## A.11 Answer: TW 11 tutorial

1. The correlations:

Plot 1: $$0.94$$ (correlation D);
Plot 2: $$-0.95$$ (correlation A);
Plot 3: $$0.12$$ (correlation B):
Plot 6: $$0.75$$ (correlation C).

Correlation is inappropriate for Plot 4) (non-linear) and Plot 5 (non-linear).

2. Examples of the direction in Plot 1: Any two variables moderately positively correlated, such as height and weight, distance lived from university and travel time, etc.

3. Examples of direction in Plot 2: Any two variables moderately negatively correlated, such as hours of weekly exercise and body weight, number of SCI110 tutorial missed and final mark, etc.

4. These are:

Plot 1: $$88.4$$%;
Plot 2: $$90.3$$%;
Plot 3: $$1.4$$%;
Plot 6: $$56.3$$%.

1. No answer. But a line is fine for Plot 1, Plot 2, Plot 3 (but very weak!) and Plot 6, but not for Plot 4 and Plot 6.
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. This interpretation is silly. A close look at the graph shows that this is extrapolation anyway: the data only start at about 3 years of age.