Solutions to Practice Problems for two sample hypothesis tests

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Solutions to Practice Problems for two sample hypothesis tests
1. A company that manufactures guitar picks in two shifts. Production supervisor Vinny
Pickman must keep tabs on both departments making sure that the number of
defectives is not particularly higher or lower in either shift. Each month he chooses a
random sample of days and counts the number of defects. Last month the morning
shift averaged 10 defects with a standard deviation of 5 defects in a sample of 30 days.
The evening shift, in a sample of 30 days averaged 18 defects with a standard deviation
of 6 defects. Should Vinny schedule a reprimand for one of the shift supervisors?
Explain your answer.
Two averages (independent samples)
First check for equal standard deviations
Ho: σ1=σ2
Ho: σ1≠σ2
TS = 62/52 =36/25=1.44
CV F.05,29,29=1.84
CV
FTR
REJ
1.44
1.84
So the standard deviations are not significantly different so we will use the pooled
variance test statistic.
Ho: μm-μe = 0 (the shifts are the same)
Ho: μm-μe ≠ 0 (the shifts are different) (if positive morning worse, and vice versa)
sp2 = (25*29 + 36*29)/58 =30.5
TS = (10-18)/ √(30.5(1/30 +1/30)
TS = -8/1.43
= -5.6
CV = t,.10,58 =1.296
Eve worse
Morning worse
-5.6
-1.296
0
1.296
Call in the evening supervisor. His shift is significantly worse.
2. A company that manufactures guitar picks in two shifts. Production supervisor Vinny
Pickman must keep tabs on both departments making sure that the number of defectives is
not particularly higher or lower in either shift. Each month he chooses a random sample
of days and counts the number of defects. His employees don’t think it is fair to count
defects because the number of picks produced differs between shifts. Vinny, understands
their complaint and changes his system. Last month the morning shift had 100 defective
picks in a sample1000 picks. The evening shift, in a sample of 800 picks had 120 defects.
Should Vinny schedule a reprimand for one of the shift supervisors? Explain your answer.
TWO Proportions
Ho: πm-πe = 0 (the shifts are the same)
Ho: πm-πe ≠ 0 (the shifts are different) (if positive morning worse, and vice versa)
TS = (.10-.15)/ √(.1*.9/1000 +.15*.85/800)
TS = -.05/.0158
= -3.17
CV = Z.10 =1.28
Eve worse
Morning worse
-3.17
-1.28
0
1.28
Call in the evening supervisor. We can be 90 percent sure that they are worse.
3. A company sells tennis racquets, and is considering decreasing the price at several
stores. For the randomly selected stores the table below reflects the number of racquets
sold in the week before and the week after a price decrease
Store Location
Before
Rockbille
100
Wentby
80
Kipgoin
70
Sandyville
100
Tommytown
50
Can we be sure the price decrease worked?
Dependent samples:
After
120
85
90
110
55
Store
Before
After
B-A
Location
Rockbille
100
120
-20
Wentby
80
85
-5
Kipgoin
70
90
-20
Sandyville
100
110
-10
Tommyto
50
55
-5
wn
avg
-12
std dev
7.582875
Std err
3.391165
Ho: μd > 0 (assume that before is more than after so the difference will be positive)
Ho: μd < 0 (if negative after is more. This is what we’d like to prove)
TS = -12/3.39 = -3.54
CV = t.10,4= -1.533
After more
Before more
-3.54
-1.533
0
We can be 90 percent sure that they sold more after the price decrease.
4. One bright McDonalds executive suggested that perhaps there are regional differences
in pickle preference. To test his idea he checked pickle removal in east and west coast
locations. On the west coast 73 percent of 100 customers removed the pickle, while on the
east coast only 63 percent of 60 customers removed the pickle. Can we be reasonably sure
the regions differ?
TWO Proportions
Ho: πw-πe = 0 (the coasts are the same)
Ho: πw-πe ≠ 0 (the coasts are different) (if positive west coast removes more pickles, and
vice versa)
TS = (.73-..63)/ √(.73*.27/100 +.63*.37/60)
TS = .10/.077 =1.30
CV = Z.10 =1.28
East
WEST
-1.28 0
1.28 1.30
We can be 90 percent sure that they are more likely to remove pickles on the west coast.
(we could not be 95% sure of this)
5. To investigate the cost side of the pickle debate, McDonalds has run a test to see if the
average cost per sandwich is less without the pickle. (It’s cheaper to produce the
burger without the pickle, but when people request pickles cost increases dramatically
because special arrangements have to be made. At several test stores 400 sandwiches
with pickles cost an average of $.59 to make with a standard deviation of $.28. At
another batch of test stores, 500 sandwiches without pickles cost an average of $.57 to
make with a standard deviation of $.38. Can we be reasonably sure that the sandwiches
without pickles cost less to produce?
Two averages (independent samples)
First check for equal standard deviations
Ho: σ1=σ2
Ho: σ1≠σ2
TS = .382/.282 =.14/.07=1.84
CV F.05,499,399=1.35
CV
FTR
REJ
1.35 1.84
So the standard deviations are significantly different so we will use the unpooled variance
test statistic.
Ho: μwith-μwithout < 0 (without pickles costs more)
Ho: μwith-μwithout > 0 (Without pickles costs less what we are trying to prove)
TS = (.59-57)/ √(.282/400 +.382/500)
TS = .02/.02
=1
CV = Z.05 =1.64
Without more
Without less
0
1
1.64
We cannot be 95 percent (or 90 percent) sure that sandwiches without pickles will cost
less to produce.
6. In a separate matter regarding sesame seed buns, a new baker claims that he can provide
an equivalent bun with more seeds for the same price. He has sent a sample of 61 buns for
McDonald's to test. In the new baker's sample each bun has an average of 140 sesame
seeds (with a standard deviation of 30 seeds). Forty-one buns from the current baker
averaged only 130 seeds (with a standard deviation of 20 seeds). Assuming that the
number of seeds is the deciding factor in the decision, should McDonalds switch buns?
Two averages (independent samples)
First check for equal standard deviations
Ho: σ1=σ2
Ho: σ1≠σ2
TS = 302/202 =2.25
CV F.05,60,40=1.64
CV
FTR
REJ
1.64 2.25
So the standard deviations are significantly different so we will use the unpooled variance
test statistic.
Ho: μnew-μold < 0 (original supplier is better)
Ho: μnew-μold > 0 (new supplier is better)
TS = (140-130)/ √(302/61 +202/41)
TS = 10/4.95
=1
CV = Z.05 =1.64
old better
new better
0
1.64
2.01
We can be 95 percent sure that the new supplier provides more seeds.
7. A study is undertaken with the following results: Each patient rates their pain on a scale
of one to ten both before and after hypnosis. If there is significant evidence of the benefits
of hypnosis, more research will be undertaken. If not, the program will be dropped.
Patient A
B
C
D
E
F
G
H
Before 6.6
6.5
9.0
10.3
11.3
8.1
6.3
11.6
After
2.4
7.4
8.5
8.1
6.1
3.4
2.0
6.8
Should research continue? Explain why or why not.
Dependent samples
Before
After
6.6
6.5
9
10.3
11.3
8.1
6.3
11.6
B-A
6.8
2.4
7.4
8.5
8.1
6.1
3.4
2
mean
std dev
std err
-0.2
4.1
1.6
1.8
3.2
2
2.9
9.6
3.125
2.911431
1.029346
Ho: μd < 0 (assume that before is Less than after so the Hyp does not work)
Ho: μd > 0 (if positive after is less pain. This is what we’d like to prove)
TS = 3.125/1.029 = 3.03
CV = t.10,7= 1.415
After more
Before more Hypnosis works
0
1.415 3.03
We can be 90 percent sure that there is less pain after hypnosis.
8. Pennsylvania is considering increasing public school funding to keep pace with private
schools. This would mean committing millions of additional dollars to the system. This
should only happen if performance on SAT scores is significantly lower in public schools.
In a random sample of one hundred public school students, the average SAT score was
950 with a standard deviation of 200 points. In a random sample of sixty-four private
school students the average was 1000 with a standard deviation of 50 points. Should the
state increase funding to public schools?
Two averages (independent samples)
First check for equal standard deviations
Ho: σ1=σ2
Ho: σ1≠σ2
TS = 2002/502 =16
CV F.05,99,63=1.47
CV
FTR
REJ
1.47
16
So the standard deviations are significantly different so we will use the unpooled variance
test statistic.
Ho: μpvt-μpub < 0 (public school better)
Ho: μpvt-μpub > 0 (public schools worse)
TS = (1000-950)/ √(2002/100 +502/64)
TS = 50/20.95= 2.39
CV = Z.05 =1.64
Public better
public worse
0
1.64
2.38
We can be 95 percent sure that the public schools are worse than private schools.
9. There is a house in New Orleans (it’s called the Rising Sun). Supposedly the gambling
games there pay off better than the casinos in Las Vegas. Vegas claims that their casinos
pay off more than anywhere else on the planet. In Vegas, a sample of 49 gamblers lost an
average of $50 each night with a standard deviation of $50. In a sample of 36 Rising Sun
Gamblers, the average loss was $30 with a standard deviation of $25. Does either place
deserve its reputation? Explain your answer.
Two averages (independent samples)
First check for equal standard deviations
Ho: σ1=σ2
Ho: σ1≠σ2
TS = 502/252 =4
CV F.05,48,35=1.69
CV
FTR
REJ
1.69
4
So the standard deviations are significantly different so we will use the unpooled variance
test statistic.
Ho: μveg-μno = 0 ( both places are the same)
Ho: μveg-μno ≠ 0 (If positive, Vegas gamblers lose more. If negative, N.O. gamblers lose
more)
TS = (50-30)/ √(502/49 +252/36)
TS = 20/8.26 = 2.41
CV = Z.05 =1.64
N.O. worse
same
vegas worse
-1.64
0
1.64
2.41
We can be 95 percent sure that Vegas gamblers lose more so NO deserves its reputation
for paying off better.
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