Comparing Two Population Parameters Equal Variance t-test for Means, Section 11.1 - 11.2 Unequal...
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Transcript of Comparing Two Population Parameters Equal Variance t-test for Means, Section 11.1 - 11.2 Unequal...
Comparing Two Population Parameters
Equal Variance t-test for Means, Section 11.1 - 11.2
Unequal Variance t-test Means, Section 11.3
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Chapter Objectives
Select and use the appropriate hypothesis test in comparing Means of two independent samples
Continuous variables
Means of two dependent samples Continuous variables
Proportions of two independent samples Discrete variables - count
Variances of two independent samples
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Differences between Means, Independent Samples The two samples are independent
Selection of one sample is in no way affected by the selection of the other sample
We are interested in whether men and women with a college education and working full-time have the same annual earnings Random variable is annual earnings; two populations Draw a sample from each population; calculate the respective sample means
Observe
Why do we observe a difference between the sample means? The samples are drawn from populations with different means, i.e., 1 2 The samples are drawn from populations with the same population means,
but because of sampling error, we observe different sample means, 1 = 2
21 XX 021 XX
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Sampling Distribution of
To answer this question, need to know how is distributed Difference between sample means is a statistic
Need to know the characteristics of the sampling distribution of the difference between the sample means
21 XX
21 XX
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Sampling Distribution of
Let the random variable, X, follow a normal distribution in the populations or the sample sizes, n1 and n2, are both greater than 30
Imagine drawing all possible samples from each of the two populations Form all possible pairs of differences in sample
means from population 1 and population 2 The distribution of the differences between these
pairs of sample means is the sampling distribution
21 XX
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Distribution is normal X follows normal
distribution in the populations or CLT
Mean of the distribution
Unbiased estimator Standard error is
Characteristics of Sampling Distribution
21 XX 1 - 2
Sampling Distribution of the Difference in Sample Means,
normally distributed
2121 )( XXE
2
22
1
21
21 nnXX
Z
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Assumptions about Population Variances Two different scenarios arise in the
comparison of the population means of independent samples
Variances of the underlying populations are either known to be or assumed to be equal Use Equal (or pooled) variance t-test
Variances are not assumed to be equal Use Unequal (or separate) variance t-test
22
21
22
21
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Equal Variance t-Test for Differences in Means Two-sided test
H0:1 - 2 = 0 There is no difference in the population means
H1:1 - 2 0 There is a significant difference in the population means
One-sided tests H0:1 - 2 ≤ 0 H1:1 - 2 > 0
H0:1 - 2 ≥ 0 H1:1 - 2 < 0
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Under the Null Hypothesis
What do we expect to observe under the null hypothesis?
The center of the sampling distribution is zero If we observe a large
absolute difference between the sample means, then it is unlikely that the samples came from populations with equal means
21 XX 0
Sampling Distribution of the Difference in Sample Means,
under the null
normal
Where does the observed lie?21 XX
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Equal Variance t-Test for Differences in Means We need to convert into a standardized
variable If we knew the population variances, the test
statistic would be
21 XX
2221
21
2121
//
)()(
nn
XXZ
Realistically, we do not know the population variances
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Equal Variance t-Test for Differences in Means Pool both samples together
Use all information to produce a more reliable estimate of the unknown population variance
Weighted average of the sample variances Weights are the respective degrees of freedom
Substitute for unknown population variances
2
)1()1(
21
222
2112
nn
SnSnS p
2pS
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Equal Variance t-Test for Differences in Means
Note: the degrees of freedom are n1 + n2 - 2 Compare the value of this t-test statistic with a critical t
value DR: if (Crit. Value t ≤ t-test statistic ≤ Crit. Value t) do not
reject Or use p-value approach
22
12
2121
2//
)()(21
nSnS
XXt
pp
nn
2221
21
2121
//
)()(
nn
XXZ
PP 5 13
Problem - Negative Income Tax Would this welfare program cause the recipients to stop working? An experiment in the 1960s was done to find out
Target population consisted of 10,000 low-income families in three New Jersey cities From these families, 400 were chosen at random for the control group Another 225 were chosen at random for the treatment group - and put
on the negative income tax. All 625 families were followed for three years
The control families averaged 7,100 hours of paid work and their standard deviation was 3,900 hours
The treatment families averaged 6,200 hours and their standard deviation was 3,400 hours
Is the difference between the sample means statistically significant at ⍺ = 0.05?
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Problem - Negative Income Tax H0: 1 - 2 = 0
The NIT has no effect on hours of work
H1: 1 - 2 0 The NIT has an effect on hours of work
Assume the population variances are equal Calculate the pooled sample variance
45.640,897,13623
)3400(224)3900(399 222
pS
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Problem - Negative Income Tax Ask how far the observed difference in the
sample means lies from the center of the sampling distribution (0) if the null hypothesis is true
Find the critical t value at ⍺ = 0.05
90.2
)225
1
400
1(45.640,897,13
200,6100,7623
t
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Problem - Negative Income Tax Degrees of freedom
n1 +n2 - 2 = 400 + 225 -2 = 623 =
Compare the t-test statistic with the critical value 2.90 (test statistic) > 1.96 (critical value) Reject null hypothesis
There is a significant difference in mean hours worked between the control and the treatment group
96.1025,. t
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Online Homework covers section 11.1-11.2 CengageNOW fifth assignment
Chapter 11:Intro to Two Samples CengageNow sixth assignment
Chapter 11:Tests about Two Means
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Unequal Variance t-Test for Differences in Means Assume variances of the two populations are
not equal
How do we estimate the standard error of the difference in sample means?
Substitute sample variances
22
21
2221
21
2121
//
)()(
nn
XXZ
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Unequal Variance t-Test for Differences in Means Test Statistic
Use an approximation for the degrees of freedom Behrens Fischer solution
2221
21
2121
//
)()(
nSnS
XXtdf
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Calculating the Degrees of Freedom
Round df down to the nearest integer Compare the test statistic with the critical t
value using estimated df degrees of freedom DR: if (Crit. Value t ≤ t-test statistic ≤ Crit. Value t)
do not reject
)]1/()/()1/()/[(
)]/()/[(
22
2221
21
21
22
221
21
nnsnns
nsnsdf
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Two Possible Tests for Differences in Means, Independent Samples Variances of the underlying populations are
either known to be or assumed to be equal Use Equal (or pooled) variance t-test
Variances are not assumed to be equal Use Unequal (or separate) variance t-test
Can test whether population variances are equal
22
21
22
21
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P-value Approach with t - Distribution Tables Calculate the p-value for the test statistic, 2.90 Look at the t table, infinite degrees of freedom
Find the table entries closest to the test statistic, 2.90 2.576 cuts off = .005
It is the value closest to our test statistic Test statistic is more extreme
So the p-value must be < .005 in one tail of the distribution or <.01 for both tails
Compare the p-value with 0.05 .01 (p-value) <.05 (level of significance)
Therefore we reject the null hypothesis
df\p 0.4 0.25 0.1 0.05 0.025 0.01 0.0051 0.32492 1.00000 3.07768 6.31375 12.70620 31.82052 63.6567430 0.25561 0.68276 1.31042 1.69726 2.04227 2.45726 2.75000
inf 0.25335 0.67449 1.28155 1.64485 1.95996 2.32635 2.57583
T-table
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t - Distribution for Infinite Degrees of Freedom
Find boundaries for your particular test statistic, 2.902.58 cuts off 0.005 of t values2.90 > 2.58 2.90 cut off less than 0.005 of t values, p-value < 0.005
0
0.10
0.05
0.025
0.01
1.28 1.64 1.96 2.33
0.005
2.58
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T Distribution for Infinite Degrees of Freedom
0
0.10
0.05
0.025
0.01
1.28 1.64 1.96 2.33
0.005
2.58
Suppose test statistics is 2.201.96 < 2.20 < 2.330.01 < area in tail < 0.025 = p-value for one-tail test0.02 < p-value < 0.05 for two-tail test - multiply area in tail by two
2.20