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:
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\% } \\ \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Applications RECORD } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143\\ \hline \end{array} \end{array} Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>. -Referring to Instruction 12-12,the 90% confidence interval for the mean change in the amount of time needed as a result of recording one additional loan application is A) narrower than [0.0109,0.0143]. B) wider than [0.0109,0.0143]. C) wider than [0.1492,0.6555]. D) narrower than [0.1492,0.6555].](https://d2lvgg3v3hfg70.cloudfront.net/TB1606/11ea942b_a8fe_1602_8fd5_012b27dd9014_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00.jpg)
![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\% } \\ \text { Intercept } & 0.4024 & 0.1236 & 3.2559 & 0.0030 & 0.1492 & 0.6555 \\ \text { Applications RECORD } & 0.0126 & 0.0008 & 15.2388 & 4.3946 \mathrm{E}-15 & 0.0109 & 0.0143\\ \hline \end{array} \end{array} Note: 4.3946E-15 is 4.3946 x 10<sup>-15</sup>. -Referring to Instruction 12-12,the 90% confidence interval for the mean change in the amount of time needed as a result of recording one additional loan application is A) narrower than [0.0109,0.0143]. B) wider than [0.0109,0.0143]. C) wider than [0.1492,0.6555]. D) narrower than [0.1492,0.6555].](https://d2lvgg3v3hfg70.cloudfront.net/TB1606/11ea942b_a8fe_1603_8fd5_4fe9479160c9_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00_TB1606_00.jpg)
-Referring to Instruction 12-12,the 90% confidence interval for the mean change in the amount of time needed as a result of recording one additional loan application is
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