Chapter 24, Problem 13E

To determine

To create and interpret a $95\%$ confidence interval for the difference in mean sugar content and check the necessary assumptions and conditions.

Answer to Problem 13E

The confidence interval is $(32.49,40.80)$ .

Explanation of Solution

The data is given in the question for the sugar content in children and adult’s cereals.

Thus, by calculating the mean and standard deviation of the data using excel function we get,

The formulas used in excel to find the mean and standard deviations are as follows:

1. AVERAGE( $\text{number1, }\left[\text{number2}\right]$, ...): Returns the average (arithmetic mean) of the arguments.
2. STDEV( $\text{number1, }\left[\text{number2}\right]$,...): Estimates standard deviation based on a sample.

Children cereal
Average	=AVERAGE(C26:C44)
S.D.	=STDEV(C26:C44)

And the result will be as:

Children cereal
Average	46.8
S.D.	6.418

Now, for the adult’s cereal we have:

Adult cereal
Average	=AVERAGE(D26:D53)
S.D.	=STDEV(D26:D53)

And the result will be:

Adult cereal
Average	10.154
S.D.	7.612

Thus, we have,

  $\begin{array}{l}{\overline{x}}_1=46.8\\ {\overline{x}}_2=10.154\\ n_1=19\\ n_2=28\\ s_1=6.418\\ s_2=7.612\\ \alpha =0.05\end{array}$

Thus, the conditions are as follows:

Random condition: It is satisfied because we assume that the cereals of children and adults were randomly selected.

Independent condition: It is satisfied because the groups of children and adultsare independent.

Normal condition: It is satisfied because we assume that the cereals of children and adultsare approximately normally distributed.

Therefore, all the conditions are met.

Now let us determine the degrees of freedom:

  $df=43$

Therefore, the t -value will be:

  $t=2.017$

Thus, the confidence interval will be:

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_2)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(46.8-10.154)-2.017\times \sqrt{\frac{{6.418}^2}{19}+\frac{{7.612}^2}{28}}\\ =32.49\\ ({\overline{x}}_1-{\overline{x}}_2)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(46.8-10.154)+2.017\times \sqrt{\frac{{6.418}^2}{19}+\frac{{7.612}^2}{28}}\\ =40.80\end{array}$

The confidence interval is $(32.49,40.80)$ .