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

question 153

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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:
$\begin{array}{lr}\hline {\text { Regression Statistics }} \\\hline \text { Multiple R } & 0.9447 \\\text { R Square } & 0.8924 \\\text { Adjusted R } & 0.8886 \\\text { Square } & \\\text { Standard } & 0.3342 \\\text { Error } & \\\text { Observations } & 30 \\\hline\end{array}$
ANOVA
$\begin{array}{lrrrrr}\hline & d f & { S S } & { M S } & F & \text { Significance } F \\\text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\\text { Residual } & 28 & 3.1282 & 0.1117 & & \\\text { Total } & 29 & 29.072 & & &\end{array}$
$\begin{array}{l}\begin{array}{lrrrrrr} & \text { Coefficients } & \begin{array}{c}\text { Standard } \\\text { Error }\end{array} & t \text { Stat } & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\\text { Applications } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143\end{array}\\\text { Recorded }\end{array}$ Note: 4.3946E-15 is 4.3946 x 10^-15.
$Instruction 12-12 The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours)it takes to record a loan application.Data are collected from a sample of 30 days,and the number of applications recorded and completion time in hours is recorded.Below is the regression output: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.9447 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R } & 0.8886 \\ \text { Square } & \\ \text { Standard } & 0.3342 \\ \text { Error } & \\ \text { Observations } & 30 \\ \hline \end{array} ANOVA \begin{array}{lrrrrr} \hline & d f & { S S } & { M S } & F & \text { Significance } F \\ \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & & \end{array} \begin{array}{l} \begin{array}{lrrrrrr} & \text { Coefficients } & \begin{array}{c} \text { Standard } \\ \text { Error } \end{array} & t \text { Stat } & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Applications } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \end{array}\\ \text { Recorded } \end{array} Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>. -Referring to Instruction 12-12,the model appears to be adequate based on the residual analyses.$ $Instruction 12-12 The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours)it takes to record a loan application.Data are collected from a sample of 30 days,and the number of applications recorded and completion time in hours is recorded.Below is the regression output: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.9447 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R } & 0.8886 \\ \text { Square } & \\ \text { Standard } & 0.3342 \\ \text { Error } & \\ \text { Observations } & 30 \\ \hline \end{array} ANOVA \begin{array}{lrrrrr} \hline & d f & { S S } & { M S } & F & \text { Significance } F \\ \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & & \end{array} \begin{array}{l} \begin{array}{lrrrrrr} & \text { Coefficients } & \begin{array}{c} \text { Standard } \\ \text { Error } \end{array} & t \text { Stat } & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Applications } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143 \end{array}\\ \text { Recorded } \end{array} Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>. -Referring to Instruction 12-12,the model appears to be adequate based on the residual analyses.$
-Referring to Instruction 12-12,the model appears to be adequate based on the residual analyses.

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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