Ten different families are tested for the number of gallons of water a day they use before and after viewing a conservation video.Construct a 90% confidence interval for the mean of the
Differences. $$\begin{array} { | l | l | l | l | l | l | l | l | l | l | l | }
\hline \text { Before } & 33 & 33 & 38 & 33 & 35 & 35 & 40 & 40 & 40 & 31 \
\hline \text { After } & 34 & 28 & 25 & 28 & 35 & 33 & 31 & 28 & 35 & 33 \
\hline
\end{array}$$

Question

Ten different families are tested for the number of gallons of water a day they use before and after viewing a conservation video.Construct a 90% confidence interval for the mean of the
Differences. $$\begin{array} { | l | l | l | l | l | l | l | l | l | l | l | } 
\hline \text { Before } & 33 & 33 & 38 & 33 & 35 & 35 & 40 & 40 & 40 & 31 \
\hline \text { After } & 34 & 28 & 25 & 28 & 35 & 33 & 31 & 28 & 35 & 33 \
\hline
\end{array}$$

Quizplus · Accepted Answer

To construct a 90% confidence interval for the mean of the differences, first, calculate the differences between the "Before" and "After" values for each family. Then, find the mean ($\bar{x}$) and standard deviation (s) of these differences. Finally, use the t-distribution with n-1 degrees of freedom (where n is the number of differences) to find the critical value for a 90% confidence interval. The formula for the confidence interval is $\bar{x} \pm t_{\frac{\alpha}{2}} \cdot \frac{s}{\sqrt{n}}$.Differences: -1, 5, 13, 5, 0, 2, 9, 12, 5, -2. Mean ($\bar{x}$) = 4.8, Standard Deviation (s) = 4.39, n = 10.Using a t-table or calculator for a 90% confidence interval with 9 degrees of freedom gives a t-value around 1.833. Plugging into the formula gives:$$4.8 \pm 1.833 \cdot \frac{4.39}{\sqrt{10}} \approx (1.8, 7.8)$$Thus, the 90% confidence interval for the mean of the differences is approximately (1.8, 7.8).

Ten Different Families Are Tested for the Number of Gallons