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ECO 72 ­ INTRODUCTION TO ECONOMIC STATISTICS
Topic 9
One­Sample
Hypothesis Tests
These slides are copyright © 2003 by Tavis Barr. This material may be distributed only subject to the terms and conditions set forth in the Open Publication License, v1.0 or later (the latest version is presently available at http://www.opencontent.org/openpub/).
Hypothesis Testing
●
The purpose of hypothesis testing is to state the validity of a claim about a parameter of the population.
–
Is the population mean really, say, 10?
–
Is the population proportion really 25%?
–
Is the population stabdard deviation really 3?
(We won't cover this last test.)
Hypothesis Testing
●
Background and Methodolgy
●
One­Sample Hypothesis Test for Population Mean
●
●
–
Large Sample
–
Small Sample
One­Sample Hypothesis Test for Population Proportion (Large Sample Only)
Type I and Type II Errors
Hypothesis Testing
●
●
●
Hypothesis tests must be set up in a certain order to be done correctly
We have to be careful that our knowledge about the data does not affect our choice of hypotheses
Otherwise we can be tempted to choose hypotheses that we already know are true or false
Hypothesis testing
The order of a hypothesis test:
1. The null hypothesis (H0) and the alternative hypothesis (H1) is stated
2. A level of significance is decided on
3. A test statistic is selected
4. A decision rule is selected, usually involving a critical value. Usually we want to develop a confidence interval for the population parameter at the critical value.
5. A sample is collected, and the statistic and its critical value is determined
The Steps of a Hypothesis Test
1. The null hypothesis (H0) and the alternative hypothesis (H1) is stated
●
●
The null hypothesis is a statement that we want to test
•
“The population mean is zero”
•
“The population proportion is 40 percent”
The alternative hypothesis is what is true when the null hypothesis is not true
•
“The population mean is not zero”
•
“The population proportion is not 40 percent”
The Steps of a Hypothesis Test
2. A level of significance is decided on
●
●
●
The level of significance: If we declare the null hypothesis false, what chance of being wrong we are willing to allow ourselves?
Rarely can a hypothesis be declared certainly true or untrue. Rather, it can be declared likely or unlikely
Larger data sets or less noisy data may allow us to be more stringent
The Steps of a Hypothesis Test
3. A test statistic is selected
–
A test statistic is a number that describes how likely our null hypothesis is
–
The test statistic usually follows a common distribution, such as the Normal or t.
–
The distribution of the test statistic is independent of sample properties such as the mean or standard deviation. It may depend on the number of observations in the sample.
The Steps of a Hypothesis Test
4.A decision rule is selected, usually involving a critical value. ●
●
●
Because the test statistic follows a known distribution, it will be above or below a given value with a known probability if the null hypothesis is true
If the value is too large or too small, the null hypothesis is improbable
How large or small the statistic needs to be depends on what significance level we chose
The Steps of a Hypothesis Test
5. A sample is collected, and the statistic and its critical value is determined. A decision is made.
●
●
●
The key point: We have decided on all of the methodological questions (significance level, test statistic, etc.) before we look at the data
This way, we know that our familiarity with the data does not influence our decision about, say, what significance level to use
Note that we never choose to accept a null hypothesis, only reject it or not reject it.
Test about a Population Mean
●
●
First, we make a null hypothesis about what the population mean is. We call this mean (H0)
Then we make a confidence interval based on the sample mean and make a decision based on where the hypothesized population mean lies:
Population Mean Lies:
Outside 90% Interval
Outside 95% Interval
Outside 99% Interval
●
Decision:
Reject at 10% significance level
Reject at 5% significance level
Reject at 1% significance level
But note that we decide on significance level before looking at data
Test About a Population Mean
●
For large samples, the statistic is:
z=
●
●
X− H0 
Std. Error
=
X− H0 
 s /n
2
If  H0  is thetruepopulation mean,then theCLT says X
isNormal with meanH 0 and standarddeviation  s /n
2
So if the null hypothesis is true, then z follows the standard normal distribution
P(­1.645<z<1.645) =0.9
­2
­1
0
1
f(z)
z
2 Test About a Population Mean
For large samples, the statistic is:
z =
X− H0 
Std. Error
=
X− H0 
 s /n
2
z is:
(H0) is:
Hypothesized mean is:
Less than ­1.645
More than 1.645
More than X­1.645(S.E.)
Less than X+1.645(S.E.)
Above 90% Conf. Int.
Below 90% Conf. Int.
Reject at 10 percent
Less than ­1.96
More than 1.96
More than X­1.96(S.E.)
Less than X+1.96(S.E.)
Above 95% Conf. Int.
Below 95% Conf. Int.
Reject at 5 percent
Less than ­2.576
More than 2.576
More than X­2.576(S.E.)
Less than X+2.576(S.E.)
Above 99% Conf. Int.
Below 99% Conf. Int.
Reject at 1 percent
Test About a Population Mean
For large samples, the statistic is:
z=
X− H0 
Std. Error
=
X− H0 
 s /n
2
­1.645 < z < 1.645: Do not reject
1.645 < |z| < 1.96: Reject at 10%
1.96 < |z| < 2.576: Reject at 5%
­3
2.576 < |z|: Reject at 1%
­2
­1
0
1
2
3
Test of Popn' Mean – Example
●
According to a survey of 7,615 clients of the Maine Addiction Treatment System from 1990­1995, the average income of a client of a client is $928.30 per month, with a standard deviation of $875.80 per month.
Source: http://www.bu.edu/econ/faculty/ma/Papers_Archive/interaction_JHE_March2004.pdf
●
●
The federal poverty line in 1995 for a household of two was $863.33 per month
Can we reject the hypothesis, at the 5 percent level, that the average income of a client was equal to the poverty line for a household of two?
Test of Popn' Mean – Example 2
●
●
●
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One hundred people in some middle­income country have their red blood cell count taken. The sample mean is 4.8 million cells per milliliter, and the sample standard deviation is 1.5 million cells/ml.
Five million cells per milliliter is considered a healthy count. At a 5% significance level, can we reject the hypothesis that the blood cell count in this country is too low?
Test of Pop'n Mean – Example 2
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Sample mean is 4.8 million, std. dev. is 1.2 million, sample size is 100
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H0: Mean is 5 million
●
z statistic: [ X −H 0 ]/  s2 / n
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X − H0 =4.8mil−5 mil=−200,000
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Standarderror :  s /n= 1,200,000 /100
2
= 14,400,000,000=120,000
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So z = ­200,000/120,000 = ­1.67
2
Test of Popn' Mean – Example 3
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A company claims its cell phones emit 1.2 W/kg of radiation. A Consumer Reports engineer wants to test this claim and measures radiation from a sample of 300 phones. The sample mean is 1.33 W/kg, and the sample standard deviation is 0.25 W/kg.
At a 5% significance level, can we reject the company's claim that the phones emit 1.2 W/kg?
Test of Pop'n Mean – Example 3
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Sample mean is 1.33, std. dev. is 0.25, sample size is 300
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H0: Mean is 1.2
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z statistic:
●
X − H0 =1.33−1.2=0.13
●
●
[ X −H 0 ]/  s / n
2
Standarderror :  s /n= 0.25 /300
= .000208=0.0144
2
So z = 0.13/0.0144 = 9.03
2
Means Test with Small Sample
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z test makes use of Central Limit Theorem
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CLT only applies to samples over 30
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With small sample, we can use t statistic
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t statistic only works when original variable follows the Normal distribution
We compute the t statistic exactly like the z:
t=
X− H0 
Std. Error
=
X−H 0 
2
s
 /n
Means Test with Small Sample
●
We compute the t statistic exactly like the z:
●
●
t=
X− H0 
Std.Error
=
X−H 0 
s /n
2
z statistic follows a standard normal distribution; t statistic follows a t distribution with n­1 degrees of freedom
This means unlike the z statistic, whose critical values are always the same (1.65, 1.96, 2.58), those of the t statistic depend on the sample size
Means Test with Small Sample
●
t =[ X −H 0 ]/ Std.Err.
=[ X −H 0 ]/  s / n
2
●
●
●
t statistic follows a t distribution with n­1 degrees of freedom
Critical values depend on the sample size
Use two­tailed test
Confidence Intervals
80%
90%
95%
Level of Significance for One­Tailed Test
df
0.100 0.050 0.025
Level of Significance for Two­Tailed Test
0.20
0.10
0.05
1
3.08
6.31
12.71
2
1.89
2.920
4.3
3
1.64
2.35
3.18
4
1.53
2.13
2.78
5
1.48
2.02
2.57
98%
99%
99.9%
0.010
0.005
0.0005
0.02
31.82
6.97
4.54
3.75
3.37
0.01
0.001
63.657 636.619
9.925 31.599
5.841 12.924
4.604 8.610
4.032 6.869
6
7
8
9
10
1.440
1.42
1.4
1.38
1.37
1.943
1.895
1.860
1.833
1.812
2.45
2.37
2.31
2.26
2.23
3.14
3
2.87
2.82
2.76
3.707
3.499
3.355
3.250
3.169
5.959
5.408
5.041
4.781
4.587
11
12
13
14
15
1.36
1.36
1.350
1.35
1.34
1.796
1.782
1.771
1.761
1.753
2.2
2.18
2.160
2.15
2.13
2.72
2.68
2.650
2.62
2.6
3.106
3.055
3.012
2.977
2.947
4.437
4.318
4.221
4.140
4.073
t Test Example
●
●
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The SAT has a mean score of 1,000 nationally.
We look at a sample of 6 students from a school and want to know if they score above or below average
Scores are: 1020, 970, 820, 740, 850, 930.
t Test Example
●
Scores are: 1020, 970, 820, 740, 850, 930.
●
First step: Find sample mean and variance.
X:
1020970820740850930
=888.33
6
1020−888.332970−888.332 820−888.332
2
2
2
2 740−888.33 850−888.33 930−888.3
 :
5
17336.116669.4444669.44422002.781469.4441736.111
=
5
=10776.67
t Test Example
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Sample mean: 888.33
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Sample variance: 10776.67
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H0:  = 1000
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6 observations, so 5 d.o.f.
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t=[ X−H 0 ]/ Std.Err.
=[ X−H 0 ]/  s /n
=888.33−1000/ 10776.67/6
=−111.67/42.38=−2.63
2
Confidence Intervals
80%
90%
95%
Level of Significance for One­Tailed Test
df
0.100 0.050 0.025
Level of Significance for Two­Tailed Test
0.20
0.10
0.05
1
3.08
6.31
12.71
2
1.89
2.920
4.3
3
1.64
2.35
3.18
4
1.53
2.13
2.78
5
1.48
2.02
2.57
98%
99%
99.9%
0.010
0.005
0.0005
0.02
31.82
6.97
4.54
3.75
3.37
0.01
0.001
63.657 636.619
9.925 31.599
5.841 12.924
4.604 8.610
4.032 6.869
6
7
8
9
10
1.440
1.42
1.4
1.38
1.37
1.943
1.895
1.860
1.833
1.812
2.45
2.37
2.31
2.26
2.23
3.14
3
2.87
2.82
2.76
3.707
3.499
3.355
3.250
3.169
5.959
5.408
5.041
4.781
4.587
11
12
13
14
15
1.36
1.36
1.350
1.35
1.34
1.796
1.782
1.771
1.761
1.753
2.2
2.18
2.160
2.15
2.13
2.72
2.68
2.650
2.62
2.6
3.106
3.055
3.012
2.977
2.947
4.437
4.318
4.221
4.140
4.073
Another t Test Example
●
The average January temperature in New York in the 1990s was:
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
35.5
29.2
30.2
30.9
20.4
32.5
25.7
27.6
35.6
29.7
Source: http://www.ncdc.noaa.gov/oa/climate/research/cag3/Y8.html
●
Can we reject at the 10% level the hypothesis that the average temperature is 32 degrees?
Another t Test Example
●
The average January temperature in New York in the 1990s was:
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
35.5
29.2
30.2
30.9
20.4
32.5
25.7
27.6
35.6
29.7
●
Can we reject at the 10% level the hypothesis that the average temperature is 32 degrees?
–
First, calculate the sample mean and variance:
X: 29.73 s2: 20.57
Another t Test Example
●
Sample mean: 29.73
●
Sample variance: 20.57
●
H0:  = 32
●
10 observations, so 9 d.o.f.
●
t=[ X−H 0 ]/ Std.Err.
=[ X−H 0 ]/  s /n
=29.73−32/  20.57/10
=−2.27/1.43=−1.58
2
Confidence Intervals
80%
90%
95%
Level of Significance for One­Tailed Test
df
0.100 0.050 0.025
Level of Significance for Two­Tailed Test
0.20
0.10
0.05
1
3.08
6.31
12.71
2
1.89
2.920
4.3
3
1.64
2.35
3.18
4
1.53
2.13
2.78
5
1.48
2.02
2.57
98%
99%
99.9%
0.010
0.005
0.0005
0.02
31.82
6.97
4.54
3.75
3.37
0.01
0.001
63.657 636.619
9.925 31.599
5.841 12.924
4.604 8.610
4.032 6.869
6
7
8
9
10
1.440
1.42
1.4
1.38
1.37
1.943
1.895
1.860
1.833
1.812
2.45
2.37
2.31
2.26
2.23
3.14
3
2.87
2.82
2.76
3.707
3.499
3.355
3.250
3.169
5.959
5.408
5.041
4.781
4.587
11
12
13
14
15
1.36
1.36
1.350
1.35
1.34
1.796
1.782
1.771
1.761
1.753
2.2
2.18
2.160
2.15
2.13
2.72
2.68
2.650
2.62
2.6
3.106
3.055
3.012
2.977
2.947
4.437
4.318
4.221
4.140
4.073
Hypothesis Test for Population Proportion
●
●
●
●
In a large enough sample, sample proportion is normally distributed
Expected value is population proportion (p); standard error is  p1−p/ n
We write population proportion as , so hypothesized proportion is written (H0)
If null hypothesis is true, and proportion is really (H0), then standard error is  H 0 1−H 0 / n
Hypothesis Test for Population Proportion
●
If null hypothesis is true, sample proportion is normally distributed with mean (H0) and standard error  H 0 1−H 0 / n
●
That means that the following statistic will be standard normally distributed:
z=
p−H 0 
 H 0 1− H0 / n
Hypothesis Test for
Proportion – Example
●
According to a June, 2007 Gallup poll of 1,007 U.S. adults, 53% of Americans believe that the theory of evolution is definitely or probably true.
http://www.usatoday.com/news/politics/2007­06­07­evolution­poll­results_n.htm
●
At the five percent significance level, can we reject the possibility that half of U.S. adults believe that evolution is definitely or probably true?
Hypothesis Test for
Proportion – Example 2
●
●
Insurance fraud rate in New York is about 6%.
The director of insurance investigations in Nassau County wants to figure out if fraud rate is above or below state average. ●
Conducts an intensive review of 500 cases.
●
Finds that in his sample, 40 cases are fraudulent.
Hypothesis Test for
Proportion – Example 2
●
Null hypothesis: (H0) = .06
●
Sample proportion: 40/500 = .08
●
Standard error of sample proportion if null hypothesis is true:  H 0 1−H 0 /n= 0.061−0.06/500=0.0106
p−H 0 
0.08−0.06
z=
−
=1.883
 H 0 1− H0 /n 0.0106
Hypothesis Test for
Proportion – Another Example
●
On January 4, 2003, Slashdot readers were asked who they thought were the dumbest people in Los Angeles. Their responses:
CyberCafeGang Violence Kids
City Councilmen demanding cybercafe security
The LAPD Who Want To Crack Down on CyberCafes
The MPAA
CowboyNeal hasn't been to LA
TOTAL
●
1619
468
1047
6175
1328
10637
Can we reject the hypothesis that half of Slashdot readers think the MPAA are the dumbest people in LA at the 1% significance level?
Hypothesis Test for
Proportion – Another Example
●
Null hypothesis: (H0) = .50
●
Sample proportion: 6175/10637 = .5805
●
Standard error of sample proportion if null hypothesis is true:  H 0 1−H 0 /n= 0.51−0.5/10636=0.0048
p−H 0 
0.5805−0.5
z=
−
=16.61
0.0048
 H 0 1− H0 /n
Type I and Type II Errors
●
●
Conclusions from statstistical tests are not made with certainty
We may make a mistake, either rejecting a hypothesis that is actually true, or not rejecting a hypothesis that is actually false
Type I Error
●
●
A Type I Error occurs when we reject a null hypothesis that is actually true
The probability of doing this, known as , is just the significance level of the test:
2.5% Prob.
X
P[(­1.96(SE) < X < ­1.96(SE)] =0.95

5% test:
P(|z|>1.96) = 0.05
2.5% Prob.
z
Type II Error
●
●
A Type II Error occurs when we fail to reject a null hypothesis that is actually false
The probability of this, known as , depends on the alternative hypothesis
Critical values
under H0
Probability of observing a value in non­rejection range if H1 is actually true 
(H0)
z
(H1)
z