Questions: Time to Practice 1. Use these data to answer Questions 1a and 1b. These data are saved as Chapter 5 Data Set 2. a. Compute the Pearson product-moment correlation coefficient by hand and show all your work. b. Construct a scatterplot for these 10 pairs of values by hand. Based on the scatterplot, would you predict the correlation to be direct or indirect? Why? Number Correct (out of a possible 20) Attitude (out of a possible 100) --------------------------------------- 17 94 13 73 12 59 15 80 16 93 14 85 16 66 16 79 18 77 19 91

Solution Steps

Given data: \[ \begin{array}{|c|c|} \hline \text{Number Correct (out of 20)} & \text{Attitude (out of 100)} \\ \hline 17 & 94 \\ \hline 13 & 73 \\ \hline 12 & 59 \\ \hline 15 & 80 \\ \hline 16 & 93 \\ \hline 14 & 85 \\ \hline 16 & 66 \\ \hline 16 & 79 \\ \hline 18 & 77 \\ \hline 19 & 91 \\ \hline \end{array} \]

Let \( x \) be the "Number Correct" and \( y \) be the "Attitude".

\[ \begin{aligned} &\text{Mean of } x: \bar{x} = \frac{17 + 13 + 12 + 15 + 16 + 14 + 16 + 16 + 18 + 19}{10} = 15.6 \\ &\text{Mean of } y: \bar{y} = \frac{94 + 73 + 59 + 80 + 93 + 85 + 66 + 79 + 77 + 91}{10} = 79.7 \\ \end{aligned} \]

\[ \begin{aligned} &\text{Covariance: } \text{cov}(x, y) = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y}) \\ &= \frac{1}{9} \left[(17-15.6)(94-79.7) + (13-15.6)(73-79.7) + (12-15.6)(59-79.7) + (15-15.6)(80-79.7) + (16-15.6)(93-79.7) + (14-15.6)(85-79.7) + (16-15.6)(66-79.7) + (16-15.6)(79-79.7) + (18-15.6)(77-79.7) + (19-15.6)(91-79.7)\right] \\ &= \frac{1}{9} \left[1.4 \cdot 14.3 + (-2.6) \cdot (-6.7) + (-3.6) \cdot (-20.7) + (-0.6) \cdot 0.3 + 0.4 \cdot 13.3 + (-1.6) \cdot 5.3 + 0.4 \cdot (-13.7) + 0.4 \cdot (-0.7) + 2.4 \cdot (-2.7) + 3.4 \cdot 11.3\right] \\ &= \frac{1}{9} \left[20.02 + 17.42 + 74.52 + (-0.18) + 5.32 + (-8.48) + (-5.48) + (-0.28) + (-6.48) + 38.42\right] \\ &= \frac{1}{9} \left[135.78\right] \\ &= 15.0867 \end{aligned} \]

\[ \begin{aligned} &\text{Standard deviation of } x: s_x = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \\ &= \sqrt{\frac{1}{9} \left[(17-15.6)^2 + (13-15.6)^2 + (12-15.6)^2 + (15-15.6)^2 + (16-15.6)^2 + (14-15.6)^2 + (16-15.6)^2 + (16-15.6)^2 + (18-15.6)^2 + (19-15.6)^2\right]} \\ &= \sqrt{\frac{1}{9} \left[1.96 + 6.76 + 12.96 + 0.36 + 0.16 + 2.56 + 0.16 + 0.16 + 5.76 + 11.56\right]} \\ &= \sqrt{\frac{1}{9} \left[42.4\right]} \\ &= 2.1726 \end{aligned} \]

\[ \begin{aligned} &\text{Standard deviation of } y: s_y = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (y_i - \bar{y})^2} \\ &= \sqrt{\frac{1}{9} \left[(94-79.7)^2 + (73-79.7)^2 + (59-79.7)^2 + (80-79.7)^2 + (93-79.7)^2 + (85-79.7)^2 + (66-79.7)^2 + (79-79.7)^2 + (77-79.7)^2 + (91-79.7)^2\right]} \\ &= \sqrt{\frac{1}{9} \left[202.5 + 44.89 + 428.49 + 0.09 + 176.89 + 27.04 + 188.49 + 0.49 + 7.29 + 126.49\right]} \\ &= \sqrt{\frac{1}{9} \left[1202.66\right]} \\ &= 11.5581 \end{aligned} \]

\[ \begin{aligned} &\text{Pearson correlation coefficient: } r = \frac{\text{cov}(x, y)}{s_x s_y} \\ &= \frac{15.0867}{2.1726 \cdot 11.5581} \\ &= \frac{15.0867}{25.1081} \\ &= 0.6008 \end{aligned} \]

The scatterplot will be constructed using the given data points.

Final Answer