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

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12

2121

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)()(21

nSnS

XXt

pp

nn

2221

21

2121

//

)()(

nn

XXZ

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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