(d)
What is the P-value if the alternative hypothesis is H 1 : µ > 30
STATISTICS 402B
P-value = 0.00003
Spring 2016
Homework Set#1 Solution
2.6. Suppose that we are testing H 0 : µ 1 = µ 2 versus H 1 : µ 1 = µ 2 with a sample size of n 1 = n 2 = 12.
Both sample variances are unknown but assumed equal. Find bounds on the P-value for the following
observed
1. Problem 2.6values of the test statistic:
(a)
t 0 = 2.30
Table P-value = 0.02, 0.05
Computer P-value = 0.0313
(b)
t 0 = 3.41
Table P-value = 0.002, 0.005
Computer P-value = 0.0025
(c)
t 0 = 1.95
Table P-value =
(d)
t 0 = -2.45
Table P-value =
0.05 , 0.1
Computer P-value = 0.0640
0.02 , 0.05 Computer P-value = 0.0227
Note that the degrees of freedom is (12 +12) – 2 = 22. This is a two-sided test
2. See attached solution to this problem below.
2.7. Suppose that we are testing H 0 : µ 1 = µ 2 versus H 1 : µ 1 > µ 2 with a sample size of n 1 = n 2 = 10.
sample
variancestoare
unknown
but assumed
equal. Find bounds on the P-value for the following
3. See Both
attached
solution
this
problem
below.
Solutions
from
Montgomery,
D. C. (2012) Design and Analysis of Experiments, Wiley, NY
observed values
of the test
statistic:
4. Problem 2.24
(a)
(b)
(c)
t 0 = 2.31
Table P-value = 0.01, 0.025
Computer P-value = 0.01648
(a) State the hypotheses that should be tested in this experiment.
Table P-value = 0.001, 0.0005
Computer P-value = 0.00102
t 0 = 3.60
t 0 = 1.95
H : µ1 ≠ µ2
H0: µ1 = µ2
Table P-value = 0.05,1 0.025
Computer P-value = 0.03346
(b) Test these hypotheses using α=0.05. What are your conclusions?
y1 = 16. 015
y2 = 16. 005
σ1 = 0. 015
2-2
σ 2 = 0. 018
n1 = 10
n2 = 10
zo =
y1 − y2
σ12
n1
+
σ 22
16. 015 − 16. 018
=
0. 0152 0. 0182
+
10
10
n2
= 1. 35
z0.025 = 1.96; do not reject
(c) What is the P-value for the test? P = 0.1770
(d) Find a 95 percent confidence interval on the difference in the mean fill volume for the two machines.
The 95% confidence interval is
y1 − y 2 − z α 2
(16.015 − 16.005) − (1.96)
σ 12
n1
2
+
σ 22
≤ µ 1 − µ 2 ≤ y1 − y 2 + z α 2
n2
σ 12
n1
+
σ 22
n2
2
0.015 0.018
0.0152 0.0182
+
≤ µ1 − µ 2 ≤ (16.015 − 16.005) + (1.96)
+
10
10
10
10
− 0.0045 ≤ µ 1 − µ 2 ≤ 0.0245
2.25. Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking
strength of this plastic is important. It is known that σ1 = σ2 = 1.0 psi. From random samples of n1 = 10
and n2 = 12 we obtain y 1 = 162.5 and y 2 = 155.0. The company will not adopt plastic 1 unless its
breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should they
use plastic 1? In answering this questions, set 1up and test appropriate hypotheses using α = 0.01.
Construct a 99 percent confidence interval on the true mean difference in breaking strength.
H0: µ1 - µ2 =10
H1: µ1 - µ2 >10
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
5. Problem 2.29
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
(a) Is there evidence to support the claim that the higher baking temperature results in wafers with a lower
mean photoresist thickness? Use α = 0.05.
(c) Test an appropriate set of hypothesis indicating that the mean score does not depend on birth
order.
H 0 : µ1 = µ 2
H1 : µ1 > µ 2
H 0 : µ1 = µ22
H0 : µd = 0
(n1 − 1) S1 + or
(n2equivalently
− 1) S 22 (8 − 1)(4.41)
+ (8 − 1)(2.54)
2
SH
=
= 3.48
p =: µ ≠ µ
H
:
µ
1
d
1
1 n1 +
2 n2 − 2
8 + 8≠−02
S p = 1.86
Minitab Output
Paired T for Birth Order: 1 - Birth Order: 2
Birth Order: 1
Birth Order: 2
Difference
N
10
10
10
y1 − y2
t =
Mean 0
6.967
7.018
-0.051t
=
9.37 − 6.89
StDev1 SE1 Mean
1 1
Sp
1.86 +
+ 0.256
0.810
n
n
8 8
20.333
1.053 1
0.441
0.139
= 1.761
= 2.65
0.05,14
= 1.761,
reject H 0 . There
appears0.264)
to be a lower mean thickness at the higher temperature.
t 0.05,14mean
95%Since
CI for
difference:
(-0.366,
T-Test
ofalso
mean
= output.
0 (vs not = 0): T-Value = -0.37 P-Value = 0.723
This is
seendifference
in the computer
Minitab
Output
Do not
reject.
The P-value is 0.723.
6. We have α Two-Sample
= .05, δ =T-Test
.9, βand
= CI:
.1,Thickness,
σ = .5 Temp
From Table
3
(for
two-sided
t-test
with
α = .05) we get approx. 2n − 1 = 14 and thus a sample size of
Two-sample T for Thick@95 vs
Thick@100
2.34.
An
article
in
the
Journal
of
Strain
Analysis
2, 1983)
for of n = 8
approx 8 reps for each type of mortar. From
JMP(vol.18,
we getno.exactly
16compares
meaningseveral
againprocedures
sample size
N
Mean for steel
StDevplateSEgirders.
Mean
predicting the shear
strength
Data for nine girders in the form of the ratio of
Thick@95 8
9.37
0.74
to observed load
for two of2.10
these procedures,
the Karlsruhe and Lehigh methods, are as follows:
7. Problempredicted
2.34
Thick@10 8
6.89
1.60
0.56
Difference
= mu Thick@95
mu Thick@100
Girder
Karlsruhe -Method
Lehigh Method
Difference
Difference^2
Estimate for difference: 2.475
1.186
1.061
0.125
0.015625
S1/1
95% lower
bound for difference: 0.833
1.151
0.992
0.159
T-Test of
difference = 0 (vs >): T-Value = 2.65 P-Value = 0.0090.025281
DF = 14
S2/1
Both useS3/1
Pooled StDev
1.322 = 1.86
1.063
0.259
0.067081
1.339
1.062
0.277
0.076729
S4/1
(b) WhatS5/1
is the P-value
0.009
1.200for the test conducted
1.065in part (a)? P =0.135
0.018225
1.402
1.178
0.224
0.050176
S2/1
(c) FindS2/2
a 95% confidence
difference in means.
Provide a practical
interpretation of this
1.365 interval on the1.037
0.328
0.107584
interval.
1.537
1.086
0.451
0.203401
S2/3
1.559
1.052
0.507
0.257049
S2/4
. This lower confidence
From the computer output the 95% lower confidence
bound
is
0.833
≤
µ
−
µ
Sum =
2.465
0.821151
1
2
= in the two
0.274
bound is greater than 0; therefore, there is Average
a difference
temperatures on the thickness of the
photoresist.
(a) Is there any evidence to support a claim that there is a difference in mean performance between the two
methods? Use α = 0.05.
H 0 : µ1 = µ2
H1 : µ1 ≠ µ2
d=
or equivalently
H0 : µd = 0
H1 : µ d ≠ 0
1 n
1
di = (2.465 ) = 0.274
∑
9
n i =1
1
1
 n 2 1  n 2 
2
1

2
 ∑ di −  ∑ di  
−
0.821151
(2.465)


n
 i =1   =
9
sd =  i =1

 = 0.135


n −1
−
9
1








2
2-18
2-27
2
Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY
d
0.274
=
= 6.08
Sd
0.135
9
n
= 2.306 , reject the null hypothesis.
t0 =
tα
, n −1
= t0.025,8
(b) What is the P-value for the test in part (a)?
P=0.0002
(c) Construct a 95 percent confidence interval for the difference in mean predicted to observed load.
d − tα
Sd
,n −1
n
0.135
≤ µ d ≤ d + tα
Sd
,n −1
n
0.135
0.274 − 2.306
≤ µ d ≤ 0.274 + 2.306
9
9
0.17023 ≤ µ d ≤ 0.37777
2
2
(d) Investigate the normality assumption for both samples.
The normal probability plots of the observations for each method follow. There are no serious concerns
with the normality assumption, but there is an indication of a possible outlier (1.178) in the Lehigh method
data.
Anderson-Darling Normality Test
A-Squared: 0.286
P-Value: 0.537
Av erage: 1.34011
StDev : 0.146031
N: 9
2-28
3
Question 2
a)
s12 = (362932 − (1900) 2 / 10) / 9 = 214.6667
s 22 = ( 456412 − ( 2134) 2 / 10) / 9 = 112.9333
s p = ( s12 + s 22 ) / 2 = ( 214.6667 + 112.9333) / 2 = 12.798 with 18 d.f.
b) H 0 : µ1 − µ 2 ≥ 0 vs. H a : µ1 − µ 2 <0
s y1 − y2 = s p
t0 =
1
1
+
= 12.798 × 2 / 10 = 5.723
n1 n 2
( y1 − y2 )
190 − 213.4
=
= −4.09
5.723
1 1
sp
+
n1 n2
Rejection Region: t> t .05,18 ; t .05,18 = 1.734 so t>1.734
t0 is in the R.R. Thus, reject H 0 at α = .05. There is
evidence that Mixture 2 provides higher flare illumination.
c) 90% CI for µ1 − µ 2 is: ( y1 − y 2 ) ± t .05,18 × s p
1 1
+
n1 n2
That is,(190-213.4) ± 1.734 × 5.723 giving (-33.32,-13.48)
Since 0 is not in this interval we reject H 0 at α = .05. We also
conclude that since the entire interval is negative, that µ1 − µ 2
is negative.
d) Using =
σ 10,=
β .1,
=
α .05,
=
δ 15 we have
|𝛿𝛿| 15
𝑑𝑑 =
=
= .75
2𝜎𝜎 20
By referring to the 𝛽𝛽 curves for the one-sided test with 𝛼𝛼 = .05
the sample size we curve is between 14 and 19. Approximating it
as 18, we have that approximately 2𝑛𝑛 − 1 = 17. Thus the sample size
needed is approximately 9 from each mixture.
From the internet calculator by Russ Lenth, by inputting the
above parameters and 𝑛𝑛1 = 𝑛𝑛2 = 9 we get a power of 0.919 so the
sample sizes satisfy the criteria. JMP also gives a sample size
of 9 (use 𝛼𝛼 = 0.1 with JMP calculator for the two-sided tests as
the above is a one-sided test)
e)
I.
II.
III.
f)
g)
The two samples are independent – the mixtures were prepare
and tested independently
The two samples are random samples from normal
distributions – the boxplots and normal probability plots
(see JMP output below) do not indicate a deviation from
this requirement
The variances are equal – the sample varainces do not
differ by more than a factor of 3, indicating that the
population variances are equal; supported by both the
boxplots (similar IQR) and the normal probability plots
(parallel)
𝑡𝑡0= -4.08831 p-value=0.0003 (left-tail test)
90% CI for 𝜇𝜇1 − 𝜇𝜇2 is (-33.325, -13.475)
JMP Output
240
64
-1.28
-0.67
0.0
230
220
210
200
190
180
170
160
Mixture2
Mixture1
Mixture
Normal Quantile
Mixture2
Mixture1
t Test
Mixture1-Mixture2
Assuming equal variances
Difference
-23.400
Std Err Dif
5.724
Upper CL Dif
-13.475
Lower CL Dif
-33.325
Confidence
0.9
t Ratio
DF
Prob > |t|
Prob > t
Prob < t
-4.08831
18
0.0007*
0.9997
0.0003*
0.67
1.28
1.64
Question 3
We’ll use subscripts 1 and 2 to signify the old and new alloys, resp.
s12 =
(5474.93 − (233.7) 2 /10) / 9 =
1.4846
a) y�1 =23.37
s22 =
(8379.93 − (288.5) 2 /10) / 9 =
6.3006
y� 2 =28.85
𝑠𝑠𝟐𝟐𝟐𝟐 is more than 4 times as large as 𝑠𝑠12 leading us to suspect that
the population variances may be different.
b) H 0 : µ1 − µ 2 ≥ 0 vs. H1 : µ1 − µ2 <0
t0 =
( y1 − y2 )
2
1
2
2
s
s
+
n1 n2
=
23.37 − 28.85
= −6.21
6.3006 1.4847
+
10
10
𝑠𝑠 2 𝑠𝑠 2
(𝑛𝑛1 + 𝑛𝑛2 )2
. 778532
. 606108
1
2
=
=
= 13.018
𝑑𝑑𝑑𝑑 = 2
(𝑠𝑠1 /𝑛𝑛1 )2 (𝑠𝑠22 /𝑛𝑛2 )2 (. 630062 + . 148472 . 046558
+
9
9
𝑛𝑛1 − 1
𝑛𝑛2 − 1
So rounding down we get 𝑑𝑑𝑑𝑑 = 13.
Thus the reject 𝐻𝐻1 if 𝑡𝑡 <-𝑡𝑡.01,13. Since 𝑡𝑡.01,13 = 2.65, 𝑡𝑡0 is in the
rejection region. Thus we reject 𝐻𝐻0 at 𝛼𝛼 = .01. Thus the mean load
capacity of beams made with new alloy is larger than those made
with the old alloy.
c) α = 0.02 𝑡𝑡.01,13 = −2.6503. So the 98% confidence interval for the
difference in population means
is:
𝑠𝑠12 𝑠𝑠22
y1 − 𝑦𝑦2 ∓ 𝑡𝑡0.01,13 � +
= [−7.82, −3.14]
𝑛𝑛1 𝑛𝑛2
d) The boxplots (different IQR’s) and the slopes of normal quantile
plot (not
variances.
parallel)
indicate
possible
e) t-statistic =-6.21081 p-value<.0001
(see attached JMP output for Problem#3.)
f) 98% CI for 𝜇𝜇1 − 𝜇𝜇2 is (-3.5741, -7.3859)
(see JMP output for Problem#3 attached.)
See attached JMP output for Problem#3.
unequal
population