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

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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:
 Regression Statistics  Multiple R 0.8691 R Square 0.7554 Adjusted R Square 0.7467 Standard Error 44.4765 Observations 30.0000\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 \\\hline \text { Standard Error } & 44.4765 \\\hline \text { Observations } & 30.0000 \\\hline\end{array}
ANOVA
df SS MSF Significance F Regression 1171062.9193171062.919386.47590.0000 Residuall 2855388.43091978.1582 Total 29226451.3503\begin{array}{|l|r|r|r|r|r|}& d f & \text { SS } & {M S} & F & \text { Significance } F \\\hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\\hline \text { Residuall } & 28 & 55388.4309 & 1978.1582 & \\\hline \text { Total } & 29 & 226451.3503 & &\end{array}

 Coefficients  Standard Eror 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}{lrrrrrrr}\hline & \text { Coefficients } & \text { Standard Eror } &{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 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000 \\ \hline \end{array}  ANOVA  \begin{array}{|l|r|r|r|r|r|} & d f & \text { SS } & {M S} & F & \text { Significance } F \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residuall } & 28 & 55388.4309 & 1978.1582 & \\ \hline \text { Total } & 29 & 226451.3503 & & \end{array}    \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Eror } &{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 null hypothesis for testing whether there is a linear relationship between revenue and number of dowloads is  There is no linear relationship between revenue and number of downloads .  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 \\ \hline \text { Standard Error } & 44.4765 \\ \hline \text { Observations } & 30.0000 \\ \hline \end{array}  ANOVA  \begin{array}{|l|r|r|r|r|r|} & d f & \text { SS } & {M S} & F & \text { Significance } F \\ \hline \text { Regression } & 1 & 171062.9193 & 171062.9193 & 86.4759 & 0.0000 \\ \hline \text { Residuall } & 28 & 55388.4309 & 1978.1582 & \\ \hline \text { Total } & 29 & 226451.3503 & & \end{array}    \begin{array}{lrrrrrrr} \hline & \text { Coefficients } & \text { Standard Eror } &{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 null hypothesis for testing whether there is a linear relationship between revenue and number of dowloads is  There is no linear relationship between revenue and number of downloads .
-Referring to Instruction 12-11,the null hypothesis for testing whether there is a linear relationship between revenue and number of dowloads is "There is no linear relationship between revenue and number of downloads".


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