Stat 401B
Lab#6 Solution Key
Fall 2014
Problem 1
a) The normal probability plot (see attached JMP output) shows that the points fall approximately
on a straight line supporting the assumption that the paired differences of PCB measurements
at the sites is a sample from a normal distribution. The box plot also supports this conclusion.
See the attached Excel output for the calculation of sd and sd .
b) Let µ d = µ1982 − µ1996 where µ1982 and µ1996 are assumed to be the means of PCB contents in
gull egg populations during those years. Assuming that the paired differences in PCB content is
a sample from normal population with mean µ d the following hypotheses can be tested:
H 0 : µ d ≤ 0 vs. H a : µ d > 0
tc =
38.80
d −0
=
= 11.23
sd
12.46 / 13
t.01,12 = 2.681 , that is R.R. is t > 2.681
Since t c is in R.R. we rewject the null hypothesis at α = .01 and conclude that
there has been a significant decrease in mean PCB content in the herring gull
egg population.
c) The p-value for this test is the probability of obtaining a value greater than 11.23 for the tstatistic if the null hypothesis is true. (or in other words, the probability of observing a difference
of greater than 38.8 if there was no difference in the population means).
Can calculate this as p-value= P[T12 > 11.23]
From the t-tables we can bound this as 0 < p < .0005
d) A 98% confidence interval is:
d ± t.01,12 × sd ; that is 38.8 ± (2.681)(3.455) or (29.53,48.06)
Using this interval to test H 0 : µ d ≤ 0 vs. H a : µ d > 0 ,
We reject the null hypothesis at α = .01 because 0 is not in the above interval.
e) From the JMP output attached, t= 11.2291 and the p-value is <.0001
f) From the JMP output attached, 98% CI is (29.538,48.067)
Problem 2
(a)
No violations, the boxplot resembles an approximately normal distribution and the normal
probability plot falls in a straight line, which represents a normal distribution.
(b)
𝐻0 : 𝜎 2 ≥ 1
𝐻𝑎 : 𝜎 2 < 1
2
𝑆 =
𝑥2 =
∑ 𝑦2−
2
(∑ 𝑦)
𝑛
𝑛−1
(𝑛−1)𝑆 2
𝜎02
= 0.79
= 15.01
2
= 10.12
𝑥0.95,19
Fail to reject 𝐻0 at 𝛼 = 0.05. There is insufficient evidence to determine that the standard
deviation meets the criterion.
(c)
(d)
(e)
(𝑛−1)𝑆2
95% 𝐶𝐼 = �
𝑥 2 = 15.02
2
𝑥𝑈
(𝑛−1)𝑆 2
<𝜎<�
𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 0.2786
95% 𝐶𝐼 = (0.68, 1.30)
𝑥𝐿2
= (0.68, 1.30)
Problem 3
(a)
Both plots fall approximately in a straight line, so the normality assumption is reasonable. The
two lines are parallel, so the assumption that the population variances are the same is
reasonable.
(b)
𝑆𝐴2
(c)
𝑆𝐵2
=
=
∑ 𝑦2−
2
(∑ 𝑦)
𝑛
∑ 𝑦2−
2
(∑ 𝑦)
𝑛
𝑛−1
𝑛−1
𝐻0 : 𝜎𝐴2 = 𝜎𝐵2
𝐻𝑎 : 𝜎𝐴2 ≠ 𝜎𝐵2
𝐹=
2
𝑆𝐴
2
𝑆𝐵
= 0.442
= 0.476
= 0.946
𝑅𝑅: 𝐹 > 𝐹0.025,9,9 = 4.03, 𝑜𝑟 𝐹 < 𝐹0.9755,9,9 =
1
4.03
= 0.248
Fail to reject 𝐻0 at 𝛼 = 0.05. There is insufficient evidence that the population variances are
different.
(d)
𝐹 = 1.056
𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 0.9366
Problem 1 a)
61.48
64.47
45.5
59.7
58.81
75.86
71.57
38.06
30.51
39.7
29.78
66.89
63.93
13.99
18.26
11.28
10.02
21
17.36
28.2
7.3
12.8
9.41
12.63
16.83
22.74
Excel calculations of s_d and s_dbar
47.49
46.21
34.22
49.68
37.81
58.5
43.37
30.76
17.71
30.29
17.15
50.06
41.19
2255.3001
2135.3641
1171.0084
2468.1024
1429.5961
3422.25
1880.9569
946.1776
313.6441
917.4841
294.1225
2506.0036
1696.6161
504.44
38.8030769
21436.626
155.2334897
12.45927324
Problem 1 d) e) f)
Analysis of Gull Eggs PCB data (JMP Output)
Test Mean=value
Hypothesized Value
Actual Estimate
DF
Std Dev
Test Statistic
Prob > |t|
Prob > t
Prob < t
0
38.8031
12
12.4593
t Test
11.2291
<.0001*
<.0001*
1.0000
Confidence Intervals
Parameter
Mean
Std Dev
Estimate
38.80308
12.45927
Lower CI
29.53867
8.429312
Upper CI
48.06748
22.84097
1-Alpha
0.980
0.980
Stat401B
Etch Uniformity
Lab#6
8
-0.67
-1.64 -1.28
0.0
Prob#2 JMP Output
1.28 1.64
0.67
7.5
7
6.5
6
5.5
5
4.5
4
0.1
0.2
0.5
0.8
0.9 0.95
Normal Quantile Plot
Moments
Mean
Std Dev
Std Err Mean
Upper 95% Mean
Lower 95% Mean
N
5.828
0.8890717
0.1988025
6.2440983
5.4119017
20
Test Standard Deviation
Hypothesized Value
Actual Estimate
DF
Test
Test Statistic
Min PValue
Prob < ChiSq
Prob > ChiSq
1
0.88907
19
ChiSquare
15.01852
0.5572
0.2786
0.7214
Confidence Intervals
Parameter
Mean
Std Dev
Estimate
5.828
0.889072
Lower CI
5.411902
0.67613
Upper CI
6.244098
1.298553
1-Alpha
0.950
0.950
Stat401B
Lab#6
44
Prob#3 JMP Output
-1.64 -1.28
-0.67
0.0
0.67
1.28 1.64
A
44
43
B
Time
43
42
42
41
0.9
0.9
0.8
0.5
0.2
B
0.1
40A
0.0
41
Machine
Normal Quantile
Means and Std Deviations
Level
A
B
Number
10
10
Mean
43.2700
42.1400
Std Dev Std Err Mean
0.665081
0.21032
0.683455
0.21613
Lower 95%
42.794
41.651
Tests that the Variances are Equal
0.8
Std Dev
0.6
0.4
0.2
0.0A
B
Machine
Level
Count
Std Dev
10
10
0.6650814
0.6834553
A
B
Test
O'Brien[.5]
Brown-Forsythe
Levene
Bartlett
F Test 2-sided
F Ratio
0.0110
0.0034
0.0006
0.0063
1.0560
MeanAbsDif to
Mean
0.5360000
0.5400000
DFNum
1
1
1
1
9
DFDen
18
18
18
.
9
Welch's Test
Welch Anova testing Means Equal, allowing Std Devs Not Equal
F Ratio
14.0404
t Test
3.7471
DFNum
1
DFDen
17.987
Prob > F
0.0015*
MeanAbsDif to
Median
0.5300000
0.5400000
p-Value
0.9176
0.9543
0.9807
0.9366
0.9366
Upper 95%
43.746
42.629