Chapter 13.1, Problem 31E

Confidence interval for the conditional mean: In Example 13.5, we constructed a 95% confidence interval for the slope ${\beta }_1$ in the model to predict the number of calories from the number of grams of fat. The 95% confidence interval is $4.8182<{\beta }_1<6.4605$. Let ${\mu }_y|15$ be the mean number of calories for food products containing 15 grams of fat, and let ${\mu }_y|20$ be the mean number of calories for food products containing 20 grams of fat. Construct a 95% confidence interval for the difference ${\mu }_y|20-{\mu }_y|15$.

To determine

To construct: a $95\%$ confidence interval for the difference of means

Answer to Problem 31E

  $15.541104<{\mu }_1-{\mu }_2<56.67889$

Explanation of Solution

Given information: the complete table.

Product	Fat(x)	Calories(y)	Product	Fat(x)	Calories(y)
$3$ muskerteers	$12.75$	$436$	Mr.Goodbar	$33.21$	$538$
kit kat	$25.99$	$518$	100 Grand	$19.33$	$468$
M & M Plain	$21.13$	$492$	Bay ruth	$21.6$	$459$
M & M Peanut	$26.13$	$515$	Bit O'Honey	$7.5$	$375$
Milky way	$17.23$	$456$	Butterfinger	$18.9$	$459$
Skittles	$4.37$	$405$	Oh henry	$23$	$462$
Snickers	$23.85$	$491$	Reeses'speices	$24.77$	$497$
Starburst	$8.36$	$408$	Toot sie roll	$3.31$	$387$
Twix	$24.85$	$502$	Twizzlers	$2.32$	$350$

Formula Used: confidence interval for the different intervals between two means, Independent samples.

  ${\overline{x}}_1-{\overline{x}}_2-t_{\alpha /2}\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}<{\mu }_1-{\mu }_2<{\overline{x}}_1-{\overline{x}}_2+t_{\alpha /2}\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}$

first we need to select the two samples which has fifteen grams of fat and twenty grams of fat .The products which do not have both is not going to be in the two samples.

	Sample   $1$			Sample   $2$
Product	$20$ grams Fat(x)	Calories(y)	Product	$15$ grams Fat(x)	Calories(y)
kit kat	$25.99$	$518$	100 Grand	$19.33$	$468$
M & M Plain	$21.13$	$492$	Butterfinger	$18.9$	$459$
M & M Peanut	$26.13$	$515$	Milky way	$17.23$	$456$
Snickers	$23.85$	$491$
Twix	$24.85$	$502$
Mr.Goodbar	$33.21$	$538$
Bay ruth	$21.6$	$459$
Oh henry	$23$	$462$
Reeses'speices	$24.77$	$497$

The sample means are calculated as shown.

Sample $1$ mean:

  $\begin{array}{l}{\overline{y}}_1=\frac{{\displaystyle \sum _{i=1}^n{y}_i}}{n}\\ {\overline{y}}_1=\frac{(518+492+\mathrm{........}+497)}{9}\\ {\overline{y}}_1=497.11\end{array}$

Sample $2$ mean:

  $\begin{array}{l}{\overline{y}}_2=\frac{{\displaystyle \sum _{i=1}^n{y}_i}}{n}\\ {\overline{y}}_2=\frac{(468+459+456)}{3}\\ {\overline{y}}_2=461\end{array}$

The sample variances are calculated as shown.

Sample $1$ variance:

  $\begin{array}{l}{s}_1{}^2=\frac{{\displaystyle \sum _{i=1}^n{(y_i-\overline{y})}^2}}{n-1}\\ s_1{}^2=\frac{{(518-497.11)}^2+{(492-497.11)}^2+\mathrm{........}+{(497-497.11)}^2}{9-1}\\ s_1{}^2=\frac{5200.9}{8}\\ s_1{}^2=650.11\\ s_1=25.497\end{array}$

Sample $2$ variance:

  $\begin{array}{l}{s}_2{}^2=\frac{{\displaystyle \sum _{i=1}^n{(y_i-\overline{y})}^2}}{n-1}\\ s_2{}^2=\frac{{(468-461)}^2+{(459-461)}^2+\mathrm{........}+{(456-461)}^2}{3-1}\\ s_2{}^2=\frac{78}{2}\\ s_2{}^2=39\\ s_2=6.24\end{array}$

Since both samples have $12$ products, we find the t value for degrees of freedom $10$ and $95\%$ confidence interval. So, the t value is $2.228$ .

  

Now we can calculate the $95\%$ confidence interval for the difference of means

  $\begin{array}{l}{\overline{y}}_1-{\overline{y}}_2-t_{\alpha /2}\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}<{\mu }_1-{\mu }_2<{\overline{y}}_1-{\overline{y}}_2+t_{\alpha /2}\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}\\ 497.11-461-t_{\alpha /2}\sqrt{\frac{650.11}{9}+\frac{39}{3}}<{\mu }_1-{\mu }_2<497.11-461+t_{\alpha /2}\sqrt{\frac{650.11}{9}+\frac{39}{3}}\\ 36.11-t_{\alpha /2}9.232<{\mu }_1-{\mu }_2<36.11+t_{\alpha /2}9.232\\ 36.11-(2.228)(9.232)<{\mu }_1-{\mu }_2<36.11+(2.228)(9.232)\\ 36.11-20.568896<{\mu }_1-{\mu }_2<36.11+20.568896\\ 36.11-20.568896<{\mu }_1-{\mu }_2<36.11+20.56889\\ 15.541104<{\mu }_1-{\mu }_2<56.67889\end{array}$