## A.11 Answer: TW 11 tutorial

1. No answer. But a line is fine for Plots 1, 2, 6 and 3 (but very weak!), but not for Plots 4 and 5.
2. My very rough slope estimates are:
Plot 1: $$(50-10)/10 \approx 4$$; Plot 2: $$(20 - 50)/15 \approx -2$$; Plot 3: $$0$$; Plot 6: $$(55 - 35)/5 = 4$$.
3. My very rough intercept estimates are: Plot 1: $$8$$; Plot 2: $$40$$; Plot 3: $$32$$; Plot 6: $$10$$.
4. My very rough estimates are: Plot 1; $$\hat{y} = 8 + 4x$$; Plot 2: $$\hat{y} = 40 - 2x$$; Plot 3: $$\hat{y} = 32$$; Plot 6: $$\hat{y} = 10 + 4x$$.

1. Approximately linear, negative, reasonably strong.
2. Condition, extras (air con, etc.), sedan/hatch, colour, when rego due, new/old tyres, location, etc.
4. Looks to be expensive, as $15,000 would be above the line (at least for the line I'd draw). 5. Probably$3900.
6. My guess is about $$b_0\approx 17$$ or $17,000. This would mean the average price of a 2014 second-hand Corolla can be expected to be about$17,000.
7. $$b_1 = (1 - 17)/(16 - 0)\approx -1$$. That is, the price reduces by about $1000 each year older the Corolla gets. 8. Using the above, we have $$\hat{y} = 17 - x$$ approximately. Guessing the regression line won't, of course, produce this level of precision, so anything close-ish to this is fine. 9. $$\hat{y} = 16.54 - 0.96x$$ (jamovi) or $$\hat{y} = 16.536 - 0.963x$$ (SPSS), where $$y$$ is the price in thousands of dollars, and $$x$$ is the age in years. 10. $$r = -0.929$$, and so $$R^2 = (-0.929)^2 = 0.863$$, or about $$86$$%, so about $$86$$% of the variation in prices can be explained by age alone. The rest can be explained by the car's condition, odometer reading, accessories, service history, etc. 11. jamovi: $$\hat{y} = 16.54 - (0.96\times 20) = -2.66$$, or -$2660.
SPSS: $$\hat{y} = 16.536 - (0.963\times 20) = -2.72$$, or -$2720. This is clearly silly, as we are extrapolating. 12. jamovi: $$\hat{y} = 16.54 - (0.96\times 6) = 10.78$$, or$10,780.
SPSS: $$\hat{y} = 16.536 - (0.963\times 6) = 10.76$$, or \$10,760.
So the price seems a bit steep, unless it is highly specified and in great condition.
13. Hardly need a test... $$H_0$$: $$\beta_1 = 0$$ vs $$H_1$$: $$\beta_1 < 0$$. From output, $$t = -15.059$$, and $$P = 0.000/2 = 0.000$$, so $$P < 0.0001$$: very strong evidence that older cars fetch lower second-hand prices, on average.
14. $$-0.963 \pm (2\times 0.064)$$, or $$-0.963\pm 0.128$$, or from $$-1.091$$ to $$-0.835$$.
15. Looks the same really, just reflected left-to-right.
• Size of $$r$$ won't change, sign from negative to positive; i.e. $$r = 0.93$$.
• Value of $$R^2$$ will be the same.
• Slope will be the same except sign will change (in both cases, the values on the horizontal axis are one year apart).
• Intercept will change a lot... it is the predicted value of the price if the line is extended to year 0 (which is, of course, meaningless).

3. $$r = \sqrt{0.431} = 0.6565$$, or about $$r = 0.66$$. Positive linear relationship, reasonably strong.
4. $$\hat{y} = (0.311 \times 20) + 1.531 = 7.751$$.
5. Slope of $$0.311$$: for each extra year of age, the stem diameter increases by about $$0.311$$ mm on average. Intercept of $$1.531$$: for stems of zero years of age, the average stem diameter is about $$1.531$$ mm, which is silly. The graph shows that this is extrapolation anyway: the data only start at about $$3$$ years of age.