A.11 Answer: TW 11 tutorial
Answers for Sect. 11.2
- No answer. But a line is fine for Plots 1, 2, 6 and 3 (but very weak!), but not for Plots 4 and 5.
- My very rough slope estimates are:
Plot 1: (50−10)/10≈4; Plot 2: (20−50)/15≈−2; Plot 3: 0; Plot 6: (55−35)/5=4. - My very rough intercept estimates are: Plot 1: 8; Plot 2: 40; Plot 3: 32; Plot 6: 10.
- My very rough estimates are: Plot 1; ˆy=8+4x; Plot 2: ˆy=40−2x; Plot 3: ˆy=32; Plot 6: ˆy=10+4x.
Answers for Sect. 11.3
- Approximately linear, negative, reasonably strong.
- Condition, extras (air con, etc.), sedan/hatch, colour, when rego due, new/old tyres, location, etc.
- No answer.
- Looks to be expensive, as $15,000 would be above the line (at least for the line I'd draw).
- Probably $3900.
- My guess is about b0≈17 or $17,000. This would mean the average price of a 2014 second-hand Corolla can be expected to be about $17,000.
- b1=(1−17)/(16−0)≈−1. That is, the price reduces by about $1000 each year older the Corolla gets.
- Using the above, we have ˆy=17−x approximately. Guessing the regression line won't, of course, produce this level of precision, so anything close-ish to this is fine.
- ˆy=16.54−0.96x (jamovi) or ˆy=16.536−0.963x (SPSS), where y is the price in thousands of dollars, and x is the age in years.
- r=−0.929, and so R2=(−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.
- jamovi: ˆy=16.54−(0.96×20)=−2.66, or -$2660.
SPSS: ˆy=16.536−(0.963×20)=−2.72, or -$2720.
This is clearly silly, as we are extrapolating. - jamovi: ˆy=16.54−(0.96×6)=10.78, or $10,780.
SPSS: ˆy=16.536−(0.963×6)=10.76, or $10,760.
So the price seems a bit steep, unless it is highly specified and in great condition. - Hardly need a test... H0: β1=0 vs H1: β1<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.
- −0.963±(2×0.064), or −0.963±0.128, or from −1.091 to −0.835.
- 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 R2 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).
Answers for Sect. 11.4

FIGURE A.1: Plots giving different correlations
Answers for Sect. 11.5.1
Answers for Sect. 11.5.2
- 'What is the relationship between stem age and stem diameter in heather in the central north island of NZ?'
- Good! Too many tick marks; horizontal-axis labels would be better horizontal.
- r=√0.431=0.6565, or about r=0.66. Positive linear relationship, reasonably strong.
- b0=1.531 and b1=0.311.
- ˆy=(0.311×20)+1.531=7.751.
- 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.