Transcript Hypothesis Test

Hypothesis Testing Large Samples
1. A random sample of 100 people in the city revealed that
tennis is played, on the average, 1.2 hours per week during
the summer. The population standard deviation is .4 hours
for all people in the United States. Test whether this
sample indicates that the number of hours tennis is played
in this city differs from the national average of 1.1 hours.
Use  =.01.
1. H O :  1.1
H A :  1.1
2. Two tail z test
n > 30 known 
3. .005
.005
.495 .495
-2.57
x   1.2 1.1
4. z 

 2.5
.4
SE
100
2.57
5. Fail to reject, there is not
enough evidence at the .01
level to show that the average
hours are different.
Hypothesis Testing Large Samples
2. The population of all minority workers has a mean
wage of $14,500 with a standard deviation of
$200.00. Test whether a sample of 100 having an
average of $14,300 and = .05 differs from the
population average.
1. H O :   $14,500 2. Two tail z test
3.
n > 30 known 
H A :   $14,500
.025
.025
.475 .475
-1.96
x   14,300 14,500
4. z 

 10
200
SE
100
1.96
5. Reject HO at the .05 level.
There is evidence that the
salaries are different.
Hypothesis Testing Large Samples
3. A new bus route has been established. For the old route,
the average waiting time was 18.3 minutes. However, a
random sample of 40 waiting times between buses using
the new route had a mean of 15.1 minutes with a sample
standard deviation of 6.2 minutes. Does this indicate that
the new route is different from the old route? Use  = .05.
1. H O :   18.3
H A :   18.3
2. Two tail z test
n > 30
3. .025
.025
.475 .475
-1.96
x   15.118.3
4. z 

 3.27
6.2
SE
40
1.96
5. Reject HO at the .05 level.
There is evidence that the
new route is different.
Hypothesis Testing Small Samples
1. A manufacturer of ball bearings have a diameter of .25
inches and a sample standard deviation of .05 inches. A
random sample of n = 25 ball bearings reveal a mean
diameter of .2670 inches. Conduct a hypothesis test at the
10 % level of significance to determine whether there is
statistical significance that the manufacturing process is
not running correctly, that is µ ≠ .25 inches.
1. H O :   .25
H A :   .25
2. Two tail t test
3. .05
n < 30 unknown 
24 d.f.
x   .2670  .25
4. t 

 1.7
.05
SE
25
.05
-1.711
1.711
5. Fail to reject HO at the .1
level. There is not enough
evidence that it is not
running incorrectly.
Hypothesis Testing Small Samples
2. A random sample of n = 10 prices of CDs gives the
following values: $24.19, $21.04, $12.34, $16.07, $19.65,
$19.36, $15.99, $15.02. $18.68, $20.60. Conduct a
hypothesis test at the 10% level to determine whether
evidence exists to support the claim that the population
average is less than $19.00.
x  18.294, s  3.45
1. H O :   19
H1 :   19
2. One tail t test
3.
n < 30 unknown
 
9 d.f.
x   18.294 19
4. t 

 .647
3.45
SE
10
.01
-1.383
5. Fail to reject HO at the .1
level. There is not
evidence that average CD
price is less than $19.00.
Hypothesis Testing
Large Sample Population Proportion
1. A telephone survey was conducted to determine viewer
response to a new show. The sponsor would like to see a
favorable response from over 65% of the respondents.
From a sample of 500 viewers, 345 of the responses were
favorable. Is there sufficient evidence to satisfy the sponsor
at the .01 level?
1. H O :   .65
H A :   .65
2. One tail z test
(500)(.65) > 5
(500)(.35) > 5
3.
.01
2.33
p
.69  .65
4. z 

 1.87 5. Fail to reject HO at the .1
level. There is not enough
SE
(.65)(.35)
evidence to satisfy the
500
sponsor.
Hypothesis Testing
Difference of Population Means
1. In a certain industry, the standard deviation for workers in Chicago
is the same as in Los Angeles  = $2.75 per hour. A random sample of
30 workers in Chicago revealed a mean wage of $22.75 per hour and a
random sample of 40 from Los Angeles revealed a mean wage of
$18.55 per hour. Perform a hypothesis test at the .01 significance level
to determine whether they are different and construct an 80 %
confidence interval.
1. H O : C   L  0
2. Two tail z test
nC, nL > 30
H A : C   L  0
xC  xL
xC  xL
4. z 

SE
 C2  L2

nC nL

22.75  18.55
2.752
30

4.20 4.20


 6.32
.44 .66
2.752
40
3.
.005
-2.58
.005
2.58
5. Reject HO at the .1
level, the wages are
different
4.2 + 1.28(.66)
(3.35, 5.04)
Hypothesis Testing
Difference of Population Means
2. In a study of heart surgery, the effect of a beta-blocker was used to
reduce pulse rate. One group of 30 received the drug and had an
average pulse rate of 65.2 beats per minute and a standard deviation of
7.8. For the control group of 30 which received the placebo, the mean
was 70.3 beats per minute with a standard deviation of 8.3. Do betablockers reduce pulse rate? Is the result significant at the .05 level?
Construct a 99% confidence interval for the difference in mean pulse
rate.
1. H O :  B   P  0
2. One tail z test
nB, nP > 30
HA : B  P  0
xB  xP
xB  xP
4. z 

SE
 B2  P2

nB nP

65.2  70.3
7.82
30

 5.1  5.1


 2.46
4.32 2.07
8.32
30
3. .05
-1.645
5. Reject HO at the .05
level, beta blockers
reduce blood pressure
-5.1 + 2.58(2.07)
(-10.46, .26)
Hypothesis Testing
Difference of Population Means
2. The prices for a certain drug at a
private and a chain drug store are
shown. Perform a hypothesis test at
the .01 level to determine if the prices
are different.
1. H O :  P  C  0
2. Two tail t test
22 d.f.
H A :  P  C  0
xP  xC
xP  xC
4. z 

SE
 P2  P2

nC nC

8.75  7.97
1.152
10

.78
.78


 1.75
.197 .444
.952
14
Group
n
x
s
Private
10
8.75
1.15
Chain
14
7.97
.95
3. .005
-2.82
2.82
5. Fail to reject HO at
the .01 level, no
difference in
prices.
Hypothesis Testing
Difference of Population Means
1. Two groups of infants were compared
with the amount of blood hemoglobin
levels at 12 months of age. Is there
evidence at the .01 level that the two
methods are different. Give a 95%
confidence interval between the two
populations of infants.
1. H O :  B   F  0
2. Two tail t test
40 d.f.
HA : B  F  0
xB  xF
xB  xF
4. z 

SE
 B2  F2

nB nF

13.3  12.4
1.7 2
23

.9
.9


 1.65
.296 .544
1.82
19
Group
n
x
s
Breast-Fed
23
13.3
1.7
Formula
19
12.4
1.8
3. .005
-2.70
2.70
5. Reject HO at the .1
level, no difference
in feeding methods.
.9 + 1.96(.54)
(-.158, 1.958)
Hypothesis Testing
Difference of Population Proportions
1. A sample of 2051 men taking a drug to reduce heart attacks had 56
heart attacks during the next five years while a sample of 2030 resulted
in 84 heart attacks during the next 5 years. Is there evidence at the .01
level that the drug reduces the possibility of a heart attack?
pd = 56/2051=.027
pp = 84/2030=.041
1. H O :  p   d  0
2. One tail z test
3. .01
HA :  p   d  0
-2.33
pd  p p
pd  p p
4. z 

SE
pd (1  pd ) p p (1  p p ) 5. Reject HO at the .1

level, the drug reduces
nd  1
np 1
heart attacks.
.027  .041
 .014


 2.47
.005
(.027)(. 973) (.041)(. 959)

2050
2029
Hypothesis Testing
Difference of Population Proportions
2. A sample of 267 white people and 230 black people were asked if
the government was doing enough in the areas of housing,
unemployment, and education. 161 of the white people surveyed and
136 of the black people said no. Is there evidence at the .05 level that
the white people and black people disagree on this issue?
p1 = 161/267=.60
p2 = 136/230=.59
1. H O :  W   B  0
2. Two tail z test
3. .025
.025
HA : W   B  0
-1.96
1.96
pW  p B
pW  pB
4. z 

SE
pW (1  pW ) p B (1  pB ) 5. Fail to reject HO at

the .05 level, there is
nW  1
nB  1
no difference between
.60  .59
.01


 .226 voting preference.
(.60)(. 40) (.59)(. 41) .044

266
229
Hypothesis Testing
Matched Pair Samples
1. Cola makers test new recipes for loss of sweetness before and after storage. The
results of ten tasters are shown before and after storage. Is their evidence at the .05
level that sweetness is greater before storage?
1
2
3
4
5
6
7
8
9
10
Before
2.3
1.6
2.3
4.1
3.2
5.2
1.5
1.7
3.2
2.5
After
0.3
1.2
1.6
2.1
3.6
3
2.8
0.5
2.1
0.2
2
.4
.7
2
-.4
2.2
-1.3
1.2
1.1
2.3
1. H O : 1   2  0
H A : 1   2  0
2. One tail t test
9 d.f.
3.
d  1.02
sd  1.196
.05
1.83
D
1.02
4. t 

 2.696
s
1.196
n
10
5. Reject HO at the .05
level, sweetness
is greater before
storage
Hypothesis Testing
Matched Pair Samples
2. A physical education teacher tested the results of jogging on a person’s
cardiovascular system. The resting pulse rates were measured before and after the
completion of a 5 week jogging program. Is there evidence at the .01 level that the
resting pulse rate decreased?
1
2
3
4
5
6
7
Before
74
78
81
83
83
87
90
After
72
74
75
81
82
84
85
-2
-4
-6
-2
-2
-3
-5
1. H O : 1   2  0
H A : 1   2  0
2. One tail t test
6 d.f.
D
 3.29
4. t 

 4.836
s
1.80
n
7
d  3.29
sd  1.80
3.
.01
-3.14
5. Reject HO at the .01
level, the resting pulse
rate decreased
Hypothesis Testing
Chi Square
The following table shows the results of boys and
Girls that get in trouble in school. Are gender and getting in
trouble independent at the .01 level?
Got in
Trouble
No Trouble
Total
Boys
46 40.97
71 76.02
117
Girls
37 42.03
83 77.97
120
Total
83
154
237
1. HO: Gender and trouble are independent
HA: Gender and trouble are dependent
2. Chi Square test (2 - 1)(2 - 1) = 1 d.f.
3. Determine rejection region
4. Compute test statistic
1.87
6.64
5. Fail to Reject HO, gender and trouble are not independent
The following table shows the Myers-Brigs personality
preference and professions for a random sample of
2408 people in the listed professions. E refers to
extroverted and I refers to introverted.
Occupa tion
Extroverted
In troverte d
Clerg y
30 8
22 6
Doctor
66 7
93 6
La wyer
11 2
15 9
Row To tal
Colu mn Total
Use the chi-squared test to determine if the listed
occupations and personality preferences are independent
at the .01 level of significance.
Occupa tion
Extroverted
In troverte d
Clerg y
30 8
241
22 6
Doctor
66 7
93 6
La wyer
11 2
1087
723
122
Colu mn Total
15 9
1321
293
879
149
Row To tal
534
1603
271
2408
Use the chi-squared test to determine if the listed
occupations and personality preferences are independent
at the .01 level of significance.
1. HO: Preferences and profession are independent
HA: Preferences and profession are dependent
2. Chi Square test (3 - 1)(2 - 1) = 2 d.f.
3. Determine rejection region
4. Compute test statistic
9.21
43.55
( f o  f e ) 2 (308  241) 2 (667  724) 2 (112 122) 2
 




fe
241
724
122
= 43.55
2
(226  293) 2 (936  879) 2 (189 149) 2


293
879
149
5. Reject HO, preferences and profession are not independent