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Complete: Chapter 15 Problem Set
Based on your scatter diagram, you would expect the correlation to be negative \( \nabla \).
The mean \( x \) score is \( M_{X}= \) \( \square \)
\( \square \) 6 , and the mean \( \mathrm{y} \) score is \( \mathrm{M}_{\mathrm{Y}}= \) \( \square \) 5. \( \square \)
A-Z
Now, using the values for the means that you just calculated, fill out the following table by calculating the deviations from the means for \( X \) and \( Y \), the squares of the deviations, and the products of the deviations.
\begin{tabular}{ccrrccc}
\multicolumn{2}{c}{ Scores } & \multicolumn{2}{c}{ Deviations } & \multicolumn{2}{c}{ Squared Deviations } & Products \\
\hline \( \mathbf{X} \) & \( \mathbf{Y} \) & \( \mathbf{X}-\mathbf{M}_{\mathbf{X}} \) & \( \mathbf{Y}-\mathbf{M}_{\mathbf{Y}} \) & \( \left(\mathbf{X}-\mathbf{M}_{\mathbf{X}}\right)^{\mathbf{2}} \) & \( \left(\mathbf{Y}-\mathbf{M}_{\mathbf{Y}}\right)^{\mathbf{2}} \) & \( \left(\mathbf{X}-\mathbf{M}_{\mathbf{X}}\right)\left(\mathbf{Y}-\mathbf{M}_{\mathbf{Y}}\right) \) \\
8 & 2 & 3 & -3 & 9 & 9 & -9 \\
7 & 3 & 2 & -2 & 4 & 4 & -4 \\
6 & 4 & 1 & -1 & 1 & 1 & -4 \\
9 & 6 & 4 & 1 & 16 & 1 & -1 \\
0 & 10 & -5 & 5 & 25 & 25 & 4 \\
\hline
\end{tabular}
The sum of squares for \( \mathrm{x} \) is \( \mathbf{S S}_{\mathrm{x}}= \) \( \square \) . The sum of squares for \( \mathrm{y} \) is \( \mathrm{SS}_{\mathrm{y}}= \) \( \square \) . The sum of products is \( \mathrm{SP}= \) \( \square \)
Because the sign of the sum of products is \( \qquad \) , the sign of the correlation coefficient \( \qquad \)
The correlation coefficient is \( r= \) \( \qquad \)
Look at your scatter diagram again. If you excluded the point \( (0,10) \), you would expect the recalculated correlation coefficient to be \( \qquad \) because \( \qquad \)
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