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

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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:
 Regression Statistics  Multiple R 0.9447 R Square 0.8924 Adjusted R 0.8886 Square  Standard 0.3342 Error 30 Observations  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  Applications  Recorded 0.40240.12363.25590.00300.14920.6555\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


Definitions:

Managed Float

A currency exchange rate policy where a currency's value is allowed to fluctuate in response to foreign exchange market mechanisms, but the central bank may intervene to stabilize the currency if necessary.

Managed Floating

An exchange rate system where a country's currency value is allowed to fluctuate in response to foreign-exchange market mechanisms, but the central bank can intervene to prevent extreme fluctuations.

Bretton Woods System

A monetary management system established post-World War II, which set up rules for commercial and financial relations among major industrial states.

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