Chapter 9.2, Problem 16E

To determine

Construct and interpret a confidence interval for the true difference between the two population means.

Answer to Problem 16E

Solution:

The value of the margin of error is 10.1847 and 90% confidence interval for the true difference between the mean distance hit with the new model and the mean distance hit with the older model, is $\left(18.1153,38.4847\right)$.

Explanation of Solution

Given Information:

The number of balls by new bat or $n_1$ is 10 and the number of balls by old bat or $n_2$ is 10.

Hitting Distance (in Feet)
New Model	Old Model
235	200
240	210
253	231
267	218
243	210
237	209
250	210
241	229
251	234
248	231

It is assumed that the population variances must be equal.

Formula used:

The confidence interval for the difference between two population means for independent data sets is given by

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

Or

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

Where ${\overline{x}}_1\text{ and }{\overline{x}}_2$ are the two sample means,

$\left({\overline{x}}_1-{\overline{x}}_2\right)$ is the point estimate for the difference between the population means, ${\mu }_1-{\mu }_2$,

and margin of error of a confidence interval for the difference between two population means is given by

$E=t_{\frac{\alpha }{2}}\sqrt{\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

Where, $t_{\frac{\alpha }{2}}$ is the critical value for the level of confidence, $c=1-\alpha$, such that the area under the $t$-distribution with $n_1+n_2-2$ degrees of freedom to the right of $t_{\frac{\alpha }{2}}$ is equal to $\frac{\alpha }{2},$

Mean of the given observations is calculated as,

$\overline{x}=\frac{{\displaystyle \sum X_i}}{n}$

n is the total number of data given.

$s_d$ is the standard deviation of the sample observations is calculated as,

$s_d=\sqrt{{\frac{{\displaystyle \sum \left(X_i-\overline{x}\right)}}{n-1}}^2}$

$s_1$ and $s_2$ are the two sample standard deviations, and

$n_1$ and $n_2$ are the two sample sizes.

Calculation:

The margin of error for given populations is,

$E=t_{\frac{\alpha }{2}}\sqrt{\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$……$\left(1\right)$

For populations where the variances are assumed to be equal, the number of degrees of freedom is calculated by substituting the value 10 for $n_1$ and 10 for $n_2$ in the given formula,

$\begin{array}{rll}df& =& n_1+n_2-2\\ & =& 10+10-2\\ & =& 18\end{array}$

Since the level of confidence is 90%, then level of significance is given as,

$\begin{array}{rll}\alpha & =& 1-0.90\\ & =& 0.1\end{array}$

Substitute 0.1 for $\alpha$ we get,

$\begin{array}{rll}{t}_{\frac{\alpha }{2}}& =& t_{\frac{0.1}{2}}\\ & =& t_{0.05}\end{array}$

For the t-distribution with 18 degrees of freedom, at 0.90 confidence level the $t$- value is,

$t_{0.05}=2.101$

S.no	New Model, $x_1$	Old Model, $x_2$	$x_1-{\overline{x}}_1$	$x_2-{\overline{x}}_2$	${\left(x_1-{\overline{x}}_1\right)}^2$	${\left(x_2-{\overline{x}}_2\right)}^2$
1	235	200	-11.5	-18.2	132.25	331.24
2	240	210	-6.5	-8.2	42.25	67.24
3	253	231	6.5	12.8	42.25	163.84
4	267	218	20.5	-0.2	420.25	0.04
5	243	210	-3.5	-8.2	12.25	67.24
6	237	209	-9.5	-9.2	90.25	84.64
7	250	210	3.5	-8.2	12.25	67.24
8	241	229	-5.5	10.8	30.25	116.64
9	251	234	4.5	15.8	20.25	249.64
10	248	231	1.5	12.8	2.25	163.84
Total	${\displaystyle \sum x_1}$ =2465	${\displaystyle \sum x_2}$ =2182	${\displaystyle \sum \left(x_1-{\overline{x}}_1\right)}$ =0	${\displaystyle \sum \left(x_2-{\overline{x}}_2\right)}$ =1.137E-13	${{\displaystyle \sum \left(x_1-{\overline{x}}_1\right)}}^2$ =804.5	${{\displaystyle \sum \left(x_2-{\overline{x}}_2\right)}}^2$ =1311.6

For mean, substitute 2465 for ${\displaystyle \sum x_1}$ and 10 for $n_1$ in the given formula as,

$\begin{array}{rll}{\overline{x}}_1& =& \frac{{\displaystyle \sum x_1}}{n_1}\\ & =& \frac{2465}{10}\\ & =& 246.5\end{array}$

Similarly substitute 2182 for ${\displaystyle \sum x_2}$ and 10 for $n_2$ in the given formula as,

$\begin{array}{rll}{\overline{x}}_1& =& \frac{{\displaystyle \sum x_1}}{n_2}\\ & =& \frac{2182}{10}\\ & =& 218.2\end{array}$

For standard deviation, substitute 804.5 for ${\displaystyle \sum \left(x_1-{\overline{x}}_1\right)}$ and 10 for $n_1$ in the given formula as,

$\begin{array}{rll}{s}_1& =& \sqrt{\frac{{\displaystyle \sum {\left(x_1-{\overline{x}}_1\right)}^2}}{n-1}}\\ & =& \sqrt{\frac{804.5}{10-1}}\\ & =& 9.45\end{array}$

For standard deviation, substitute 1311.6 for ${\displaystyle \sum \left(x_2-{\overline{x}}_2\right)}$ and 10 for $n_2$ in the given formula as,

$\begin{array}{rll}{s}_2& =& \sqrt{\frac{{\displaystyle \sum {\left(x_1-{\overline{x}}_1\right)}^2}}{n-1}}\\ & =& \sqrt{\frac{1311.6}{10-1}}\\ & =& 12.07\end{array}$

Substitute the values 2.101 for $t_{0.05}$, 9.45 for $s_1$, 12.07 for $s_2$, 10 for $n_1$ and 10 for $n_2$ in equation $\left(1\right)$.

$\begin{array}{rll}E& =& 2.101\times \sqrt{\frac{\left(10-1\right){9.45}^2+\left(10-1\right){12.07}^2}{10+10-2}}\times \sqrt{\frac{1}{10}+\frac{1}{10}}\\ & =& 10.1847\end{array}$

The confidence interval is $\left(\left({\overline{x}}_1-{\overline{x}}_2\right)-E,\left({\overline{x}}_1-{\overline{x}}_2\right)+E\right)$, that is,

$\begin{array}{rll}\text{Confidence interval}& =& \left(\left(246.5-218.2\right)-10.1847,\left(246.5-218.2\right)+10.1847\right)\\ & =& \left(18.1153,38.4847\right)\end{array}$

Interpretation:

The 90% confidence interval for the true difference between the mean distance hit with the new model and the mean distance hit with the older model, ranges from 18.1153 to 38.4847.