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

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Instruction 12.35
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:
 Regression statistics \text { Regression statistics }
 MultipleR 0.8691 R Square 0.7554 Adjusted R  Square 0.7467 Standard Error 44.4765 Observations 30.0000\begin{array}{|l|l|}\hline \text { MultipleR } & 0.8691 \\\hline \text { R Square } & 0.7554 \\\hline \begin{array}{l}\text { Adjusted R } \\\text { Square }\end{array} & 0.7467 \\\hline \text { Standard Error } & 44.4765 \\\hline \text { Observations } & 30.0000\\\hline \end{array}

 ANOVA \text { ANOVA }
dfSSMSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residual 2855388.43091978.1582 Total 29226451.3503\begin{array}{|l|l|l|l|l|l|}\hline & d f & S S & M S & F & \begin{array}{l}\text { Significance } \\F\end{array} \\\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}

 Coefficients  Standard  Error  t Stat p-value  Lower 95%  Upper 95%  Intercept 95.061426.91833.53150.0015150.200939.9218 Download 3.72970.40119.29920.00002.90824.5513\begin{array}{|l|l|l|l|l|l|l|}\hline & \text { Coefficients } & \begin{array}{l}\text { Standard } \\\text { Error }\end{array} & \text { t Stat } & p \text {-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 \\\hline\end{array}


 Instruction 12.35 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:   \text { Regression statistics }   \begin{array}{|l|l|} \hline \text { MultipleR } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000\\ \hline  \end{array}    \text { ANOVA }   \begin{array}{|l|l|l|l|l|l|} \hline & d f & S S & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \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}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { t Stat } & p \text {-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 \\ \hline \end{array}         -Referring to Instruction 12.35,what are the lower and upper limits of the 95% confidence interval estimate for the mean change in revenue as a result of a 1 thousand increase in the number of downloads?  Instruction 12.35 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:   \text { Regression statistics }   \begin{array}{|l|l|} \hline \text { MultipleR } & 0.8691 \\ \hline \text { R Square } & 0.7554 \\ \hline \begin{array}{l} \text { Adjusted R } \\ \text { Square } \end{array} & 0.7467 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000\\ \hline  \end{array}    \text { ANOVA }   \begin{array}{|l|l|l|l|l|l|} \hline & d f & S S & M S & F & \begin{array}{l} \text { Significance } \\ F \end{array} \\ \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}{|l|l|l|l|l|l|l|} \hline & \text { Coefficients } & \begin{array}{l} \text { Standard } \\ \text { Error } \end{array} & \text { t Stat } & p \text {-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 \\ \hline \end{array}         -Referring to Instruction 12.35,what are the lower and upper limits of the 95% confidence interval estimate for the mean change in revenue as a result of a 1 thousand increase in the number of downloads?
-Referring to Instruction 12.35,what are the lower and upper limits of the 95% confidence interval estimate for the mean change in revenue as a result of a 1 thousand increase in the number of downloads?

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