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Unformatted text preview: Fall 2009 Hansen
Assignment 12
Due 12/1/2009 9:10 AM 1. Problem 8.5 in Ross.
: = 200
: < 200
=5
=8
We can find = 199.11 5
− = 8 199.11 − 200
= −0.502
5 = 5% level of significance:
= . = −1.645
Since
> −1.645, we fail to reject
at the 5% level
10% level of significance:
= . = −1.28
Since
> −1.28, we fail to reject
at the 10% level either
2. Problem 8.20 in Ross.
: = 30
: < 30
unknown
= 10
We can find = 26.4 and
= = 3.5024
− = 10 26.4 − 30
= −3.25
3.5024 Could test at 5% level of significance: − ,
= − . , = −1.833
However, students might have also tested at other levels of significance: 1%, 5%, or 10% are
all reasonable values.
− . , = −2.821
− . , = −1.383
Since
< −1.833, −2.821,
− 1.383, we reject
at the 5%, 1% and 10% levels.
Based on the sample, we cannot believe the advertisement. Fall 2009 Hansen We are assuming that average gas mileage rating is normally distributed.
3. Problem 8.24 in Ross.
: = 30
: < 30
unknown
= 16
We can find = 28.25 and = 3.7907 = − = 16 28.25 − 30
= −1.8466
3.7907 5% level of significance: − ,
= − . , = −1.7531
Since
< −1.7531, we reject
at the 5% level. Based on the sample, the claim is
unjustified.
4. Graduate students.
= 14,
= 21
= 31.64,
= 24.38
= 11.44,
= 12.488
= = 146.07 = a)
, : Since
: b) , Since
c)
, : Since = = vs.
. , : −
1 + 1 = vs.
. , : 1
1
146.07 ∗ 14 + 21 at the 5% level. ≠
, 10% significance
= 1.6924 > 1.6924 we reject
= 31.64 − 24.38 ≠
, 5% significance
= 2.0345 < 2.0345 we cannot reject
= = at the 10% level. :>
, 5% significance
= . , = 1.6924
> 1.6924 we reject
at the 5% level.
vs. = 1.741 ...
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This note was uploaded on 09/19/2011 for the course CE 93 taught by Professor Hansen during the Fall '10 term at University of California, Berkeley.
 Fall '10
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