Week 4 Solutions - updated - Week 4 Solutions 9.19 a H0 = 16 versus b For =.01 z Ha 16 z 2 z.005 2.576 x 16 When.1 36 x = 16.05 z = 3 reject H0 readjust

# Week 4 Solutions - updated - Week 4 Solutions 9.19 a H0 =...

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Week 4 Solutions 9.19 a. 0 H : = 16 versus a H : 16. b. For = .01, 576 . 2 005 . 2 / z z 36 1 . 16 x z When x = 16.05, z = 3, reject 0 H ; readjust. p -value = 2(1 – .9987) = .0026<.01, reject 0 H , readjust. CI : [16.007, 16.093], readjust When x = 15.96, z = –2.4, cannot reject 0 H : do not readjust. p -value = 2(1 – .9918) = .0164, not less than .01, cannot reject 0 H , do not readjust. CI : [15.917, 16.003], do not readjust
When x = 16.02, z = 1.2, cannot reject 0 H ; do not readjust. p -value=2(1 – .8849) = .2302, not less than .01, cannot reject 0 H , do not readjust. CI : [15.977, 16.063], do not readjust When x = 15.94, z = -3.6, reject 0 H ; readjust. p -value is approximately 0.000 < .001 reject 0 H , readjust. CI : [15.897, 15.983], readjust 9.32 0 H  = 4 versus  a H    4 78 . 4 4 32 . 4 800 67 . t Since t 0.05 =1.660 & t 0.025 =1.984 & t 0.005 =2.626 & t 0.0005 =3.392 < 4.78 reject 0 H with extremely strong evidence. Since the sample mean is greater than 4 we estimate  μ  > 4.
9.41 H 0 : p = 0.73    H a : p does not equal 0.73 z = ((141/200 – 0.73)/sqrt(0.73 * 0.27 / 200) = -0.80 p-value = 2*P(Z > 0.80) = 0.4238 Insufficient evidence against H 0 .
9.61 a.  0 H   1.19%,  a H  > 1.19%. 917

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• Fall '15
• 11:11, H0, Green Day, Billie Joe Armstrong, extremely strong evidence

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