lecture 8 (hypothesis testing 2)

8/2/2019 Lecture 8 (Hypothesis Testing 2)

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Hypothesis Testing 2

Testing Percentages and

differences of means


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Test of Population Proportion

You will often have sample survey datathat shows that a percentage of peoplesaid yes to a question

A test of a population proportion will allowyou to test whether your sample evidenceis likely to represent the true population

percentage who would say yes to thequestion


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Test of Population Proportion Testing a population proportion (), using

a sample proportion (P)

Only difference is the standard error ofproportions SE(p) measured using:

SE(p) = ({(1- )}/n)

z calculation using: z = (P - )/SE(p)


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Example

Career unit claims 50% of population ofgraduates have a job by June. Samplen=30 gives p = 33%

Test the hypothesis that careers unitclaims are correct.

Ho: 0.5 Ha: < 0.5

Significance level = 0.05


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Critical value z = -1.65 SE(p) = [(0.5)(0.5)/30] = 0.09 as p = 0.33, converts to z score

z = (0.33-0.5)/0.09 = -1.88 z = -1.88 < -1.65 so is in the rejection

region

Reject Ho. Accept Ha: less than 50% ofgraduates have a job by June


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= 0.5

= 0

Sampling Dist Proportions

Standard Normal

1

SEp

ps

Zs

-1.65

p=0.33

5% of lowest ps inrejection region

-1.88


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Tests of differences of means

Suppose have sample wage data on twodifferent companies and the two samplemean wages are different

A test of difference of means lets you testwhether the observed difference betweensample means indicates the samples were

drawn from the same population ordifferent populations with different means


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Tests of significant difference

Test for differences between twosample means

Requires SE(differences of means) If both samples are drawn from the

same population, a large difference

between sample parameters is unlikely


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0

= 0

Sampling Dist Diff of means

Standard Normal

1

SE (x1 -x2)

(x1 -x2)s

Zs

-1.96

2.5% of highest (x1 -x2)sin the rejection region

1.96


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A test of differences of means

Sample data on daily wages from 2 firms(1 and 2) shows:

n1 = 30, x1 = 180, s1 = 14

n2 = 40, x2 = 170, s2 = 10 Does the evidence show firm 1 pays a

higher mean daily wage

This difference could be due to chancerather than a distribution difference


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Doing the test

Set up the hypotheses as a 2 tailed test

The population means for each sampleare either the same (both sample meanscome from the same population) or theyare not the same and hence come fromdifferent populations with different

population means soHo: 1 = 2 H1: 1 2


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The critical value for the test

Choose a significance level

If the level is 5%, = 0.05

For a 2 tailed test, the probability in eachtail 0.025

As both samples, n > 30, we can use a Z

score for the test Critical value of Z = 1.96 from the tables


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The test statistic

We want to test whether the differencebetween the means is significantlydifferent from zero

The test statistic is therefore

Z = (x1 -x2)(1 - 2) / SE(x1 -x2)

Z = (x1 -x2) / SE(x1 -x2)

SE(x1 -x2) is the SE of the difference ofmeans


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To calculate SE(x1 -x2)

SE(x1 -x2) = (21 + 22)n1 n2

If n1 and n2 > 30 use sample standarddeviation values if 21

22 are unknown

SE = [(14)2/30 + (102)/40]SE = 3.01


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Calculating the test statistic

Z = (x1 -x2) / SE(x1 -x2)

Z = (180 170)/3.01 = 10/3.01 = +3.32

Compare Z score to the critical value =+1.96

+3.32 > + 1.96

Z score is in the top rejection region of thedistribution

Ho is rejected


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0

= 0

Sampling Dist Diff of means

Standard Normal

1

SE (x1 -x2)

(x1 -x2)s

Zs

-1.96

2.5% of highest (x1 -x2)sin the rejection region

1.96 3.32


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Interpretation

Rejecting Ho: 1 = 2

The observed difference of sample meanwages is not due to chance

We reject the idea that our two samples ofdaily wages were drawn from the samepopulation of wages with only 1 populationmean wage


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Interpretation

Accepting H1: 1 2 means that thesamples were drawn from 2 differentpopulations of data

The observed difference of sample meansis significant and reflects a difference inthe population means from which theywere drawn

It is likely that firm 1 pays more than firm 2on average


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USING EXCEL

Use Data Analysis add-in under tools

Choose z test: two sample for means

or if either sample is smaller than 30 choose

t test:two sample assuming equalvariances

Highlight variable range for first sample

Highlight variable range for second sample

Choose sig level.


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Difference of proportions

A similar test can be done when you havedata on 2 samples which give 2 sampleproportions or percentages

You can test whether the observeddifference in proportions is due to chanceor due to differences in the populationsfrom which the samples were drawn, theonly difference is the standard error of thedifference of proportions


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0

= 0

Sampling Dist Diff ofproportions

Standard Normal

1

SE(p1-p2)

(p1 -p2)s

Zs

2.5% of lowest (p1 - p2)sin the rejection region


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The standard error of difference ofproportions

Standard error of diff proportions= SE (p1-p2)

p1 = sample 1 proportion

p2 = sample 2 proportion = estimated population proportion

SE (p1-p2) = [ (1- ) + (1- ) ]

n1 n2 = (n1p1 + n2p2)/ (n1 + n2)


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An example

Data is available from 2 surveys of the %of households with smokers

n1 = 50 n2 = 50 and p1 = 20% p2 = 30%

Survey 2 shows a higher % of smokers butis this difference due to chance or due to adifference in the populations from whichthe 2 surveys were drawn?


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The hypotheses and critical value

The null hypothesis, as before, is thatthere is no difference between thepopulation proportions from which the 2

samples were drawn (they are drawn fromthe same population

Ho: 1 = 2 H1: 1 2 For significance level = 0.01 (2 tailed) Z

critical value = + or 2.58


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Calculating the SE (p1-p2)

= (n1p1 + n2p2)/ (n1 + n2)

= (50(0.20) + 50(0.30)/ (50 + 50) = 0.25

SE (p1-p2) = [ (1- ) + (1- ) ]n1 n2

SE (p1-p2) = [ 0.25(0.75) + 0.25(0.75) ]50 50

SE (p1-p2) = [ 0.00375 + 0.00375 ] = 0.087


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Calculating test statistic

The test statistic Z = (p1 p2)/ SE (p1-p2)

Z = (0.20 0.30)/0.087= -0.10 / 0.087 = -1.15

Compare Z score to the critical Z value

-1.15 > -2.58, so Z test score is not in therejection region

We cannot reject Ho


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0

= 0

Sampling Dist Diff ofproportions

Standard Normal

1

SE(p1-p2)

(p1 -p2)s

Zs

-2.58

0.5% of lowest (p1 - p2)sin the rejection region

-1.15


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Interpretation

We accept the null Ho: 1 = 2 There is no difference between the

population proportions from which the

samples were drawn The apparent difference of proportions is

due to chance

Both samples are drawn from a populationof people with an estimated population of25% smokers


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Final Points

You now know that observed sampledata values do not necessarily mean thatthe population also has this value

You also know that observed differencesbetween 2 sample values does notnecessarily mean that the difference issignificant

Be careful, observed patterns from sampledata may be due to chance. Test them.


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Final Points





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Final Points




lecture 8 (hypothesis testing 2)

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