## D.32 Answers: Regression

Answers to exercises in Sect. 35.13.

**Answer to Exercise 35.1**:

**1.**\(b_0 = 97.499\) (the intercept); \(b_1=0.0764\) (the slope).

**2.**\(\hat{y} = 97.499 + 0.0764x\): \(x\) is the inlet temperature (in \(^\circ\)C) and \(y\) is removal efficiency (in %).

**3.**When inlet temperature increases by 1 degree C, on average the removal efficiency

*increases*by 0.076 percentage points.

**4.**\(H_0\): \(\beta=0\); \(H_1\): \(\beta\ne 0\) (two-tailed;RQ implies two-tailed test).

**5.**\(t = 10.742\), which is huge; \(P<0.001\).

**6.**\(0.076\pm (2\times 0.007)\), or \(0.076\pm 0.014\), or \(0.062\) to \(0.090\).

**Answer to Exercise 35.2**:

**1.**Intercept

*not*about 110; that’s where the line ‘stops,’ but the intercept is the predicted value of \(y\) when \(x=0\). We have to extend the line quite a bit. Using rise-over-run, guess slope is \((190-110)/(180-110) =1.14\).

**2.**\(\hat{y} = -3.69 + 1.04x\), where \(y\) is punting distance (in feet), and \(x\) is right leg strength (in pounds).

**3.**For each extra pound of leg strength, the punting distance increases, on average, by about 1 foot.

**4.**\(H_0\): \(\beta=0\); \(H_1\): \(\beta\ne 0\). (You could answer in terms of correlations.) The question is stated as a two-tailed question, but testing if stronger legs

*increase*kicking distance seems sensible.

**5.**\(t = 6.16\), which is huge; \(P = 0.0001\) (two-tailed).

**6.**\(1.0427 \pm (2\times 0.1692)\), or \(1.0427 \pm 0.3384\), or 0.70 to 1.4.

**7.**Very strong evidence in the sample (\(t = 6.16\); \(P = 0.0001\) (two-tailed)) that punting distance is related to leg strength (slope: \(1.0427\); \(n=13\)).

**Answer to Exercise 35.3**:

**1.**

*Way*too many decimal places. \(r\) is not relevant as relationship is non-linear.

**2.**Regression is inappropriate: the relationship is non-linear.

**3.**\(y\) should be \(\hat{y}\); the slope and intercept have been

*swapped*(from the plot, the intercept for their line is about 0.4, which they give as the slope).

**4.**The whole thing is as dodgy-as…

**Answer to Exercise 35.4**:

**1.**\(\hat{y} = 17.47 - 2.59x\), where \(x\) is the percentage bitumen by weight, and \(y\) is the percentage air voids by volume.

**2.**

*Slope*: an increase in the bitumen weight by one percentage point

*decreases*the average percentage air voids by volume by 2.59 percentage points.

*Intercept*: dodgy (extrapolation); in principle 0% bitumen content by weight, the percentage air voids by volume is about 17.47%.

**3.**\(t=-74.9\): Massive! Extremely strong evidence (\(P<0.001\)) of a relationship.

**4.**\(\hat{y} = 17.4712 - (2.5937\times 5) = 4.5027\), or about 4.5%. Expected good prediction, as relationship is strong.

**5.**\(\hat{y} = 17.4712 - (2.5937\times 6) = 1.909\), or about 1.9%. Might be a poor prediction, since this is extrapolation.

**Answer to Exercise 35.5**:

**1.**\(b_0\): When someone spends

*no*time on sunscreen application, an average of 0.27g has been applied; nonsense. \(b_1\): Each extra minute spent on application adds an average of 2.21g of sunscreen: sensible.

**2.**The value of \(\beta_0\) could be zero… which would make sense.

**3.**\(\hat{y} = 0.27 + (2.21\times 8) = 17.95\); an average of about 18g.

**4.**About 64% of the variation in sunscreen amount applied can be explained by the variation in the time spent on application.

**5.**\(r = \sqrt{0.64} = 0.8\), and need a positive value of \(r\). A strong and positive correlation between the variables.

**Answer to Exercise 35.6**:

**1.**No.

**2.**Possibly; no idea of accuracy of predictions really.

**3.**Intercept: Weight of infant with chest circumference zero; silly. Slope: average increase in birth weight (in g) for each increase in chest circumference by one cm.

**4.**Intercept: cm; slope: cm/gram.

**5.**\(\hat{y}=2538.7\)g.

**6.**Too many decimal place! Regression equation implies predicting to 0.0001 of a gram. \(r\) has too many decimal places too.