}{n_1}\right)}^2+\frac{1}{n_2-1}{\left(\frac{s_2^2}{n_2}\right)}^2}\\ =\frac{{\left(\frac{{2.8}^2}{115}+\frac{{0.7}^2}{220}\right)}^2}{\frac{1}{115-1}{\left(\frac{{2.8}^2}{115}\right)}^2+\frac{1}{220-1}{\left(\frac{{0.7}^2}{220}\right)}^2}\\ =121.50\end{array}$

The degrees of freedom using software can be obtained by using the below mentioned steps.

Step 1: Go to Stat > Basic Statistics > 2-Sample t.

Step 2: Choose the option of ‘Summarized data’ from the dropdown and put the data of sample size, sample mean and sample standard deviation of both the samples.

Step 3: Substitute the value of confidence level, test difference and alternative with 95.0, 0, and not equal respectively.

Step 4: Click OK twice.

The degrees of freedom is obtained as 121 in Minitab output.

The below steps are followed in Minitab software to calculate the pooled standard deviation.

Step 1: Go to Stat > Basic Statistics > 2-Sample t.

Step 2: Choose the option of ‘Summarized data’ from the dropdown and put the data of sample size, sample mean and sample standard deviation of both the samples. And select the option “Assume equal variances”.

Step 3: Substitute the value of confidence level, test difference and alternative with 95.0, 0, and not equal respectively.

Step 4: Click OK twice.

The pooled standard deviation is obtained as 1.734.

The standard error of the difference of the mean, when assumption of the equal variances is not made, can be calculated as:

$\begin{array}{c}S{E}_1=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\sqrt{\frac{{2.8}^2}{115}+\frac{{0.7}^2}{220}}\\ =0.2653\end{array}$

The standard error of the difference of the mean, when assumption of the equal variances is made, can be calculated as:

$\begin{array}{c}S{E}_2=s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\\ =1.734\times \sqrt{\frac{1}{115}+\frac{1}{220}}\\ =0.1995\end{array}$

To test hypothesis for the difference of population means proceed as follows:

First of all, define the null and alternate hypotheses for a two-tailed test as:

$\begin{array}{c}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1\ne {\mu }_2\end{array}$

Here, the alternate hypothesis is two-sided as the provided problem does not provide any specific information regarding the group, which is more exposed to dust than the other one.

To obtain the value of the test statistic, Minitab is used. Below mentioned steps are followed to obtain the result.

Step 1: Go to Stat > Basic Statistics > 2-Sample t.

Step 2: Choose the option of ‘Summarized data’ from the dropdown and put the data of sample size, sample mean and sample standard deviation of both the samples. And select the option “Assume equal variances”.

Step 3: Substitute the value of confidence level, test difference and alternative with 95.0, 0, and not equal respectively.

Step 4: Click OK twice.

The value of the test statistic, degrees of freedom and the P-value is obtained as 24.56, 333 and 0.000 respectively. Moreover, the 95% confidence interval is obtained as $\left(4.508,5.292\right)$.

Interpretation: The degrees of freedom is obtained as 121 by using the manual method and software. The pooled standard deviation is obtained as 1.734. The obtained pooled standard deviation is closer to the sample standard deviation of the first group when the large sample is taken. The standard error for the cases where the assumption of the equal variances is not made and assumption of the equal variances is made are obtained as 0.2653 and 0.1995.

There is adequate evidence to reject the null hypothesis because P-value of 0.000 is less than the level of significance, that is, $\alpha =0.05$. Therefore, it can be concluded that ${\mu }_1\ne {\mu }_2$, and it can be claimed there is a significant difference in the mean exposure to dust between the two groups. The result is same with the result obtained in the exercise 7.78.