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

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SCENARIO 12-12
The manager of the purchasing department of a large saving 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:  Regression Statistics  Multiple R 0.9447 R Square 0.8924 Adjusted R 0.8886 Square  Standard 0.3342 Error  Observations 30 ANOVA  df  SS  MS  F  Significance F Regression 125.943825.9438232.22004.3946E15 Residual 283.12820.1117 Total 2929.072 Coefficients  Standard Error t Stat  P-value  Lower 95%  Upper 95%  Intercept 0.40240.12363.25590.00300.14920.6555 Applications 0.01260.000815.23880.00000.01090.0143 Recorded \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 } & \\\text { Observations } & 30 \\\hline\end{array}\\\text { ANOVA }\\\begin{array} { l r r r r r } \hline & { \text { df } } & { \text { SS } } & { \text { MS } } & \text { F } & \text { Significance } F \\\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 } & \text { Standard Error } & 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 & 0.0000 & 0.0109 & 0.0143 \\\text { Recorded } & & & & & & \\\hline\end{array}\end{array} 12-46 Simple Linear Regression  SCENARIO 12-12 The manager of the purchasing department of a large saving 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 } & \\ \text { Observations } & 30 \\ \hline \end{array}\\ \text { ANOVA }\\ \begin{array} { l r r r r r }  \hline & { \text { df } } & { \text { SS } } & { \text { MS } } & \text { F } & \text { Significance } F \\ \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 } & \text { Standard Error } & 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 & 0.0000 & 0.0109 & 0.0143 \\ \text { Recorded } & & & & & & \\ \hline \end{array} \end{array}  12-46 Simple Linear Regression   Simple Linear Regression 12-47 -Referring to Scenario 12-12, the value of the measured t-test statistic to test whether the amount of time depends linearly on the number of loan applications recorded is A) 0.8924 B) 3.2559 C) 15.2388 D) 232.2200 Simple Linear Regression 12-47
-Referring to Scenario 12-12, the value of the measured t-test statistic to test whether the amount of time depends linearly on the number of loan applications recorded is


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