TECH 4433 Quality Control

TECH 4433 Quality ControlNAME________________________
Chapter 4 Homework Answer Sheet (PART 2)
(1 pt for each blank) 77 points
4.17(a)H0: u1 = u2
H1: u1 – u2 ≠ 0
TS: sp = 0.1204
tcalc = 0.106
RR: tcrit = 2.160
CC: Fail to reject H0. The claim is correct.
(b)Implications are that the mean surface measurements made by the two technicians are equal.
Suppose the null hypothesis was rejected, we would conclude that the mean surface measurements are significantly different.
(c)0.128 < u1 – u2 < 0.141
(d)H0: o-21 = o-22
H1: o-21 ≠ o-21
TS: Fcalc = 1.18
RR: Fcrit = 3.81
CC: Fail to reject H0.
Implications are that if we null hypothesis is rejected, a type 1 error will be committed.
(e)__________ < o-21 / o-22 < _____________
(f)__________ < o-22 < _____________
4.20(a)H0: p = 0.10
H1: p ≠ 0.10
TS: zcalc = -0.47
4.20RR: zcrit = -1.96
CC: Fail to reject H0.
P = 0.6374
(b)0.05 < p < 0.134.22(a)p1 = 0.05
p2 = 0.067
(b)H0: p1 = p2
H1: p1 ≠ p2
TS: zcalc = -0.77
RR: zcrit = -0.96
CC: Fail to reject the null hypothesis
(c)-0.051< p1 – p2 < 0.0184.25H0: ud = 0
H1: ud ≠ 0
TS: tcalc = -1.101
RR: tcrit = -3.11
CC: Do not reject the null hypothesis. There is no significant mean difference.
4.27(a)H0: ᴊ => 1
H1: o-2 < 1
TS: X2 calc = 44.95
RR: X2 crit = 30.14
CC: Do not reject null hypothesis.
4.27(b)1.17 < o- < 2.25
4.35(a)H0: All treatment effects are zero…