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
H1: C2F6 flow rate has no effect on etch uniformity
dfTreatments = 2
dfE = 16

dfT = 18
SSTreatments =
SSE = __________
SST = __________
MSTreatments = __________
MSE = __________
TS: Fcalc = 3.45RR: Fcrit = __________
CC: ___________________
4.43(a)B0 = 143.82
B1 = 1.879
Strength = 143.82 + 1.879*(% hardwood)
(b)H0: B1 = 0
H1: B1 ≠ 0
TS: tcalc = 16.125
RR: tcrit = 2.31
CC: We reject the null hypothesis. The slope is significantly different from zero.
4.47(b)H0: B1 = B2 = B3 = 0
H1: At least two Beta coefficients are different from zero
TS: Fcalc = 11.12
RR: P = 0.0032
CC: We reject the null hypothesis. The model is significant.
4.47(c)All the three independent variables are significant. There is no need to remove any from the model
H0: B1 = 0
H1: B1 ≠ 0
TS: tcalc =
RR: P = ___________
CC: ______________________________