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

Question

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

Quizplus · Accepted Answer

To estimate the daily average store sales based on $3000 advertising expenditures, we first calculate the predicted sales using the regression equation. The regression equation is given by the intercept and the coefficient for advertising. The intercept is 63.3333 (in thousands of dollars) and the coefficient for advertising is 6.6667.Predicted sales = Intercept + (Coefficient for Advertising × Advertising Expenditure)= 63.3333 + (6.6667 × 3) [since the advertising expenditure is $3000 or 3 in thousands]= 63.3333 + 20 = 83.3333 (in thousands of dollars)To calculate the 95% confidence interval for the prediction, we use the standard error of the forecast (SEF) and the t-value for the desired confidence level. However, the exact calculation for the SEF and the application of the t-value is not directly provided in the question. Instead, we are given a "distance value" of .164, which seems to be related to the calculation of the confidence interval around the prediction.Given the predicted value (83.3333) and the distance value (.164), we calculate the confidence interval as follows:Lower bound = Predicted value - Distance valueUpper bound = Predicted value + Distance valueHowever, the distance value provided does not directly translate to the bounds given in the options. The correct approach involves adjusting the predicted sales by the distance value, but the distance value seems to be a part of the calculation that involves the standard error of the prediction and the critical value from the t-distribution. Without the exact method to apply the distance value, we focus on the calculation based on the regression output and the provided advertising expenditure.Given the misunderstanding with the distance value, we look for an option that reasonably reflects a confidence interval around a prediction for $3000 in advertising expenditures. The correct calculation involves more detailed steps including the prediction error and the t-distribution for the confidence interval, which are not fully detailed in the provided information.Option C) $71,312 to $95,356 seems to be a reasonable estimate for the confidence interval around the predicted sales of 83.3333 thousand dollars (or $83,333.30), considering typical confidence interval calculations around a prediction. However, without the exact application of the distance value, this explanation relies on interpreting the provided data and understanding that the confidence interval should encompass the predicted value based on the given advertising expenditure.

A Local Grocery Store Wants to Predict the Daily Sales