A.11 Answers to Teaching Week 11 tutorial

Answers to Sect. 11.2

  1. The correlations:

    Plot 1: \(0.94\) (correlation A);
    Plot 2: \(-0.95\) (correlation D);
    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: \(56.3\)%.

  5. The answers:

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

Answers to Sect. 11.3

  1. Approximately linear, negative, reasonably strong.
  2. Condition, extras (air con, etc.), sedan/hatch, colour, when rego due, new/old tyres, location, etc.
  3. No answer.
  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 \(P=0.000/2=0.000\): 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).

A.11.1 Answers to Sect. 11.5.

A.11.2 Answers to Sect. 11.6

1. 'What is the relationship between stem age and stem diameter in heather in the central north island of NZ?' 2. Pretty good! Too many tick marks; horizontal-axis labels would be better horizontal.
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.