SMAM 314
Exam 3
Name____________
1. Mark the following statements True (T) or False(F) (6 points)
___F__A. Suppose that students who had a statistics course in high school took a test on
statistics before and after completing a statistics course at RIT. In order to determine
whether there was significant improvement a t test on independent samples would be
appropriate.
__T__B. Suppose you fail to reject a null hypothesis at α = .10 . You will also fail to
reject the hypothesis at α = .01.
_F___C. A 95% confidence interval will be narrower than both a 90% confidence
interval and a 99% confidence interval.
2. Make the best choice.(4 points)
_b____A. A test of hypothesis with null hypothesis H0 µ = 10 vs. alternative hypothesis
H1 µ ≠ 10 is performed for a normal population with known standard deviation. The
calculated value of the test statistic is Z=2. The correct course of action would be:
a. Fail to reject H0 at α = .05 and α=.10 . b. Reject H0 at α = .05 but fail to reject H0
at α = .01 . c. Reject H0 at α = .05 but fail to reject H0 at α = .10 . d. Fail to reject H0
at α = .05 but reject H0 at α = .02 . e. None of the previous four choices.
_c____B. The p value for a test of hypothesis is .035. Which of the following is NOT
correct. a. Reject H0 at α = .05 and α=.10 . b. Reject H0 at α = .04 but fail to reject H0
at α = .03 . c. Reject H0 at α = .04 and α = .03 d. Fail to reject H0 at α = .02 and
α = .01. e. None of statements a-d. (That means all of the statements in a-d are true)
3. The fraction of defective integrated circuits produced in a photolithography process is
being studied. A random sample of 300 circuits is tested revealing 13 defectives.
A. Calculate a 98% two sided confidence interval on the fraction of defective circuits
produced by this process.(10 points)
 13
p=
= .0433
300
.0433 ± 2.326
(.0433)(.9567)
300
.0433 ± .0273
(.016,.0706)
B. Find a 99% upper confidence bound.(5 points)
UCB = .0706
C. What sample size is necessary to compute a 98% confidence interval with error bound
at most 0.01?(10 points)
E = zα
2
⎛ zα
N=⎜ 2
⎜⎝ E


p(1− p)
N
2
⎞


⎟ p(1− p)
⎟⎠
2
⎛ 2.326 ⎞
N=⎜
(.0433)(.9567) = 2241.21
⎝ .01 ⎟⎠
N = 2242
or
⎛ zα
N=⎜ 2
⎜⎝ E
2
⎞
⎟ (.25) = 13525.69
⎟⎠
N = 13526
4. A machine produces metal rods used in an automobile suspension system. A random
sample of six rods is selected. The resulting diameter in millimeters are as follows.
8.24 8.21
8.23
8.25 8.26
8.23
A. Find a 95% confidence interval for the true mean diameter of the rods.(10 points)
x = 8.237,s = .0175
.0175
8.237 ± 2.571
6
8.236 ± .018
(8.218,8.254)
B. Interpret the confidence interval you found in A.(5 points)
The probability that the interval contains the true mean diameter of the rods is 95%
C. Based only on your answer to A if you performed a test of hypothesis H0 µ = 8 vs H1
µ ≠ 8 would you reject H0 or fail to reject H0. Explain.(5 points)
Reject H0 because 8 is outside of the confidence interval.
5. In semiconductor manufacturing, wet chemical etching is often used to remove silicon
from the backs of wafers prior to metallization. The etch rate is an important
characteristic of this process and is known to follow a normal distribution. Two different
etching solutions have been compared using two random samples of 10 wafers for each
solution. The observed etch rates are as follows (in mils per minute).
Row
1
2
3
4
5
6
7
8
9
10
Solution 1
9.9
9.4
9.3
9.6
10.2
10.6
10.3
10.0
10.3
10.1
Solution 2
10.2
10.6
10.7
10.4
10.5
10.0
10.2
10.7
10.4
10.3
Consider the normal probability plots
Probability Plot of Solution 1, Solution 2
Normal - 95% CI
99
Variable
Solution 1
Solution 2
95
90
Mean StDev N
AD
P
9.97 0.4218 10 0.269 0.595
10.4 0.2309 10 0.211 0.804
Percent
80
70
60
50
40
30
20
10
5
1
8.5
9.0
9.5
10.0
Data
10.5
11.0
11.5
A. Is it reasonable to assume that the data is normally distributed based on the normal
probability plots? Explain.(3 points)
The data appears to be normally distributed because the points lie reasonably close to a
straight line.
B. A test was performed to determine if the assumption of equal variances was
reasonable. The result was
F-Test (Normal Distribution)
Test statistic = 3.34, p-value = 0.087
Based on the p value of the F test would you reject the null hypothesis H 0 σ12 = σ 22 in
favor of the alternative hypothesis H1 σ12 ≠ σ 22 at α = .10 ? Explain.(4 points)
The hypothesis H0 would be rejected at α = .10 because the p value is 0.087 which is
less.
C. A test of hypothesis was performed to determine whether there is a difference in the
average etch rate for the two solutions. The Minitab output follows.
Solution 1
Solution 2
N
10
10
Mean
9.970
10.400
StDev
0.422
0.231
SE Mean
0.13
0.073
Difference = mu (Solution 1) - mu (Solution 2)
Estimate for difference: -0.430
95% CI for difference: (-0.759, -0.101)
T-Test of difference = 0 (vs not =): T-Value = -2.83
P-Value = 0.014
DF = 13
Use the Minitab output above together with your understanding of hypothesis testing to
answer the following questions. No computations are needed. (18 points)
(1) What is the null and the alternative hypothesis?
H0 µ 1 = µ 2
H1 µ 1 ≠ µ 2
(2) What assumptions were made in performing this test?
Independent random samples
Normal populations
(3) What is the region of rejection based on 13 df?
T>2.160 T<-2.160
(4) From the Minitab output what is the value of the test statistic and the p value?
T=-2.83 pvalue =.014
(5) At α = .05 would you reject H0 or fail to reject H0?
Reject H0
(6) Make a statement in ordinary English concerning the difference between the etch
rates?
There is strong evidence of a significant difference between the two etch rates.
6. A computer scientist is investigating the usefulness of two different design languages
in improving programming tasks. Twelve expert programmers familiar with both
languages are asked to code a standard function in both languages and the time in
minutes is recorded. The data follow.
Programmer
1
2
3
4
5
6
7
8
9
10
11
12
Design
Language 1
17
16
21
14
18
24
16
14
21
23
13
18
Design
Language 2
18
14
19
11
23
21
10
13
19
24
15
20
Difference
-1
2
2
3
-5
3
6
1
2
-1
-2
-2
Summary Statistics
Design Language 1
Design Language 2
Difference
N
12
12
12
Mean StDev SE Mean
17.92
3.63
1.05
17.25
4.59
1.33
0.667 2.964
0.856
Perform an appropriate test of hypothesis to determine whether the average time for
coding is greater for Design Language 1 than for Design Language 2. Use α = .05 .
Write the complete report. (20 points)
H0 µ d = 0
H1 µ d > 0
Assumptions
Paired data
Differences are normally distributed
Region of rejection
reject H0 if T>1.796
Computation
0.667
T=
= .7795
2.964 / 12
Reject or fail to reject H0
Fail to reject H0
Conclusion –Statement in ordinary English
There does not appear to be a significant difference in the coding time for the two
languages.