Chapter 7.4, Problem 8P

(a)

To determine

Identify the distribution that ${\overline{x}}_1-{\overline{x}}_2$ follows if ${\sigma }_1$ and ${\sigma }_2$ are known.

Explanation of Solution

Requirements:

• When the distribution of $x_1$ (population 1) and $x_2$ (population 2) has the normal distribution with known population standard deviations ${\sigma }_1$, ${\sigma }_2$ then sampling distribution of ${\overline{x}}_1-{\overline{x}}_2$ has the normal distribution for any sample sizes of $n_1,n_2$. The z distribution is used.
• When the distribution of $x_1$ (population 1) or $x_2$ (population 2) are not normally distributed with known population standard deviations ${\sigma }_1$, ${\sigma }_2$ then sampling distribution of ${\overline{x}}_1-{\overline{x}}_2$ has the normal distribution for sample sizes of $n_1\ge 30,n_2\ge 30$.

The variable ${\overline{x}}_1-{\overline{x}}_2$ has the normal distribution with mean ${\mu }_1-{\mu }_2$ and standard deviation $\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$.

Two independent mound-shaped distributions are considered in the study. A random sample of size $n_1=36$ is taken from first distribution with sample mean ${\overline{x}}_1=15$. Also, a random sample of size $n_2=40$ is taken from second distribution with sample mean ${\overline{x}}_1=14$. The values of ${\sigma }_1,{\sigma }_2$ are known in the study.

The distributions of population are mounded shaped. The size of first sample and second samples is greater than 30. This shows that, the variable ${\overline{x}}_1-{\overline{x}}_2$ has normal distribution.

(b)

To determine

Find the 95% confidence interval for ${\mu }_1-{\mu }_2$.

Answer to Problem 8P

The 95% confidence interval for ${\mu }_1-{\mu }_2$ is $-0.580<{\mu }_1-{\mu }_2<2.580$.

Explanation of Solution

Confidence interval:

The confidence interval for ${\mu }_1-{\mu }_2$ when both ${\sigma }_1$ and ${\sigma }_2$ are known is,

$\left({\overline{x}}_1-{\overline{x}}_2\right)-E<{\mu }_1-{\mu }_2<\left({\overline{x}}_1-{\overline{x}}_2\right)+E$

In the formula, ${\sigma }_1$ and ${\sigma }_2$ are population standard deviations of populations 1 and 2, ${\overline{x}}_1$ and ${\overline{x}}_2$ are sample means from populations 1 and 2, $n_1$ and $n_2$ are sample sizes of population 1 and 2, $E=z_c\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$, c is confidence level, and $z_c$ is the critical value for confidence level c based on the standard normal distribution.

The confidence level is 95%.

Critical value:

Use the Appendix II: Tables, Table 5 (b): Confidence interval, Critical Values $z_c$.

• In the level of confidence c column, locate the value of 0.95, or 95%.
• The corresponding Critical value $z_c$ is 1.96.

The value of $z_c$ is 1.96.

Substitute 36 for $n_1$, 40 for $n_2$, 3 for ${\sigma }_1$, 4 for ${\sigma }_2$, 1.96 for $z_c$ in the margin of error formula (E).

$\begin{array}{c}E=1.96\sqrt{\frac{3^2}{36}+\frac{4^2}{40}}\\ =1.96\times \sqrt{0.65}\\ =1.96\times 0.806\\ =1.580\end{array}$

The margin of error E is 1.580.

Substitute 15 for ${\overline{x}}_1$, 14 for ${\overline{x}}_2$, 1.580 for E in the confidence formula.

$\begin{array}{c}\left(15-14\right)-1.580<{\mu }_1-{\mu }_2<\left(15-14\right)+1.580\\ 1-1.580<{\mu }_1-{\mu }_2<1+1.580\\ -0.580<{\mu }_1-{\mu }_2<2.580\end{array}$

Hence, the 95% confidence interval for ${\mu }_1-{\mu }_2$ is $-0.580<{\mu }_1-{\mu }_2<2.580$.

(c)

To determine

Identify the distribution that ${\overline{x}}_1-{\overline{x}}_2$ follows if ${\sigma }_1$ and ${\sigma }_2$ are not known.

Find the degrees of freedom.

Answer to Problem 8P

The degrees of freedom are 35.

Explanation of Solution

Calculation:

The degrees of freedom formula is,

$d.f.=smaller\left(n_1-1,n_2-1\right)$

In the formula $n_1$ and $n_2$ are sample sizes of population 1 and 2.

Requirements:

• When the distribution of $x_1$ (population 1) and $x_2$ (population 2) has the normal distribution or mounded shaped with unknown population standard deviations ${\sigma }_1$, ${\sigma }_2$ then sampling distribution of ${\overline{x}}_1-{\overline{x}}_2$ has the normal distribution for any sample sizes of $n_1,n_2$. The student’s t distribution is used.
• When the distribution of $x_1$ (population 1) or $x_2$ (population 2) are not normally distributed with unknown population standard deviations ${\sigma }_1$, ${\sigma }_2$ then sampling distribution of ${\overline{x}}_1-{\overline{x}}_2$ has the normal distribution for sample sizes of $n_1\ge 30,n_2\ge 30$. Also, ${\sigma }_1$ is replaced with $s_1$ and ${\sigma }_2$ is replaced with $s_2$.

The variable ${\overline{x}}_1-{\overline{x}}_2$ has the normal distribution with mean ${\mu }_1-{\mu }_2$ and standard deviation $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$.

Two independent mound-shaped distributions are considered in the study. A random sample of size $n_1=20$ is taken from first distribution with sample mean ${\overline{x}}_1=12$. Also, a random sample of size $n_2=25$ is taken from second distribution with sample mean ${\overline{x}}_1=14$. The values of ${\sigma }_1,{\sigma }_2$ are not known in the study.

The population distributions are mound-shaped and the population standard deviations are not known. Also, the size of first sample and second samples is greater than 30. This shows that, the variable ${\overline{x}}_1-{\overline{x}}_2$ has student’s t distribution.

Substitute 36 for $n_1$, 40 for $n_2$ in the degrees of freedom formula.

$\begin{array}{c}d.f.=smaller\left(36-1,40-1\right)\\ =smaller\left(35,39\right)\\ =35\end{array}$

Hence, the degrees of freedom are 35.

(d)

To determine

Find the 95% confidence interval for ${\mu }_1-{\mu }_2$.

Answer to Problem 8P

The 95% confidence interval for ${\mu }_1-{\mu }_2$ is $-0.636<{\mu }_1-{\mu }_2<2.636$.

Explanation of Solution

Confidence interval:

The confidence interval for ${\mu }_1-{\mu }_2$ when both ${\sigma }_1$ and ${\sigma }_2$ are unknown is,

$\left({\overline{x}}_1-{\overline{x}}_2\right)-E<{\mu }_1-{\mu }_2<\left({\overline{x}}_1-{\overline{x}}_2\right)+E$

In the formula, $s_1$ and $s_2$ are sample standard deviations of populations 1 and 2, ${\overline{x}}_1$ and ${\overline{x}}_2$ are sample means from populations 1 and 2, $n_1$ and $n_2$ are sample sizes of population 1 and 2, $E=t_c\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$, c is confidence level, and $t_c$ is the critical value for confidence level c.

The confidence level is 95%.

Critical value:

Use the Appendix II: Tables, Table 6: Critical Values for Student’s t Distribution:

• In d.f. column locate the value 35.
• In c row of locate the value 0.95.
• The intersecting value of row and columns is 2.030.

The critical value is 2.030.

Substitute 36 for $n_1$, 40 for $n_2$, 3 for $s_1$, 4 for $s_2$, 2.030 for $t_c$ in the margin of error formula (E).

$\begin{array}{c}E=2.030\sqrt{\frac{3^2}{36}+\frac{4^2}{40}}\\ =2.030\times \sqrt{0.65}\\ =2.030\times 0.806\\ =1.636\end{array}$

The margin of error E is 1.636.

Substitute 15 for ${\overline{x}}_1$, 14 for ${\overline{x}}_2$, 1.636 for E in the confidence formula.

$\begin{array}{c}\left(15-14\right)-1.636<{\mu }_1-{\mu }_2<\left(15-14\right)+1.636\\ 1-1.636<{\mu }_1-{\mu }_2<1+1.636\\ -0.636<{\mu }_1-{\mu }_2<2.636\end{array}$

Hence, the 95% confidence interval for ${\mu }_1-{\mu }_2$ is $-0.636<{\mu }_1-{\mu }_2<2.636$.

(e)

To determine

Find the 95% confidence interval for ${\mu }_1-{\mu }_2$ using degrees of freedom based on Satterthwaite’s approximation.

Answer to Problem 8P

The 95% confidence interval for ${\mu }_1-{\mu }_2$ using degrees of freedom based on Satterthwaite’s approximation is $-0.607<{\mu }_1-{\mu }_2<2.607$.

Explanation of Solution

Confidence interval:

Use Ti 83 calculator to obtain the confidence interval as follows:

• Select STAT, take the arrow to the TESTS menu, and then ‘0’ numbered key.
• Select Stats under Inpt.
• Enter ${\overline{x}}_1$ as 15, $S_{x_1}$ as 3, and $n_1$ as 36.
• Enter ${\overline{x}}_2$ as 14, $S_{x_2}$ as 4, and $n_2$ as 40.
• Enter C-Level as 0.95
• Select No under Pooled.
• Click Calculated.

Output using Ti 83 calculator is given below:

From Ti 83 calculator output, the confidence interval is $\left(-0.607,2.607\right)$.

Hence, the 95% confidence interval for ${\mu }_1-{\mu }_2$ using degrees of freedom based on Satterthwaite’s approximation is $-0.607<{\mu }_1-{\mu }_2<2.607$.

(f)

To determine

Identify whether there is 95% confident that ${\mu }_1$ is larger than ${\mu }_2$.

Explanation of Solution

All the confidence intervals calculated have both negative and positive values. The 95% confidence interval calculated for difference of means $\left({\mu }_1-{\mu }_2\right)$ includes negative and positive values, then it cannot be determined whether ${\mu }_1-{\mu }_2>0$ or ${\mu }_1-{\mu }_2<0$. The relation between two population means cannot be either ${\mu }_1<{\mu }_2$ or ${\mu }_1>{\mu }_2$. This shows that, there would be 95% sure that there is no difference between the average of first population $\left({\mu }_1\right)$ and average of second population $\left({\mu }_2\right)$. Also, ${\mu }_1$ cannot be larger than ${\mu }_2$.

Hence, there cannot be 95% confidence that ${\mu }_1$ is larger than ${\mu }_2$.