Basic Statistics

240 ANSWERS TO SELECTED PROBLEMS

Chapter 11

11.1 (a) x2 = 8.22 with 1 d.f. and P = ,004 so reject the null hypothesis. (b) RR =

2.071. (c) OR = 3.32. The 95% confidence limits of the odds ratio go from

1.44 to 7.65.

11.2 (a) OR = 2.33. (b) OR = 2.33. (c) x2 for (a) is 7.883 or P = .0050. x2 for

(c) is 11 .OOO or P = .0009.

11.3 (a) The McNemar’s x2 is 1.26 with 1 d.f., or nonsignificant. (b) The OR =

1.438 and the confidence interval goes from .732 to 2.721. Since the confidence

limit covers 1, it also shows that the chemical was not shown to have an effect.

In addition, the confidence limit has an approximately 95% chance of including

the true odds ratio.

11.4 x2 = 23.27 with 1 d.f. and P < .0005. OR = .26 of having been assaulted if

currently living with spouse.

11.5 x2 = 5.084 with 1 d.f. and P = .0241. Reject the null hypothesis.

11.6 x2 = 20.8232 with 6 d.f. and P = ,002. Reject the null hypothesis of equal

proportions.

11.7 Combine the results for moderate and severe heart failure.

11.8 Reject the null hypothesis with P < .001.

Chapter 12

12.1 (a) Fixed-X. (b) Rate = -1518.8 + ,7745year. (c) The 95% confidence

interval goes from .4637 to 1.0854. t = 5.6360 with 9 d.f., so P = .0003.

12.2 (a) Single sample. (b) Both negative. (c) Approximate 95% confidence limits

are -.92 to -.15. (d) t = -3.0026 and P = .0170 in both cases.

12.3 (a) T = ,9236. (b) From Problem 7.9 we could only conclude that the hypothesis

of equal means could not be rejected. From the scatter diagram and T = .9236,

we also learned that the two observers read each plate with a similar outcome.

12.4 (a) Single sample. (b) CBR = 80.32 - .857LifeExp. (c) Yes. (d) Decrease by

8.57. (e) t = -6.651 and P < .00005. Yes.

12.5 (a) LifeExp = 63.48 + .0005GDP. The line does not fit the points at all well.

(b) T = .71. (c) The line fits much better after the transformation. T = .88.

LifeExp = 28.28 + 11.3461og(GDP). (d) The value of T is increased since T

measures the linear association between X and Y.

12.6 (a) T = -.7114. CBR = 26.6317 - .0005GPD. (b) If we use log(GDP),

T = -.8788. The magnitude of T increased and the points fit a straight line