Basic Statistics

ANSWERSTO SELECTED PROBLEMS 239

Chapter 8

8.1 t = 4.38 with 80 d.f. and P < .0005, so the null hypothesis is rejected.

Pressure does appear to increase bleeding time.

8.2 t = 3.62 with 11 d.f. and P = ,004, so reject the null hypothesis. The male

rats appear to be heavier.

8.3 t = 9.158 with 18 d.f. and P < .0005, so reject the null hypothesis. Yes, it

was effective.

8.4 (a) t = -3.2004 with 8 d.f. and P = .013. Conclude that the oral and rectal

temperatures are different. (b) t = 4.30 with 8 d.f. and P < ,005, so reject the

null hypothesis that the mean temperature is 37°C.

8.5 The sample size should be 161 patients in each group.

8.6 z = 1.56, so we cannot reject the null hypothesis of equal means.

Chapter 9

9.1 The variances are not significamtly different. F = 1.68.

9.2 F = 1.957, so the variances are not significantly different.

9.3 The samples are not independent since they were taken for littermates.

9.4 F = 3.00 with q = 20 and 1/2 = 31. Reject the null hypothesis of equal

variances.

Chapter 10

10.1 Expect T = .5. The variance of an observation is ~(1- T) = .25. The variance

of the mean of 10 observations is .025.

10.2 Expect 95.45%.

10.3 z = 2.21 and P = .03, so reject the null hypothesis. The 95% confidence

interval is .013 to .167 and thus does not cover 0. The confidence interval gives

us an interval that we are 95% certain will cover the true difference.

10.4 Less than .00005.

10.5 (a) The 95% confidence for the difference is -.039 to ,239. There may be no

difference between the remedies. (b) The sample size in the text was larger,

and the confidence interval computed there did not cover 0.

10.6 z = 1.903, so P > .05 and there is not a significant difference.

10.7 We need n = 294 if do not use the correction factor and 3 14 with the correction

factor.