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Instruction 12-12
the Manager of the Purchasing Department of a Large

question 188

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Instruction 12-12
The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours) it takes to record a loan application.Data are collected from a sample of 30 days,and the number of applications recorded and completion time in hours is recorded.Below is the regression output:
$\begin{array}{l}\begin{array} { l r } \hline { \text { Regression Statistics } } \\\hline \text { Multiple R } & 0.9447 \\\text { R Square } & 0.8924 \\\text { Adjusted R } & 0.8886 \\\text { Square } & \\\text { Standard } & 0.3342 \\\text { Error } & 30 \\\text { Observations } & \\\hline\end{array}\\\text { ANOVA }\\\begin{array} { l r r r r r } \hline & d f & { \text { SS } } & { \text { MS } } & F & { \begin{array} { c } \text { Significance } \\F\end{array} } \\\hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm { E } - 15 \\\text { Residual } & 28 & 3.1282 & 0.1117 & & \\\text { Total } & 29 & 29.072 & & & \\\hline\end{array}\\\begin{array} { l r r r r r r } \hline & \text { Coefficients } & \begin{array} { c } \text { Standard } \\\text { Error }\end{array} & t \text { Stat } & P \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \begin{array} { l } \text { Intercept } \\\text { Applications } \\\text { Recorded }\end{array} & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\\hline\end{array}\end{array}$ Note: 4.3946E-15 is 4.3946 x 10^-15.
$Instruction 12-12 The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours) it takes to record a loan application.Data are collected from a sample of 30 days,and the number of applications recorded and completion time in hours is recorded.Below is the regression output: \begin{array}{l} \begin{array} { l r } \hline { \text { Regression Statistics } } \\ \hline \text { Multiple R } & 0.9447 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R } & 0.8886 \\ \text { Square } & \\ \text { Standard } & 0.3342 \\ \text { Error } & 30 \\ \text { Observations } & \\ \hline \end{array}\\ \text { ANOVA }\\ \begin{array} { l r r r r r } \hline & d f & { \text { SS } } & { \text { MS } } & F & { \begin{array} { c } \text { Significance } \\ F \end{array} } \\ \hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm { E } - 15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & & \\ \hline \end{array}\\ \begin{array} { l r r r r r r } \hline & \text { Coefficients } & \begin{array} { c } \text { Standard } \\ \text { Error } \end{array} & t \text { Stat } & P \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \begin{array} { l } \text { Intercept } \\ \text { Applications } \\ \text { Recorded } \end{array} & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \hline \end{array} \end{array} Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>. -Referring to Instruction 12-12,the estimated mean amount of time it takes to record one additional loan application is A) 0.4024 more hours. B) 0.0126 more hours. C) 0.0126 fewer hours. D) 0.4024 fewer hours.$ $Instruction 12-12 The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours) it takes to record a loan application.Data are collected from a sample of 30 days,and the number of applications recorded and completion time in hours is recorded.Below is the regression output: \begin{array}{l} \begin{array} { l r } \hline { \text { Regression Statistics } } \\ \hline \text { Multiple R } & 0.9447 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R } & 0.8886 \\ \text { Square } & \\ \text { Standard } & 0.3342 \\ \text { Error } & 30 \\ \text { Observations } & \\ \hline \end{array}\\ \text { ANOVA }\\ \begin{array} { l r r r r r } \hline & d f & { \text { SS } } & { \text { MS } } & F & { \begin{array} { c } \text { Significance } \\ F \end{array} } \\ \hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm { E } - 15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & & \\ \hline \end{array}\\ \begin{array} { l r r r r r r } \hline & \text { Coefficients } & \begin{array} { c } \text { Standard } \\ \text { Error } \end{array} & t \text { Stat } & P \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \begin{array} { l } \text { Intercept } \\ \text { Applications } \\ \text { Recorded } \end{array} & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \hline \end{array} \end{array} Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>. -Referring to Instruction 12-12,the estimated mean amount of time it takes to record one additional loan application is A) 0.4024 more hours. B) 0.0126 more hours. C) 0.0126 fewer hours. D) 0.4024 fewer hours.$
-Referring to Instruction 12-12,the estimated mean amount of time it takes to record one additional loan application is

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Instruction 12-12 the Manager of the Purchasing Department of a Large

Definitions:

Instruction 12-12
the Manager of the Purchasing Department of a Large