A Local Grocery Store Wants to Predict the Daily Sales

Multiple Choice

A local grocery store wants to predict the daily sales in dollars. The manager believes that the amount of newspaper advertising significantly affects the store sales. The manager randomly selects 7 days of data consisting of daily grocery store sales (in thousands of dollars) and advertising expenditures (in thousands of dollars) . The Excel/Mega-Stat output given below summarizes the results of fitting a simple linear regression model using this data.
Regression Analysis
$\begin{array}{rc}\mathrm{r}^{2} 0.762 & \mathrm{n} 7 \\\mathrm{R} 0.873 & \mathrm{k} \text { 1 } \\\text { Std. Error 11.547 } & \text { Dep. Var. Sales }\end{array}$
ANOVA
table
$\begin{array}{rrrrrr}\hline \text { Source } & S S & d f & M S & F & p \text {-value } \\\hline \text { Regression } & 2,133.3333 & 1 & 2,133.3333 & 16.00 & .0103 \\\text { Residual } & 666.6667 & 5 & 133.3333 & & \\\hline \text { Total } & 2,800.0000 & 6 & & &\\\hline\end{array}$

$\text { Regression output }$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\text { Confidence interval }$
$\begin{array}{rrrrrrrr}\hline\text { Variables} & \text { Coefficients } & \text { std. error } & t(d f=5) & \text { p-value } & 95 \% &95 \% \text { upper } \\& & & & & \text { lower } & \\\hline \text { Intercep } & 63.3333 & 7.9682 & 7.948 & .0005 & 42.8505 & 83.8162 \\\text { Advertising } & 6.6667 & 1.6667 & 4.000 & .0103 & &\end{array}$
-Determine a 95% confidence interval estimate of the daily average store sales based on $3000 advertising expenditures? The distance value for this particular prediction is reported as .164.

A) $64,496 to $102.170
B) $33,108 to $133,558
C) $71,312 to $95,356
D) $51,314 to $115,353
E) $42,851 to $83,816