chapter 1

Chapter 8: Tests of Hypotheses Based on a Single Sample45.a. p-value = .084 > .05 = α, so don’t reject H o .b. p-value = .003 < .001 = α, so reject H o .c. .498 >> .05, so H o can’t be rejected at level .05d. 084 < .10, so reject H o at level .10e. .039 is not < .01, so don’t reject H o .f. p-value = .218 > .10, so H o cannot be rejected.46. In each case the p-value = 1 −Φ( z)a. .0778b. .1841c. .0250d. .0066e. .456247.a. .0358b. .0802c. .5824d. .1586e. 048.a. In the df = 8 row of table A.5, t = 2.0 is between 1.860 and 2.306, so the p-value isbetween .025 and .05: .025 < p-value < .05.b. 2.201 < | -2.4 | < 2.718, so .01 < p-value < .025.c. 1.341 < | -1.6 | < 1.753, so .05 < P( t < -1.6) < .10. Thus a two-tailed p-value: 2(.05 < P( t< -1.6) < .10), or .10 < p-value < .20d. With an upper-tailed test and t = -.4, the p-value = P( t > -.4) > .50.e. 4.032 < t=5 < 5.893, so .001 < p-value < .005f. 3.551 < | -4.8 |, so P(t < -4.8) < .0005. A two-tailed p-value = 2[ P(t < -4.8)] < 2(.0005),or p-value < .001.251