Instruction 12-11 a Computer Software Developer Would Like to UseANOVA-Referring to Instruction 12-11,the Normality of Error Assumption Appears to the Number

The Answer of Instruction 12-11 a Computer Software Developer Would Like to UseANOVA-Referring to Instruction 12-11,the Normality of Error Assumption Appears to the Number

Instruction 12-11 a Computer Software Developer Would Like to UseANOVA-Referring to Instruction 12-11,the Normality of Error Assumption Appears to the Number

The Answer of Instruction 12-11 a Computer Software Developer Would Like to UseANOVA-Referring to Instruction 12-11,the Normality of Error Assumption Appears to the Number

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Instruction 12-11
a Computer Software Developer Would Like to Use

question 23

True/False

Instruction 12-11
A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed:
$\begin{array}{lr}\hline {\text { Regression Statistics }} \\\hline \text { Multiple R } & 0.8691 \\\hline \text { R Square } & 0.7554 \\\hline \text { Adjusted R Square } & 0.7467 \\\text { Standard Error } & 44.4765 \\\text { Observations } & 30.0000\end{array}$
ANOVA
$\begin{array}{|l|r|r|r|r|r|}\hline&df&SS&MS&F&\text {significance F}\\\hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\\hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\\hline \text { Total } & 29 & 226451.3503 & &\\\hline\end{array}$
$\begin{array}{lrrrrrrr}\hline & \text { Coefficients } & \text { Standard Error } & t \text { Stat } & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\\hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\\hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513\end{array}$ $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \text { Standard Error } & 44.4765 \\ \text { Observations } & 30.0000 \end{array} ANOVA \begin{array}{|l|r|r|r|r|r|} \hline&df&SS&MS&F&\text {significance F}\\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & &\\\hline \end{array} \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Error } & t \text { Stat } & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \end{array} -Referring to Instruction 12-11,the normality of error assumption appears to have been violated.$ $Instruction 12-11 A computer software developer would like to use the number of downloads (in thousands)for the trial version of his new shareware to predict the amount of revenue (in thousands of dollars)he can make on the full version of the new shareware.Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different sharewares that he has developed: \begin{array}{lr} \hline {\text { Regression Statistics }} \\ \hline \text { Multiple R } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \text { Adjusted R Square } & 0.7467 \\ \text { Standard Error } & 44.4765 \\ \text { Observations } & 30.0000 \end{array} ANOVA \begin{array}{|l|r|r|r|r|r|} \hline&df&SS&MS&F&\text {significance F}\\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residual } & 28 & 55388.4309 & 1978.1582 & & \\ \hline \text { Total } & 29 & 226451.3503 & &\\\hline \end{array} \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Error } & t \text { Stat } & \text { P-value } & \text { Lower 95\% } & \text { Upper 95\% } \\ \hline \text { Intercept } & -95.0614 & 26.9183 & -3.5315 & 0.0015 & -150.2009 & -39.9218 \\ \hline \text { Download } & 3.7297 & 0.4011 & 9.2992 & 0.0000 & 2.9082 & 4.5513 \end{array} -Referring to Instruction 12-11,the normality of error assumption appears to have been violated.$
-Referring to Instruction 12-11,the normality of error assumption appears to have been violated.

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