Question

Calculate the p-value for the following conditions and determine whether or not to reject the null hypothesis a) one-tail (upper) test, zp = 2.82, and a = 0.10 b) one-tail (lower) test, zp = -1.28, and a = 0.05 c) two-tail test, zp = -1.51, and a = 0.05 d) two-tail test, zp = 1.89, and a = 0.02 Click here to view the first page of the standard normalitable. Click here to view the second rage of the standard normal

Answer 1

c) Here the test is a two-tailed test and given that top -1.51 and x=0.05 CH op-value Z > 1.51 1:51] 2xp[z>lzpl] 2xp[za -20f0-5-plex zein 2x {0.5 -0.4 345 } 1.51) 1.0 F2 0 -1.51 1.51 0.131 Here p-value 0.131 > 0.05 z R. So we fail to reject the hull hypothesis

given that x = 0.02 d) Here, the test is a two-tailed test and Zp= 1.89 and .. p-value= ax Plz > 1 2p 1] 2 x Plz> 1.89 2 x 0.5-plocz<1.89)) 2 x {0.5 -0.4706} )) = 0.0588 -1.89 0 1. 89 Here , p-value = 0,0588 > 0.02 So, we fail to reject the null hypothesis