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CH%205%20Solutions[1]

# CH%205%20Solutions[1] - Chapter 5 Solutions 5.1(a The slope...

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Chapter 5 Solutions 5.1. (a) The slope is 0.882; this means that, on the average, reading score increases by 0.882 for each one-point increase in IQ. (b) The intercept ( 33.4) would correspond to the expected reading score for a child with an IQ of 0; neither that reading score nor that IQ has any meaningful interpretation. (c) The predicted scores for x = 90 and x = 130 are 33 . 4 + 0 . 882 × 90 = 45 . 98 and 33 . 4 + 0 . 882 × 130 = 81 . 26. (d) This is most easily done by plotting the points ( 90 , 45 . 98 ) and ( 130 , 81 . 26 ) , and then drawing the line connecting them. 45 50 55 60 65 70 75 80 85 90 90 100 110 120 130 140 Reading score IQ 5.2. If the reading score increases by 1 for each IQ point, the professor’s line has slope 1. In order for an IQ of 100 to correspond to a reading score of 50, the equation must be reading score = IQ 50. 5.3. (a) Minitab output is below; regardless of the software or calculator used, the formula should agree with the one given in the text (except for differences due to rounding). (b) The means and standard deviations are x . = 324 . 75 and s x . = 257 . 66 calories, y . = 2 . 3875 and s y . = 1 . 1389 kg. The correlation is r . = − 0 . 7786. This yields b = r · s y / s x . = − 0 . 00344 and a = y b x . = 3 . 505. Minitab output The regression equation is fat = 3.51 - 0.00344 nea Predictor Coef Stdev t-ratio p Constant 3.5051 0.3036 11.54 0.000 nea -0.0034415 0.0007414 -4.64 0.000 s = 0.7399 R-sq = 60.6% R-sq(adj) = 57.8% 5.4. See also the solution to Exercise 4.4. (a) The correlation is r . = − 0 . 7485. (b) The regression equation is ˆ y = 31 . 9 0 . 304 x . (c) The slope ( 0.304) tells us that, on the average, for every one-percent increase in returning birds, the number of new birds join- ing the colony decreases by 0.304. (d) When x = 60, we predict ˆ y . = 13 . 69 new birds will join the colony. 3 6 9 12 15 18 21 35 40 45 50 55 60 65 70 75 80 Count of new birds Percent of returning birds 96

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Solutions 97 Minitab output The regression equation is New = 31.9 - 0.304 PctRtn Predictor Coef Stdev t-ratio p Constant 31.934 4.838 6.60 0.000 PctRtn -0.30402 0.08122 -3.74 0.003 s = 3.667 R-sq = 56.0% R-sq(adj) = 52.0% 5.5. (a) Scatterplot at right. Regression gives ˆ y = 132 . 45 + 0 . 402 x (Minitab output below). The plot suggests a curved pattern, so a linear formula is not appropriate for making predictions. (b) r 2 = 0 . 0182. This confirms what we see in the graph: This line does a poor job of summarizing the relationship. (c) The mean planting rate is x = 18 . 82 thousand plants per acre, and the mean yield is y = 140 . 02 bushels / acre, which is approximately equal to 132 + 0 . 402 × 18 . 82 (except for rounding error). 100 110 120 130 140 150 160 170 12 16 20 24 28 Yield (bushels per acre) Plants per acre (thousands) Minitab output The regression equation is Yield = 132 + 0.402 Plants Predictor Coef Stdev t-ratio p Constant 132.45 14.91 8.89 0.000 Plants 0.4020 0.7625 0.53 0.606 s = 16.57 R-sq = 1.8% R-sq(adj) = 0.0% 5.6. Correlations close to 1 or 1 mean that a line is a good match for the scatterplot, while correlations close to 0 mean that the scatterplot is spread widely about any line we attempt to use for prediction. Therefore, we expect that prediction of gas used will be quite accurate, while stock return predictions are much more uncertain.
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CH%205%20Solutions[1] - Chapter 5 Solutions 5.1(a The slope...

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