chapter 1

Chapter 8: Tests of Hypotheses Based on a Single Sample67. N = 47, x = 215 mg, s = 235 mg. Range 5 mg to 1,176 mg.a. No, the distribution does not appear to be normal, it appears to be skewed to the right. Itis not necessary to assume normality if the sample size is large enough due to the centrallimit theorem. This sample size is large enough so we can conduct a hypothesis testabout the mean.b.1 Parameter of interest: µ = true daily caffeine consumption of adultwomen.2 H o : µ = 2003 H a : µ > 2004x − 200z =s / n5 RR: z ≥1. 282 or if p-value ≤ . 106215−200= = . 44235/ 471 − Φ .44 = .z ; p-value = ( ) 337 Fail to reject H o . because .33 > .10. The data does not indicate that dailyconsump tion of all adult women exceeds 200 mg.68. At the .05 significance level, reject H o because .043 < .05. At the level .01, fail to reject H obecause .043 > .01. Thus the data contradicts the design specification that sprinkler activationis less than 25 seconds at the level .05, but not at the .01 level.69.a. From table A.17, when = 9. 51.25, df = 9, and β ≈ . 20 .µ , d = .625, df = 9, and β ≈ . 60 , when = 9. 0µ , d =b. From Table A.17, β = . 25 , d = .625, n ≈ 2870. A normality plot reveals that these observations could have come from a normally distributedpopulation, therefore a t-test is appropriate. The relevant hypotheses are H o : µ = 9. 75 vsH a : µ > 9. 75 . Summary statistics are n = 20, x = 9. 85259.8525 − 9.75test statistic t == 4. 75.0965/ 20, and s = .0965, which leads to a, from which the p-value = .0001. (From MINITABoutput). With such a small p-value, the data strongly supports the alternative hypothesis. Thecondition is not met.256