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16. The coefficient of determination is the square of the correlation, or r².
For this data, r² = (–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 = 0.32² = 0.1024. This is the amount of variation in first-year 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 first-year 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.

19. H₀: 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 t₂₉ 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 t₂₄ 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.

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.

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))

27. The predicted value for y is: $\widehat{y}$ = 2.3 – 0.1(4.1) = 1.89. The value of 2.32 is more than two standard deviations from the predicted value, so it qualifies as an outlier.
Residual for (4.1, 2.34): 2.32 – 1.89 = 0.43 (0.43 > 2(0.13))

29. H₀: μ1 = μ2 = μ3 = μ4
H_a: At least two of the group means μ1, μ2, μ3, μ4 are not equal.

30. The independent samples t-test can only compare means from two groups, while one-way ANOVA can compare means of more than two groups.

31. Each sample appears to have been drawn from a normally distributed populations, the factor is a categorical variable (method), the outcome is a numerical variable (test score), and you were told the samples were independent and randomly selected, so those requirements are met. However, each sample has a different standard deviation, and this suggests that the populations from which they were drawn also have different standard deviations, which is a violation of an assumption for one-way ANOVA. Further statistical testing will be necessary to test the assumption of equal variance before proceeding with the analysis.

32. One of the assumptions for a one-way ANOVA is that the samples are drawn from normally distributed populations. Since two of your samples have an approximately uniform distribution, this casts doubt on whether this assumption has been met. Further statistical testing will be necessary to determine if you can proceed with the analysis.