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

question 34

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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.
$Instruction 12-12 The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours) it takes to record a loan application.Data are collected from a sample of 30 days,and the number of applications recorded and completion time in hours is recorded.Below is the regression output: \begin{array}{l} \begin{array} { l r } \hline { \text { Regression Statistics } } \\ \hline \text { Multiple R } & 0.9447 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R } & 0.8886 \\ \text { Square } & \\ \text { Standard } & 0.3342 \\ \text { Error } & 30 \\ \text { Observations } & \\ \hline \end{array}\\ \text { ANOVA }\\ \begin{array} { l r r r r r } \hline & d f & { \text { SS } } & { \text { MS } } & F & { \begin{array} { c } \text { Significance } \\ F \end{array} } \\ \hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm { E } - 15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & & \\ \hline \end{array}\\ \begin{array} { l r r r r r r } \hline & \text { Coefficients } & \begin{array} { c } \text { Standard } \\ \text { Error } \end{array} & t \text { Stat } & P \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \begin{array} { l } \text { Intercept } \\ \text { Applications } \\ \text { Recorded } \end{array} & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \hline \end{array} \end{array} Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>. -Referring to Instruction 12-12,the error sum of squares (SSE) of the above regression is A) 0.1117. B) 29.0720. C) 25.9438. D) 3.1282.$ $Instruction 12-12 The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours) it takes to record a loan application.Data are collected from a sample of 30 days,and the number of applications recorded and completion time in hours is recorded.Below is the regression output: \begin{array}{l} \begin{array} { l r } \hline { \text { Regression Statistics } } \\ \hline \text { Multiple R } & 0.9447 \\ \text { R Square } & 0.8924 \\ \text { Adjusted R } & 0.8886 \\ \text { Square } & \\ \text { Standard } & 0.3342 \\ \text { Error } & 30 \\ \text { Observations } & \\ \hline \end{array}\\ \text { ANOVA }\\ \begin{array} { l r r r r r } \hline & d f & { \text { SS } } & { \text { MS } } & F & { \begin{array} { c } \text { Significance } \\ F \end{array} } \\ \hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm { E } - 15 \\ \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \text { Total } & 29 & 29.072 & & & \\ \hline \end{array}\\ \begin{array} { l r r r r r r } \hline & \text { Coefficients } & \begin{array} { c } \text { Standard } \\ \text { Error } \end{array} & t \text { Stat } & P \text {-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \begin{array} { l } \text { Intercept } \\ \text { Applications } \\ \text { Recorded } \end{array} & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \hline \end{array} \end{array} Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>. -Referring to Instruction 12-12,the error sum of squares (SSE) of the above regression is A) 0.1117. B) 29.0720. C) 25.9438. D) 3.1282.$
-Referring to Instruction 12-12,the error sum of squares (SSE) of the above regression 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