Chapter 9.2, Problem 15E

To determine

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

Answer to Problem 15E

Solution:

The value of the margin of error is 0.7880 and 99% confidence interval for the true difference between the mean numbers of cavities for children whose diets are high in sugar and those whose diets are low in sugar, is $\left( 0.4,2.0\right)$.

Explanation of Solution

Given Information:

The mean number of cavities whose diet is high in sugar or. ${\overline{x}}_1$ is 3.0 and mean number of cavities whose diet is low in sugar or ${\overline{x}}_2$ is 1.8. The standard deviation of children whose diet is high in sugar or. $s_1$ is 0.7 and standard deviation of children whose diet is low in sugar or. $s_2$ is 0.6. The number of children whose diet is high in sugar or $n_1$ is 10 and number of children whose diet is low in sugar or. $n_2$ is 12.

It is assumed that 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 ${\overline{x}}_1$ 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 $\frac{\alpha }{2},$

$s_1$ and $s_2$ are the two population 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 as,

$df=n_1+n_2-2$

Substitute 10 for $n_1$ and 12 for $n_2$ in the above equation.

$\begin{array}{rll}df& =& 10+12-2\\ & =& 20\end{array}$

Since the level of confidence is 99%, then the level of significance is given as,

$\begin{array}{rll}\alpha & =& 1-0.99\\ & =& 0.01\end{array}$

Substitute 0.01 for $\alpha$ we get,

$\begin{array}{rll}{t}_{\frac{\alpha }{2}}& =& t_{\frac{0.01}{2}}\\ & =& t_{0.005}\end{array}$

For the $t$-distribution with 20 degrees of freedom, at 0.99 confidence level the $t$- value is,

$t_{0.005}=2.845$

Substitute the values 2.845 for $t_{0.005}$, 0.7 for $s_1$, 0.6 for $s_2$, 10 for $n_1$ and 12 for $n_2$ in equation $\left(1\right)$.

$\begin{array}{rll}E& =& 2.845\times \sqrt{\frac{\left(10-1\right){0.7}^2+\left(12-1\right){0.6}^2}{10+12-2}}\times \sqrt{\frac{1}{10}+\frac{1}{12}}\\ & =& 2.845\times \sqrt{\frac{\left(9\right)(0.49)+\left(11\right)(0.36)}{20}}\times \sqrt{\frac{10+12}{120}}\\ & =& 2.845\times \sqrt{\frac{4.41+3.96}{20}}\times \sqrt{\frac{22}{120}}\\ & =& 2.845\times \sqrt{0.4185}\times \sqrt{0.1833}\\ & =& 2.845\times 0.6469\times 0.4281\\ & =& 2.845\times 0.2770\\ E& =& 0.7880\end{array}$

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,

$\begin{array}{rll}\text{Confidence interval}& =& \left(\left(3.0-1.8\right)-0.7880,\left(3.0-1.8\right)+0.7880\right)\\ & =& \left(0.412,1.988\right)\\ & \approx & \left(0.4,2.0\right)\end{array}$

Interpretation:

The 99% confidence interval for the true difference between the mean numbers of cavities for children whose diets are high in sugar and the children, whose diets are low in sugar, ranges from $0.4$ to 2.0. Thus a diet low in sugary foods can reduce the number of cavities in children.