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Instruction 12 $\text { Regression statistics }$ $\text { ANOVA }$

question 70

Multiple Choice

Instruction 12.11
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:
$\text { Regression statistics }$
$\begin{array}{|l|l|}\hline\text { MultipleR } & 0.9447 \\\hline \text { R Square } & 0.8924 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.8886 \\\hline \text { Standard Error } & 0.3342 \\\hline \text { Observations } & 30 \\\hline\end{array}$

$\text { ANOVA }$
$\begin{array}{|l|l|l|l|l|l|}\hline & d f & S 5 & M S & F & \begin{array}{l}\text { Significance } \\F\end{array} \\\hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\\hline \text { Residual } & 28 & 3.1282 & 0.1117 & & \\\hline \text { Total } & 29 & 29.072 & & & \\\hline\end{array}$

$\begin{array}{|l|l|l|l|l|l|l|}\hline & \text { Coefficients } & \begin{array}{l}\text { Standard } \\\text { Error }\end{array} & \text { tStat } & \text { p-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\\hline \begin{array}{l}\text { Applications } \\\text { Recorded }\end{array} & 0.0126 & 0.0008 & 15.2388 & \begin{array}{l}4.3946 \mathrm{E}- \\15\end{array} & 0.0109 & 0.0143 \\\hline\end{array}$
Note: 4.3946E-15 is 4.3946 × 10^-¹⁵. $Instruction 12.11 The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours) it takes to record a loan application. Data are collected from a sample of 30 days, and the number of applications recorded and completion time in hours is recorded. Below is the regression output: \text { Regression statistics } \begin{array}{|l|l|} \hline\text { MultipleR } & 0.9447 \\ \hline \text { R Square } & 0.8924 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.8886 \\ \hline \text { Standard Error } & 0.3342 \\ \hline \text { Observations } & 30 \\ \hline \end{array} \text { ANOVA } \begin{array}{|l|l|l|l|l|l|} \hline & d f & S 5 & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \hline \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \hline \text { Total } & 29 & 29.072 & & & \\ \hline \end{array} \begin{array}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { tStat } & \text { p-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \hline \begin{array}{l} \text { Applications } \\ \text { Recorded } \end{array} & 0.0126 & 0.0008 & 15.2388 & \begin{array}{l} 4.3946 \mathrm{E}- \\ 15 \end{array} & 0.0109 & 0.0143 \\ \hline \end{array} Note: 4.3946E-15 is 4.3946 × 10-15. -Referring to Instruction 12.11,the estimated mean amount of time it takes to record one additional loan application is A) 0.4024 more hours. B) 0.0126 more hours. C) 0.0126 fewer hours. D) 0.4024 fewer hours.$ $Instruction 12.11 The manager of the purchasing department of a large savings and loan organization would like to develop a model to predict the amount of time (measured in hours) it takes to record a loan application. Data are collected from a sample of 30 days, and the number of applications recorded and completion time in hours is recorded. Below is the regression output: \text { Regression statistics } \begin{array}{|l|l|} \hline\text { MultipleR } & 0.9447 \\ \hline \text { R Square } & 0.8924 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.8886 \\ \hline \text { Standard Error } & 0.3342 \\ \hline \text { Observations } & 30 \\ \hline \end{array} \text { ANOVA } \begin{array}{|l|l|l|l|l|l|} \hline & d f & S 5 & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \hline \text { Regression } & 1 & 25.9438 & 25.9438 & 232.2200 & 4.3946 \mathrm{E}-15 \\ \hline \text { Residual } & 28 & 3.1282 & 0.1117 & & \\ \hline \text { Total } & 29 & 29.072 & & & \\ \hline \end{array} \begin{array}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { tStat } & \text { p-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \hline \begin{array}{l} \text { Applications } \\ \text { Recorded } \end{array} & 0.0126 & 0.0008 & 15.2388 & \begin{array}{l} 4.3946 \mathrm{E}- \\ 15 \end{array} & 0.0109 & 0.0143 \\ \hline \end{array} Note: 4.3946E-15 is 4.3946 × 10-15. -Referring to Instruction 12.11,the estimated mean amount of time it takes to record one additional loan application is A) 0.4024 more hours. B) 0.0126 more hours. C) 0.0126 fewer hours. D) 0.4024 fewer hours.$
-Referring to Instruction 12.11,the estimated mean amount of time it takes to record one additional loan application is

Definitions:

Flexor

The muscle of a muscle team that bends a part.

Skeletal Muscle

A type of muscle that is connected to bones and is involved in the movement of body parts voluntarily.

Extensor

The muscle of a muscle team that extends a part, allowing the joint to straighten.

Skeletal Muscle

A type of striated muscle that is attached to the bones of the skeleton and is under voluntary control, responsible for movement.

Instruction 12 Regression statistics \text { Regression statistics } Regression statistics ANOVA \text { ANOVA } ANOVA

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Instruction 12 $\text { Regression statistics }$ $\text { ANOVA }$