chapter 1

Chapter 8: Tests of Hypotheses Based on a Single SampleSection 8.559.a. The formula for β isn = 900, and .0014 for n = 2500.⎛ n ⎞1− Φ⎜⎟− 2.33 +⎝ 9.4 ⎠, which gives .8980 for n = 100, .1049 forb. Z = -5.3, which is “off the z table,” so p-value < .0002; this value of z is quite statisticallysignificant.c. No. Even when the departure from H o is insignificant from a practical point of view, astatistically significant result is highly likely to appear; the test is too likely to detectsmall departures from H o .60.⎛ − . 01+.9320 / n ⎞ ⎛ ( − .01 n + .9320 ⎞a. Here = Φ⎜⎟ = Φ⎜⎟)⎟⎝ .4073 / n ⎠ ⎝ .4073 ⎠β = .9793, .8554, .4325,.0944, and 0 for n = 100, 2500, 10,000, 40,000, and 90,000, respectively.b. Here z = . 025 n which equals .25, 1.25, 2.5, and 5 for the four n’s, whence p-value =.4213, .1056, .0062, .0000, respectively.c. No; the reasoning is the same as in 54 (c).Supplementary Exercises61. Because n = 50 is large, we use a z test here, rejecting H o : = 3. 2if either ≥ z .= 1. 96. The computed z value isz025or z ≤ −1. 963.05 − 3.20z == −3.12. Since –3.12 is −1. 96.34 / 50µ in favor of H a : µ ≠ 3. 2≤ , H o should be rejected in favor of H a .62. Here we assume that thickness is normally distributed, so that for any n a t test is appropriate,and use Table A.17 to determine n. We wish ( 3 ) = . 95inspection, n = 20 satisfies this requirement, so n = 50 is too large.3.2 − 3π when = = .667..3d By254