0.2 Practice tests (1-4) and final exams  (Page 26/36)

12.4: the regression equation

13 . $r\left(\frac{s_y}{s_x}\right)=0.73\left(\frac{9.6}{4.0}\right)=1.752\approx 1.75$

14 . $a=\overline{y}-b\overline{x}=141.6-1.752(68.4)=21.7632\approx 21.76$

15 . $\widehat{y}=21.76+1.75(68)=140.76$

12.5: correlation coefficient and coefficient of determination

16 . The coefficient of determination is the square of the correlation, or r ² .
For this data, r ² = (–0.56)2 = 0.3136 ≈ 0.31 or 31%. This means that 31 percent of the variation in fuel efficiency can be explained by the bodyweight of the automobile.

17 . The coefficient of determination = 0.32 ² = 0.1024. This is the amount of variation in freshman college GPA that can be explained by high school GPA. The amount that cannot be explained is 1 – 0.1024 = 0.8976 ≈ 0.90. So about 90 percent of variance in freshman college GPA in this data is not explained by high school GPA.

18 . $r=\sqrt{r^2}$
$\sqrt{0.5}=0.707106781\approx 0.71$
You need a correlation of 0.71 or higher to have a coefficient of determination of at least 0.5.

12.6: testing the significance of the correlation coefficient

19 . H ₀ : ρ = 0
H _a : ρ ≠ 0

20 . $t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}=\frac{0.33\sqrt{30-2}}{\sqrt{1-{0.33}^2}}=1.85$
The critical value for α = 0.05 for a two-tailed test using the t ₂₉ distribution is 2.045. Your value is less than this, so you fail to reject the null hypothesis and conclude that the study produced no evidence that the variables are significantly correlated.
Using the calculator function tcdf, the p -value is 2tcdf(1.85, 10^99, 29) = 0.0373. Do not reject the null hypothesis and conclude that the study produced no evidence that the variables are significantly correlated.

21 . $t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}=\frac{0.45\sqrt{25-2}}{\sqrt{1-{0.45}^2}}=2.417$
The critical value for α = 0.05 for a two-tailed test using the t ₂₄ distribution is 2.064. Your value is greater than this, so you reject the null hypothesis and conclude that the study produced evidence that the variables are significantly correlated.
Using the calculator function tcdf, the p-value is 2tcdf(2.417, 10^99, 24) = 0.0118. Reject the null hypothesis and conclude that the study produced evidence that the variables are significantly correlated.

12.7: prediction

22 . $\widehat{y}=25+16(5)=105$

23 . Because the intercept appears in both predicted values, you can ignore it in calculating a predicted difference score. The difference in grams of fiber per serving is 6 – 3 = 3 and the predicted difference in grams of potassium per serving is (16)(3) = 48.

12.8: outliers

24 . An outlier is an observed value that is far from the least squares regression line. A rule of thumb is that a point more than two standard deviations of the residuals from its predicted value on the least squares regression line is an outlier.

25 . An influential point is an observed value in a data set that is far from other points in the data set, in a horizontal direction. Unlike an outlier, an influential point is determined by its relationship with other values in the data set, not by its relationship to the regression line.

26 . The predicted value for y is: $\widehat{y}=5+0.3x=5.6$ . The value of 6.2 is less than two standard deviations from the predicted value, so it does not qualify as an outlier.
Residual for (2, 6.2): 6.2 – 5.6 = 0.6 (0.6<2(0.4))