TABLE 13-12
the Manager of the Purchasing Department of a Large

TABLE 13-12
The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output:
$$\begin{array} { l c } { \text { Regression } \text { Statistics } } \\\hline \text { Multiple R } & 0.9947 \\\text { R Square } & 0.8924 \\\text { Adjusted R Square } & 0.8886 \\\text { Standard Error } & 0.3342 \\\text { Observations } & 30 \\\hline\end{array}$$ $$\begin{array}{l}\text { ANOVA }\\\begin{array} { l c c c c c } & d f & \text { SS } & \text { MS } & F & \text { Significance } F \\\text { Regression } & 1 & 25.9438&25 .9438 & 232.2200&4 .3946 \mathrm { E } - 15 \\\text { Residual } & 28 & 3.12820 .1117 & \\\text { Total } & 29 & 29.072 \\\hline\end{array}\end{array}$$ $$\begin{array} { l r r r r r r } \hline & \text { Coefficients } & \text { Standard Enror } & t \text { Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\\text { Processed } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm { E } 15 & 0.0109 & 0.0143\end{array}$$ $\begin{array}{rrrrrrr} & \text { Coefficients } & \text { Standard Enor } & \text { t Stat } & p \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Invoices } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555\end{array}$

-Referring to Table 13-12, the 90% confidence interval for the average change in the amount of time needed as a result of processing one additional invoice is

A) narrower than [0.0109, 0.0143].
B) wider than [0.0109, 0.0143].
C) narrower than [0.1492, 0.6555].
D) wider than [0.1492, 0.6555].