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Chapter 2
Simple Comparative Experiments
Solutions
2-1 The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that
the standard deviation of breaking strength is σ = 3 psi. A random sample of four specimens is tested. The
results are y 1 =145, y 2 =153, y 3 =150 and y 4 =147.
(a) State the hypotheses that you think should be tested in this experiment.
H 0 : μ = 150 H 1 : μ > 150
(b) Test these hypotheses using α = 0.05. What are your conclusions?
n = 4, σ = 3, y = 1/4 (145 + 153 + 150 + 147) = 148.
o o
y z
n
μ σ
Since z 0.05 = 1.645, do not reject.
(c) Find the P -value for the test in part (b).
From the z- table: P ≅ 1 −[ 0_._ 7967 +( 23 )( 0_._ 7995 − 0_._ 7967 )] = 0_._ 2014
(d) Construct a 95 percent confidence interval on the mean breaking strength.
The 95% confidence interval is
- 75 ( 1. 96 )( 32 ) 148. 75 ( 1. 96 )( 32 )
2 2 − ≤ ≤ +
μ
σ μ σ α α n
y z n
y z
2-2 The viscosity of a liquid detergent is supposed to average 800 centistokes at 25°C. A random sample of 16 batches of detergent is collected, and the average viscosity is 812. Suppose we know that the
standard deviation of viscosity is σ = 25 centistokes.
(a) State the hypotheses that should be tested.
H 0 : μ = 800 H 1 : μ ≠ 800
(b) Test these hypotheses using α = 0.05. What are your conclusions?
o o
y z
n
= = = = Since^ z^ α/2^ =^ z 0.025^ = 1.96, do not reject.
(c) What is the P -value for the test? P = 2 0 0274(. ) =0 0549.
(d) Find a 95 percent confidence interval on the mean.
The 95% confidence interval is
n
y z n
y z
σ μ
σ − α 2 ≤ ≤ + α 2
( )( ) ( )( )
2-3 The diameters of steel shafts produced by a certain manufacturing process should have a mean
diameter of 0.255 inches. The diameter is known to have a standard deviation of σ = 0.0001 inch. A
random sample of 10 shafts has an average diameter of 0.2545 inches.
(a) Set up the appropriate hypotheses on the mean μ.
H 0 : μ = 0.255 H 1 : μ ≠ 0.
(b) Test these hypotheses using α = 0.05. What are your conclusions?
n = 10, σ = 0.0001, y = 0.
o o
y z
n
μ σ
Since z 0.025 = 1.96, reject H 0.
(c) Find the P -value for this test. P = 2.6547x10-
(d) Construct a 95 percent confidence interval on the mean shaft diameter.
The 95% confidence interval is
n
y z n
y z
σ μ
σ − α 2 ≤ ≤ + α 2
( ) ( )
μ
2-4 A normally distributed random variable has an unknown mean μ and a known variance σ^2 = 9. Find
the sample size required to construct a 95 percent confidence interval on the mean, that has total length of 1.0.
(c) Find the P -value for the test in part (b). P =0.
(d) Construct a 99 percent confidence interval on the mean shelf life.
The 99% confidence interval is (^2) , n 1 , 1
S
y t y t n n
− α − ≤ μ≤ + 2 n
S
α − with^ α^ = 0.01.
( ) ( )
μ
2-6 Consider the shelf life data in Problem 2-5. Can shelf life be described or modeled adequately by a normal distribution? What effect would violation of this assumption have on the test procedure you used in solving Problem 2-5?
A normal probability plot, obtained from Minitab, is shown. There is no reason to doubt the adequacy of the normality assumption. If shelf life is not normally distributed, then the impact of this on the t -test in problem 2-5 is not too serious unless the departure from normality is severe.
86 96 106 116 126 136 146 156 166 176
99
95 90 80 70 60 50 40 30 20 10 5
1
Data
Percent
AD* 1.
Goodness of Fit
Normal Probability Plot for Shelf Life ML Estimates
Mean StDev
131
ML Estimates
2-7 The time to repair an electronic instrument is a normally distributed random variable measured in hours. The repair time for 16 such instruments chosen at random are as follows:
Hours 159 280 101 212 224 379 179 264 222 362 168 250 149 260 485 170
(a) You wish to know if the mean repair time exceeds 225 hours. Set up appropriate hypotheses for investigating this issue.
H 0 : μ = 225 H 1 : μ > 225
(b) Test the hypotheses you formulated in part (a). What are your conclusions? Use α = 0.05.
y = 247. S^2 =146202 / (16 - 1) = 9746.
S = 9746.8 =98.
o o
y t S n
since t 0.05,15 = 1.753; do not reject H 0
Minitab Output T-Test of the Mean
Test of mu = 225.0 vs mu > 225.
Variable N Mean StDev SE Mean T P Hours 16 241.5 98.7 24.7 0.67 0.
T Confidence Intervals
Variable N Mean StDev SE Mean 95.0 % CI Hours 16 241.5 98.7 24.7 ( 188.9, 294.1)
(c) Find the P -value for this test. P =0.
(d) Construct a 95 percent confidence interval on mean repair time.
The 95% confidence interval is (^2) , n 1 2 , n 1
S S
y t y t n n
( ) ( )
μ
2-8 Reconsider the repair time data in Problem 2-7. Can repair time, in your opinion, be adequately modeled by a normal distribution?
The normal probability plot below does not reveal any serious problem with the normality assumption.
The 95% confidence interval is
2
2 2 1
2 1 1 2 1 2 2
2 2 1
2 1 y 1 y 2 z (^2) n n y y z (^2) n n
− − α + ≤ − ≤ − + α +
10
018 10
015 ( 16. 015 16. 005 ) ( 19. 6 ) 10
018 10
015 ( 16. 015 16. 005 ) ( 19. 6 )
2 2 1 2
2 2 − − + ≤μ −μ ≤ − + +
− 0. 0045 ≤μ 1 −μ 2 ≤ 0. 0245
2-10 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 σ 1 = σ 2 = 1.0 psi. From random samples of n 1 = 10
and n 2 = 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 up and test appropriate hypotheses using α = 0.01.
Construct a 99 percent confidence interval on the true mean difference in breaking strength.
H 0 : μ 1 - μ 2 =10 H 1 : μ 1 - μ 2 >
1
1
1
n
y.
2
2
2
n
y.
z
y y
n n
o =^
1
2
1
2
2
2
2 2
z 0.01 = 2.225; do not reject
The 99 percent confidence interval is
2
2 2 1
2 1 1 2 1 2 2
2 2 1
2 1 (^1 22 )
n n
y y z
n n
y y z
2 2 1 2
2 2 − − + ≤μ −μ ≤ − + +
2-11 The following are the burning times (in minutes) of chemical flares of two different formulations. The design engineers are interested in both the means and variance of the burning times.
Type 1 Type 2 65 82 64 56 81 67 71 69 57 59 83 74 66 75 59 82 82 70 65 79
(a) Test the hypotheses that the two variances are equal. Use α = 0.05.
2 2 0 1 2 2 1 1 2
:
:
H
H 2
σ σ
σ σ
=
≠
S
S
1 2
F
S
S
0 1
2
2
2
F 0 025 9 9. , , = 4 03. F
F
0 975 9 9 0 025 9 9
. , ,.^
= = =. Do not reject.
(b) Using the results of (a), test the hypotheses that the mean burning times are equal. Use α = 0.05.
What is the P -value for this test?
S
n S n S n n S
t
y y
S n n
p
p
p
2 1 1
2 2 2
2
1 2
0
1 2
1 2
t 0 025 18. , = 2 101. Do not reject.
From the computer output, t =0.05; do not reject. Also from the computer output P =0.
Minitab Output Two Sample T-Test and Confidence Interval
Two sample T for Type 1 vs Type 2
N Mean StDev SE Mean Type 1 10 70.40 9.26 2. Type 2 10 70.20 9.37 3.
95% CI for mu Type 1 - mu Type 2: ( -8.6, 9.0) T-Test mu Type 1 = mu Type 2 (vs not =): T = 0.05 P = 0.96 DF = 18 Both use Pooled StDev = 9.
(c) Discuss the role of the normality assumption in this problem. Check the assumption of normality for both types of flares.
The assumption of normality is required in the theoretical development of the t -test. However, moderate departure from normality has little impact on the performance of the t -test. The normality assumption is more important for the test on the equality of the two variances. An indication of nonnormality would be of concern here. The normal probability plots shown below indicate that burning time for both formulations follow the normal distribution.
Minitab Output Two Sample T-Test and Confidence Interval
Two sample T for Uniformity
Flow Rat N Mean StDev SE Mean 125 6 3.317 0.760 0. 200 6 3.933 0.821 0.
95% CI for mu (125) - mu (200): ( -1.63, 0.40) T-Test mu (125) = mu (200) (vs not =): T = -1.35 P = 0.21 DF = 10 Both use Pooled StDev = 0.
(b) What is the P -value for the test in part (a)? From the computer printout, P =0.
(c) Does the C 2 F 6 flow rate affect the wafer-to-wafer variability in etch uniformity? Use α = 0.05.
2 2 0 1 2 2 2 1 1 2 0.05,5,
0
:
:
H
H F
F
σ σ
σ σ
=
≠
= =
Do not reject; C 2 F 6 flow rate does not affect wafer-to-wafer variability.
(d) Draw box plots to assist in the interpretation of the data from this experiment.
The box plots shown below indicate that there is little difference in uniformity at the two gas flow rates. Any observed difference is not statistically significant. See the t -test in part (a).
125 200
5
4
3
Flow Rate
Uniformity
2-13 A new filtering device is installed in a chemical unit. Before its installation, a random sample
yielded the following information about the percentage of impurity: y (^) 1 = 12.5, S (^) 12 =101.17, and n 1 = 8.
After installation, a random sample yielded y (^) 2 = 10.2, S (^) 22 = 94.73, n 2 = 9.
(a) Can you concluded that the two variances are equal? Use α = 0.05.
2 2
2 1 0
002578
2 2
2 1 1
2 2
2 0 1
S
S
F
F.
H :
H :
. ,,
= = =
Do Not Reject. Assume that the variances are equal.
(b) Has the filtering device reduced the percentage of impurity significantly? Use α = 0.05.
00515
1 2
1 2 0
1 2
2 2 2
2 (^211)
1 1 2
0 1 2
t.
n n
S
y y t
S.
n n
(n )S (n )S S
H :
H :
. ,
p
p
p
Do not reject. There is no evidence to indicate that the new filtering device has affected the mean
2-14 Photoresist is a light-sensitive material applied to semiconductor wafers so that the circuit pattern can be imaged on to the wafer. After application, the coated wafers are baked to remove the solvent in the photoresist mixture and to harden the resist. Here are measurements of photoresist thickness (in kÅ) for eight wafers baked at two different temperatures. Assume that all of the runs were made in random order.
95 ºC 100 ºC 11.176 5. 7.089 6. 8.097 7. 11.739 7. 11.291 8. 10.759 7. 6.467 3. 8.315 8.
(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.
Thickness
Temp 95 100
Dotplot of Thickness vs Temp
(e) Check the assumption of normality of the photoresist thickness.
5 10 15
1
5
10
20
30
40
50
60
70
80
90
95
99
Data
Percent
AD*^ 1.
Goodness of Fit
Normal Probability Plot for Thick@ ML Estimates - 95% CI
Mean StDev
ML Estimates
2 7 12
1
5
10
20
30
40
50
60
70
80
90
95
99
Data
Percent
AD*^ 1.
Goodness of Fit
Normal Probability Plot for Thick@ ML Estimates - 95% CI
Mean StDev
ML Estimates
There are no significant deviations from the normality assumptions.
(f) Find the power of this test for detecting an actual difference in means of 2.5 kÅ. Minitab Output Power and Sample Size
2-Sample t Test
Testing mean 1 = mean 2 (versus not =) Calculating power for mean 1 = mean 2 + difference Alpha = 0.05 Sigma = 1.
Sample Difference Size Power 2.5 8 0.
(g) What sample size would be necessary to detect an actual difference in means of 1.5 kÅ with a power of at least 0.9?.
Minitab Output Power and Sample Size
2-Sample t Test
Testing mean 1 = mean 2 (versus not =) Calculating power for mean 1 = mean 2 + difference Alpha = 0.05 Sigma = 1.
Sample Target Actual Difference Size Power Power 1.5 34 0.9000 0.
Ranking
C 10 sec 20 sec
Dotplot of Ranking vs C
(e) Check the assumption of normality for the data from this experiment.
0 4 8
1
5
10
20
30
40
50
60
70
80
90
95
99
Data
Percent
AD*^ 1.
Goodness of Fit
Normal Probability Plot for 10 seconds ML Estimates - 95% CI
Mean StDev
ML Estimates
2 3 4 5 6 7 8 9 10 11
1
5
10
20
30
40
50
60
70
80
90
95
99
Data
Percent
AD*^ 0.
Goodness of Fit
Normal Probability Plot for 20 seconds ML Estimates - 95% CI
Mean StDev
ML Estimates
There are no significant departures from normality.
2-16 Twenty observations on etch uniformity on silicon wafers are taken during a qualification experiment for a plasma etcher. The data are as follows:
5.34 6.65 4.76 5.98 7. 6.00 7.55 5.54 5.62 6. 5.97 7.35 5.44 4.39 4. 5.25 6.35 4.61 6.00 5.
(a) Construct a 95 percent confidence interval estimate of σ^2.
( ) ( )
( )( ) ( )( )
2 2 2 2 2 2 , 1 (1^2 ),^1 2 2 2
2
n n
n S n S α α
− − −
(b) Test the hypothesis that σ^2 = 1.0. Use α = 0.05. What are your conclusions?
H
H
0
2
1
2
σ
σ
χ σ 0 2 0
SS
(a) Is there a significant difference between the means of the population of measurements represented by
the two samples? Use α = 0.05.
H
H
0 1 2 1 1 2
or equivalently 0
1
0 ≠
d
d H :
H :
Minitab Output Paired T-Test and Confidence Interval
Paired T for Caliper 1 - Caliper 2
N Mean StDev SE Mean Caliper 12 0.266250 0.001215 0. Caliper 12 0.266000 0.001758 0. Difference 12 0.000250 0.002006 0.
95% CI for mean difference: (-0.001024, 0.001524) T-Test of mean difference = 0 (vs not = 0): T-Value = 0.43 P-Value = 0.
(b) Find the P -value for the test in part (a). P =0.
(c) Construct a 95 percent confidence interval on the difference in the mean diameter measurements for the two types of calipers.
, 1 (^1 2 ) , 1 2 2 0.002 0. 0.00025 2.201 0.00025 2. 12 12 0.00102 0.
d d n D n
d
d
S S
d t d t n n
α μ^ μ^ μ α
2-18 An article in the Journal of Strain Analysis (vol.18, no. 2, 1983) compares several procedures for predicting the shear strength for steel plate girders. Data for nine girders in the form of the ratio of predicted to observed load for two of these procedures, the Karlsruhe and Lehigh methods, are as follows:
Girder Karlsruhe Method Lehigh Method Difference Difference^ S1/1 1.186^ 1.061^ 0.125^ 0. S2/1 1.151^ 0.992^ 0.159^ 0. S3/1 1.322^ 1.063^ 0.259^ 0. S4/1 1.339^ 1.062^ 0.277^ 0. S5/1 1.200^ 1.065^ 0.135^ 0. S2/1 1.402^ 1.178^ 0.224^ 0. S2/2 1.365^ 1.037^ 0.328^ 0. S2/3 1.537^ 1.086^ 0.451^ 0. S2/4 1.559^ 1.052^ 0.507^ 0. Sum = 2.465 0. Average = 0.
(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
H
0 1 2 1 1 2
or equivalently 0
1
0 ≠
d
d H :
H :
( ) 1
n i i
d d n (^) =
= (^) ∑ = =
2 12 (^12) (^2 ) 1 1
n n i i i i d
d d n s n
= =
⎢ − ⎜ ⎟ ⎥ ⎡^ − ⎤
⎢ ⎝^ ⎠ ⎥ ⎢^ ⎥
⎢ ⎥ ⎢^ ⎥
⎢ ⎥ ⎣^ ⎦
∑ ∑
0
d
d t S n
t (^) α (^2) ,n − 1 = t 0_._ 025 , 9 = 2_._ 306 , reject the null hypothesis.
Minitab Output Paired T-Test and Confidence Interval
Paired T for Karlsruhe - Lehigh
N Mean StDev SE Mean Karlsruh 9 1.3401 0.1460 0. Lehigh 9 1.0662 0.0494 0. Difference 9 0.2739 0.1351 0.
95% CI for mean difference: (0.1700, 0.3777) T-Test of mean difference = 0 (vs not = 0): T-Value = 6.08 P-Value = 0.
(b) What is the P -value for the test in part (a)? P =0.
(c) Construct a 95 percent confidence interval for the difference in mean predicted to observed load.
2 1 2 1
n
S
d t n
S
d t
d
d
d d ,n
d ,n
α μ^ α
(d) Investigate the normality assumption for both samples.
P-Value: 0. A-Squared: 0.
Anderson-Darling Normality Test N: 9 StDev : 0.
Av erage: 1.
1.15 1.25 1.35 1.45 1.
. . . . . . . . .
Probability
Karlsruhe
Normal Probability Plot
