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A Local Grocery Store Wants to Predict the Daily Sales

question 103

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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
r20.762n7R0.873k 1  Std. Error 11.547  Dep. Var. Sales \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
 Source SSdfMSFp-value  Regression 2,133.333312,133.333316.00.0103 Residual 666.66675133.3333 Total 2,800.00006\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}

 Regression output \text { Regression output }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad Confidence interval \text { Confidence interval }
 Variables Coefficients  std. error t(df=5)  p-value 95%95% upper  lower  Intercep 63.33337.96827.948.000542.850583.8162 Advertising 6.66671.66674.000.0103\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}
-What are the limits of the 99% prediction interval of the daily sales in dollars of an individual grocery store that has spent $3000 on advertising expenditures? The distance value for this particular prediction is reported as .164.

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