Transcript Hwk 2 Soln

Hw 2 Prob 1
•
Two new methods for producing a tire have been proposed. To ascertain
which is superior, a tire manufacturer produces a sample of 10 tires using the
first method and a sample of 8 using the second. The first set are to be road
tested at location A and the second at location B. It is known from past
experience that the lifetime of a tire that is road tested at one of these
locations is normally distributed with a mean life due to the tire but with a
variance due (for the most part) to the location. Specifically, it is known that
the lifetimes of tires tested at location A are normal with standard deviation
equal t0 4000 kilometers, whereas those tested at location B are normal with
σ = 6000 kilometers. If the manufacturer is interested in testing the
hypothesis that there are no appreciable difference in the mean life of tires
produced by either method, what conclusion should be drawn at the 5% level
of significance if the resulting data are as given in the table below?
Tire Lives in Units of 100 Kilometers
Tires tested at A
Tires tested at B
61.1
58.2
62.3
64
59.7
66.2
57.8
61.4
62.2
63.6
X A  6,165
62.2
56.6
66.4
56.2
57.4
58.4
57.6
65.4
X B  6,002
Hw 2 Prob 1
•
Two new methods for producing a tire have been proposed. To ascertain
which is superior, a tire manufacturer produces a sample of 10 tires using the
first method and a sample of 8 using the second. The first set are to be road
tested at location A and the second at location B. It is known from past
experience that the lifetime of a tire that is road tested at one of these
locations is normally distributed with a mean life due to the tire but with a
variance due (for the most part) to the location. Specifically, it is known that
the lifetimes of tires tested at location A are normal with standard deviation
equal t0 4000 kilometers, whereas those tested at location B are normal with
σ = 6000 kilometers. If the manufacturer is interested in testing the
hypothesis that there are no appreciable difference in the mean life of tires
produced by either method, what conclusion should be drawn at the 5% level
of significance if the resulting data are as given in the table below?
X A  6,165
X B  6,002
H o :  A  B
H A :  A  B
z
XA  XB

2
A
nA


2
B

nB
Since Z=0.066 is neither in
the right tail nor the left tail,
we have no statistical
evidence they are different
and we accept Ho.
6,165  6,002
2
4000 6000

10
8
.025
1.96
2
 0.066
0.95
0
.025
1.96
Hw 2 Prob 2
•
A public health official claims that the mean home water use is 350 gallons a
day. To verify this claim, a study of 20 randomly selected homes was
instigated with the result that the average daily water uses of these 20 homes
were as follows:
340
344
362
375
356
386
354
364
332
402
340
355
362
322
372
324
318
360
338
370
Do the data contradict the official’s claim?
X  353.8 , s  21.85
H o :   350
H A :   350
T
X  o 353.8  350.0

 0.78
s
21.85
n
20
t19
Z=0.78 lies between
+/- 2.145. We do not .025
have enough
evidence to conclude
2.145
our hypothesis is
wrong. Accept Ho.
0.95
0
.025
2.145
Hw 2 Prob 3
•
An industrial safety program was recently instituted in the computer chip
industry. The average weekly loss (averaged over one month) in man-hours
due to accidents in 10 similar plants both before and after the program are as
follows:
Plant
Before
After
1
30.5
23
2
18.5
21
3
24.5
22
4
32
28.5
5
16
14.5
6
15
15.5
7
23.5
24.5
8
25.5
21
9
28
23.5
10
18
16.5
Determine at the 5 percent level of significance, whether the safety program has
been proven to be effective.
d  2.15 , sd  3.00
H o :  post  pre  0
H A :  post  pre  0
T
d
sd
 2.15

  2.266
3.00
10
n
t9
.05
0.95
2.015
0
T = -2.266 is left of -2.015. Therefore, we have enough evidence to
conclude there is an improvement. Reject Ho, accept HA.
Hw 2 Prob 4
•
A machine that automatically controls the amount of ribbon on a tape has
recently been installed. This machine will be judged to be effective if the
standard deviation σ of the amount of ribbon on a tape is less than 0.15 cm.
If a sample of 20 tapes yields a sample variance of S2 = .025 cm2, are we
justified in concluding that the machine is ineffective?
H o :   0.15
H A :   0.15
( n  1)s 2 19(.025)
 

 21.11
 o2
.152
2
X219
21.11 < 30.14
21.11<30.14; therefore we
do not have enough
evidence to reject the null
hypothesize. Accept Ho,
machine is ineffective.
.05
30.14
Hw 2 Prob 5
•
There are two different choices of a catalyst to stimulate a certain chemical
process. To test whether the variance of the yield is the same no matter
which catalyst is used, a sample of 10 batches id produced using the first
catalyst, and 12 using the second. If the resulting data is S21 = .14 and S22 =
.28, can we reject, at the 10 percent level, the hypothesis of equal variance?
H o :  12   22
Note that I changed a so we could
look up two tail.
s12
F  2  0.5 F9 ,11
s2
.322 < F=.5 < 2.90;
therefore, we do not
have enough evidence
to reject Ho. Variances
are equal.
F9,11
.05
F.95,9,11=1/F.95,11,9
=1/3.2
= .322
.05
F.05,9,11=2.90