Gm533 Week 5

GM533 Week 5 Checkpoint Solution to Question 1

b. z = (42.954 – 42) / (2.64/sqrt(65)) = 2.91

Reject Points

z.001 =3.09

z..01=2.33

z.05 = 1.645

z.10 = 1.28

Since 1.28<1.645<2.33<2.91<3.09, reject H0 with  =.10, .05, .01, but not with .001.

c. p-value 1.9982  .0018

Since p-value=.0018 is less than .10, .05, and .01, reject H0 at those levels of α, but not

with α =.001.

d. There is very strong evidence.

GM533 Week 5 Checkpoint Solution to Question 2

a. H0 : µ = 16 versus Ha : µ ≠ 16.

b. For α = .01, z α/2 = z .005 = 2.576

z = (x-bar – 16)/(1/sqrt(36))

When x = 16.05, z = 3, reject H0 ; readjust.

p-value = 2(1 – .9987) = .0026<.01, reject H0 , readjust.

CI: [16.007, 16.543], readjust

When x = 15.96, z = –2.4, cannot reject H0 : do not readjust.

p-value = 2(1 – .9918) = .0164, not less than .01, cannot reject H0 , do not readjust.

CI: [15.917, 16.003], do not readjust

When x = 16.02, z = 1.2, cannot reject H0 ; do not readjust.

p-value=2(1 – .8849) = .2302, not less than .01, cannot reject H0 , do not readjust.

CI: [15.977, 16.063], do not readjust

When x = 15.94, z = -3.6, reject H0 ; readjust.

p-value is approximately 0.000 < .001 reject H0 , readjust.

CI: [15.897, 15.983], readjust

GM533 Week 5 Checkpoint Solution to Question 3

a. H0 : µ = 42, HA: µ > 42, α = .05

b. t =( 42.95 - 42)/( 2.6424 / 65) =2.899

t.01 = 2.33 so reject H0 .

The p-value of .0019 means we would reject at α = .1, .05 and .01 but would not reject at

α = .001 so we have very strong evidence.

GM533 Week 5 Checkpoint Solution to Question 4

a. H0: μ = 750 Ha: μ does not equal 750

b. t = (x-bar - µ)/(s/sqrt(n)) = (811 – 750)/( 19.65 / sqrt(5)) = 6.94

t.01/2 = 4.604

Reject H0  = .01. p-value = 0.002261 so there is very strong evidence against H0 .

GM533 Week 5 Checkpoint Solution to Question 5

a. H0 : p = .95 versus Ha : p < .95.

b. . pˆ = 316/400 = .7979

z = (.79 - .95)/(sqrt(.95(.05)/400) = -14.68...