Checkpoint 5

Complete Exercise 9.13 (The Video Game Satisfaction Case) on page 357 in your textbook

A. H0 ≤ 42 versus Ha >42

b: z = (42.954 - 42)/(2.64/√64) = 2.891

Z values at
0.10 = 1.28 - Reject
0.05 = 1.64 - Reject
0.01 = 2.33 - Reject
0.001 = 3.09 - Accept

Since we are dealing with the right side of the curve the z values are positive.

c:

Z = 2.891 therefore right side = 0.0019
Left side = 0.9981

Since p is = 0.0019 it less than 0.10, 0.05. 0.01 but greater than 0.001

D. Very strong evidence

Complete Exercise 9.19 on page 358 in your textbook. (

A. H0 = 16 versus Ha ≠ 16

b. α = 0.01
S = 0. 1
b.1 Xbar = 16.05 b.2 Xbar = 15.96
b.3 Xbar = 16.02
b.4 Xbar = 15.94

two tailed critical value rule requires we split α in 2, Therefore

(Z 0.01/2)= 2.576

Z = (xbar – 16)/( 0.1/6)

B.1 Critical Value
Z = (16.05 – 16)/( 0.1/6)
Z = 3.0 – therefore readjust

P value = 2 time the area to the right of z
2(1- Z) = 2(1- 0.99865) = 0.027
0.027 < 0.01 – therefore reject and readjust

CI
Xbar ± (2.576)( 0.01/6)
16.05 ± 0.0429 = [16.007, 16.093] readjust

B.2

Critical Value
Z = (15.96 – 16)/( 0.1/6)
Z = -2.40 – therefore do not reject do not readjust

P value = 2 time the area to the right of z
2(1- Z) = 2(1- 0.9918) = 0.0164
0.0164 > 0.01 – therefore do not reject and do not readjust

CI
Xbar ± (2.576)( 0.1/6)
15.96 ± 0.0429 = [15.917,16.003] – reject and readjust

B.3

Critical Value
Z = (16.02 – 16)/( 0.1/6)
Z = 1.2 – therefore do not reject do not readjust

P value = 2 time the area to the right of z
2(1- Z) = 2(1- 0.8849) = 0.2302
0.2302 > 0.01 – therefore do not reject and do not readjust

CI
Xbar ± (2.576)( 0.1/6)
16.02 ± 0.0429 = [15.97,16.06] – do not readjust

B.4

Critical Value
Z = (15.94 – 16)/( 0.1/6)
Z = - 3.6 – therefore reject