chapter 1

Chapter 8: Tests of Hypotheses Based on a Single Sample63.a. H o : = 3. 2µ vs H a : µ ≠3. 2 (Because H a : >3. 2µ gives a p-value of roughly .15)b. With a p-value of .30, we would reject the null hypothesis at any reasonable significancelevel, which includes both .05 and .10.64.a. H o : µ = 2150 vs H a : µ > 2150b.t =x − 2150s / n2160 − 2150107.5c. t == = 1. 3330/16t , p-value > .10 (actually . 10d. Since 1. 341. 10,15=≈ )e. From d, p-value > .05, so H o cannot be rejected at this significance level.65.a. The relevant hypotheses are H o : µ = 548 vs H a : ≠ 548rejected if either ≥ t .= 2. 228t or t ≤ −t= 2. 228587 − 548025,10393.02µ . At level .05, H o will be. 025,10−. The test statisticvalue is t = = = 12. 9 . This clearly falls into the upper tail of the10/11two-tailed rejection region, so H o should be rejected at level .05, or any other reasonablelevel).b. The population sampled was normal or approximately normal.66. n = 8 , x = 30.7875, s = 6. 53001 Parameter of interest: µ = true average heat-flux of plots covered with coal dust2 H o : µ = 29. 03 H a : µ > 29. 04x − 29.0t =s / nt ≥ tα or t ≥1. 8955 RR:, n−130.7875 − 29.06 t == . 77426.53/ 87 Fail to reject H o . The data does not indicate the mean heat-flux for pots covered withcoal dust is greater than for plots covered with grass.255