## A.11 Answer: TW 11 tutorial

### Answers for Sect. 11.2

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).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.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.These are:

**Plot 1**: \(88.4\)%;

**Plot 2**: \(90.3\)%;

**Plot 3**: \(1.4\)%;

**Plot 6**: \(56.3\)%.The answers:

- 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**. - 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**: \((55 - 35)/5 = 4\). - 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**; \(\hat{y} = 8 + 4x\);

**Plot 2**: \(\hat{y} = 40 - 2x\);

**Plot 3**: \(\hat{y} = 32\);

**Plot 6**: \(\hat{y} = 10 + 4x\).

- No answer. But a line is fine for

### 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 \(b_0\approx 17\) or $17,000. This would mean the average price of a 2014 second-hand Corolla can be expected to be about $17,000.
- \(b_1 = (1 - 17)/(16 - 0)\approx -1\). That is, the price reduces by about $1000 each year older the Corolla gets.
- 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.
- \(\hat{y} = 16.54 - 0.96x\) (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.
- \(r = -0.929\), 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.
- 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. - 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. - 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.
- \(-0.963 \pm (2\times 0.064)\), or \(-0.963\pm 0.128\), 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 \(R^2\) 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).

### Answers for Sect. 11.5

### Answers for Sect. 11.6

- 'What is the relationship between stem age and stem diameter in heather in the central north island of NZ?'
- Pretty good! Too many tick marks; horizontal-axis labels would be better horizontal.

- \(r = \sqrt{0.431} = 0.6565\), or about \(r = 0.66\). Positive linear relationship, reasonably strong.
- \(\hat{y} = (0.311 \times 20) + 1.531 = 7.751\).
- 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. 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.