chapter 1

Chapter 8: Tests of Hypotheses Based on a Single Sample19.σa. Reject H o if either z ≥ 2. 58 or z ≤ −2. 58 ; = 0. 3n94.32 − 95= = −2.270.3, soz . Since –2.27 is not < -2.58, don’t reject H o .⎛⎜⎝1 ⎞ ⎛⎟ − Φ⎜0.3 ⎠ ⎝1 ⎞⎟0.3 ⎠b. β ( 94 ) = Φ 2.58 − − 2.58 − = Φ( − .75) − Φ( − 5.91) = . 2266c.( 2.58 + 1.28)⎡1.20⎤n == 21.46 , so use n = 22.⎢ 95 94 ⎥⎣ − ⎦220. With H o : µ = 750 , and H a : < 750µ and a significance level of .05, we reject H o if z < -1.645; z = -2.14 < -1.645, so we reject the null hypothesis and do not continue with thepurchase. At a significance level of .01, we reject H o if z < -2.33; z = -2.14 > -2.33, so wedon’t reject the null hypothesis and thus continue with the purchase.21. With H o : µ = . 5 , and H a : ≠ . 5a. 1.6 < t .025,12 = 2.179, so don’t reject H oµ we reject H o if t > tα / 2, n−1or t < −tα/ 2, n−1b. -1.6 > -t .025,12 = -2.179, so don’t reject H oc. – 2.6 > -t .005,24 = -2.797, so don’t reject H od. –3.9 < the negative of all t values in the df = 24 row, so we reject H o in favor of H a .22.a. It appears that the true average weight could be more than the production specification of200 lb per pipe.b. H o : µ = 200 , and H a : µ > 200 we reject H o if > t .= 1. 699206.73−200 6.73t == = 5.80 > 1.6996.35/ 30 1.16substantiate the statement in part a.t .05,29, so reject H o . The test appears to23.x −360H o : µ = 360 vs. H a : µ > 360 ; t = ; reject H o if > t . 05,25s / n= 1. 708370.69 − 360t == 2.24 > 1.70824.36/ 26contradiction of the prior belief.t ;. Thus H o should be rejected. There appears to be a244