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Lecture 6Hypothesis Testing II: Proportions and 2 Populations

Graduate SchoolQuantitative Research Methods

Gwilym Pryceg.pryce@socsci.gla.ac.uk

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Notices: Register

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Aims & Objectives Aim

• the aim of this lecture is to continue with our introduction of the method of hypothesis testing and to demonstrate a number of applications

Objectives– by the end of this lecture students should

be able to carry out hypothesis tests on:• two population means• one population proportion• two population proportions

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Plan: 1. Review of Significance 2. Review of one sample tests on the mean 3. Hypothesis tests about Two population means

• Homogenous variances

• Heterogeneous variances

4. Deciding on whether variances are equal 5. Hypothesis tests about proportions

– One population– Two populations

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Macro commands:Confidence Intervals (CI) Hypothesis tests

Macro command

Definition Macro Command

Definition

CI_L1M Large sample CI for one mean H_L1M Large sample significance test on one mean

CI_S1M Small sample CI for one mean H_S1M Small sample significance test on one mean

CI_S2MP Small independent samples CI for difference between 2 means (pooled variance)

H_S2MP Small independent samples significance test for equality of 2 means (pooled variance)

CI_S2MD Small independent samples CI for difference between 2 means (different variances)

H_S2MD Small independent samples significance test for equality of 2 means (different variances)

CI_L1P Large sample CI for one proportion (presents output for both Traditional and Wilson methods of calculation)

H_L1P Large sample significance test on one proportion

CI_L2P Large sample CI for comparing two proportions (presents output for both Traditional and Wilson methods of calculation)

H_L2P Large samples significance test on two proportions

H_S2VF Simple small sample F-test on equality of two variances (see also Levene’s test in the SPSS help menu for more sophisticated test of homogenous variances).

N_L1M Sample size for desired margin or error for the mean

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1. Review of Significance P = significance level = chances of our

observed sample mean occurring given that our assumption about the population (denoted by “H0”) is true.

So if we find that this probability is small, it might lead us to question our assumption about the population mean.– I.e. if our sample mean is a long way from our

assumed population mean then it is:• either a freak sample • or our assumption about the population mean is wrong.

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If we draw the conclusion that it is our assumption re that is wrong and reject H0

then we have to bear in mind that there is a chance that H0 was in fact true.– In other words:

• when P = 0.05, for every twenty times we reject H0, then on one of those occasions we would have rejected H0 when it was in fact true.

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2. Review of one sample tests on the mean

We introduced a common framework for hypothesis testing:

4 Steps of Hypothesis testing:

Step (1) state H0 and H1

Step (2) state and formula

Step (3) state decision rule

Step (4) compute P & decide

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We also looked at 2 specific tests: Large sample sig. Test on one mean:

• Formula:

• Macro syntax:

H_L1M n=(?) x_bar=(?) m=(?) s=(?).

Small sample sig. Test on one mean:• Formula:

• Macro syntax:H_S1M n=(?) x_bar=(?) m=(?) s=(?).

ns

xz

/

1,/

ndfns

xti

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3. Hypothesis tests about two population means

In SPSS: this is called the “Independent Sample t-test”

• go to Analyse, Compare Means...

Two different formulas for computing t:

]1,1min[

)()(

21

2

22

1

21

2121

nndf

ns

ns

xxtc

1

11

)()(

21

21

2121

nndf

nns

xxt

p

c

Equal Variances(formula has an exact t-distribution)

Unequal Variances(does not have an exact t-distribution)

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Example where variances are different:As part of your PhD, you want to test whether the new “Fun Phonics” reading method is better than the “Letterland” method. You examine the reading power of 6 year old children from two similar schools.

– The first used the FP method and you found that this produced

an average reading proficiency score of 53.7 (based on a sample of 22 children; s.d. = 11.5).

– The second school used the Letterland method and you found that this produced an average reading proficiency score of 42.51 (sample = 24; s.d. = 16.9).

Test whether the FP method produces higher results at the 1% significance level.

9.16,24,51.42

5.11,22,7.53

LLL

FPFPFP

snx

snx

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Use the 4 steps and the following formula to test whether the FP method produces higher results at the 1% significance level.

Can you use the canned SPSS procedure to do this problem?

9.16,24,51.42

5.11,22,7.53

LLL

FPFPFP

snx

snx

]1,1min[

)()(

21

2

22

1

21

2121

nndf

ns

ns

xxtc

4 Steps of Hypothesis testing:

Step (1) state H0 and H1

Step (2) state and formula

Step (3) state decision rule

Step (4) compute P & decide

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(1) H0: FP = L (means are equal)

H1: FP > L (upper tail test)

(2) = 0.01 (implies critical t value of 2.528),

(3) Reject H0 iff P < I.e. if P < 0.01

(4) P = Prob(t > 2.644) = 0.0076, so reject H0

9.16,24,51.42

5.11,22,7.53

LLL

FPFPFP

snx

snx

21]1,1min[

2.644

2461.285

2225.132

51.427.53

21

2

22

1

21

21

nndf

ns

ns

xxtc

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Doing the calculation in SPSS: You cannot use the canned SPSS

procedure unless you have the original data.

But you can use the following macro commands:– Homogenous variances:

• H_S2Mp n1=(?) n2=(?) x_bar1=(?) x_bar2=(?) s1=(?) s2=(?).

– Heterogeneous variances:• H_S2Md n1=(?) n2=(?) x_bar1=(?)

x_bar2=(?) s1=(?) s2=(?).

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For the Letterland/FP example we would use the diff. Variances syntax: H_S2Md n1=(22) n2=(24) x_bar1=(53.7) x_bar2=(42.51)

s1=(11.5) s2=(16.9).

The upper tail sig. = 0.007588 I.e. less than 1% chance of false rejection, therefore reject H0 of equal

means in favour of the alternative hypothesis that Fun Phonics results in higher reading scores on average than Letterland.

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4. How do we decide on whether the variances are similar? Where variances are hugely

different or exactly the same, the decision is simple.

When there is any ambiguity, we can use one of two tests to help us:

• Simple Ratio of Variances Test• Levene’s Test

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Simple Ratio of Variances test: If we divide the ratio of variances of samples

from two independent populations we find that that ratio has an F distribution in repeated samples:

where the denominator degrees of freedom calculated as n1–1 and the numerator degrees of freedom calculated as n2–1.

• NB Because the critical values for the F distribution are only calculated for the upper tail, if the F value you are have calculated is less than one, you need to invert it (i.e. swap round the numerator and denominator).

F = s12 / s2

2

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This is the formula behind the following command:H8_S2VF n1=(?) n2=(?) s1(?) s2(?)

E.g. For the Letterland/FP example:H8_S2VF n1=(22) n2=(24) s1=(11.5) s2=(16.9).

Which tells us that there is less than a 5% chance of false rejection if we reject the null of equal variances. So reject the null

• I.e. we can be sure that the population variances are indeed different.

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The Levene’s test If we have the original data we can use

Levene’s test which is a canned routine in SPSS.

The Levene’s test is more sophisticated & robust than the simple ratio of variances test: – If P (I.e. “sig.”) from the Levene’s test is small

reject the H0 of equal variances & use the 1st t-formula.

– If P from the Levene’s test is large, accept H1 & use the 2nd t-formula to compute the test statistic.

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SPSS Output from test equal purchase prices between Cumberland and Durham (Nationwide data):

Group Statistics

307 65514.21 38347.03 2188.58

249 62646.33 39379.40 2495.57

House Countyfrom Postcode Cumber

Durham

Purchase PriceN Mean

Std.Deviation

Std. ErrorMean

Independent Samples Test

.030 .864 .866 554 .387 2867.88 3310.10

.864 524.653 .388 2867.88 3319.30

Equal variancesassumed

Equal variancesnot assumed

Purchase PriceF Sig.

Levene's Test forEquality of Variances

t dfSig.

(2-tailed)Mean

DifferenceStd. ErrorDifference

t-test for Equality of Means

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Two tails from one:

Along with the Levene’s test results, SPSS automatically supplies t-test results for both the equal and different variances formulas.

One problem with the SPSS t-test, however, is that it only gives the 2 tail sig., but you can work out the one tail sigs as follows:

• The two tailed significance is twice that of the smallest one tailed significance:

  2 tailed sig. = 2 min[lower tail sig., upper tail sig.]

• But it can be a bit confusing working out which one tail significance level is the one you want (see notes).

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Testing for 2 means summary:

If you’ve got the original data, • First do the Levene’s test in SPSS

Analyze, Compare Means, Independent Samples

• Then do the appropriate macro t-test to avoid confusion.H_S2Mp for equal variances or H_S2Md for different variances

If you don’t have the original data,• First do the ratio of variances test

H8_S2VF

• Then do the appropriate macro t-testH_S2Mp for equal variances or H_S2Md for different variances

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5.1 Hypothesis tests on proportions One population

(large samples only) So far looked at:

– how to make inferences about the population mean from our sample mean.

But sometimes the variable of interest is categorical

• household has or has not insurance; • person is homosexual or not homosexual; • a person has Aids or does not have Aids

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In such cases, what we are interested in is the proportion of cases that fall into a particular category:

• the proportion of households with insurance; • the proportion of people who are homosexual; • the proportion of people with Aids

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Calculating the sample proportion:

p = x / n– where:

• x = cases with the attribute of interest e.g. the number of households with insurance

• n = sample size

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CLT and Proportions: Q/ Does the Central Limit Theorem

apply to sample proportions? A/ Yes.

– Proportions from repeated random samples will be normally distributed around the population

proportion .

– We can then translate any sample proportion onto the standard normal curve by calculating its z score:

n

pzi

)1(

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Example:

– E.g. 1 As a historian, you want to find the proportion of citizens in medieval Scotland that contracted the plague. From a sample of 400 parish records, you find that 22 died of the plague. The assumption in the literature has been that 10% of the population had died. Test whether this assumption is valid using both 2 and 1 tailed tests.

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Summary of data:n = 400x = 22

= 0.1

– (1) H0: = 10%

H1: 10% (2-tailed test)

– (2) = 0.02, for example.

n

pzi

)1(

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– (3) Reject H0 iff P < I.e. if P < 0.02(this will happen if zc < - 2.33 or if zc > 2.33, where 2.33 is the z value associated with = 0.02. Since zc = -3.948, we know we can safely reject H0).

– (4) Calculate z:

P = 2x(Prob(z < -3.00))

= 2x 0.0013 = 0.0026since P < 0.02 (I.e. less than one in 50 chance of type I error) we can reject H0.

In fact, the chances of incorrect rejection of H0 are less than one in 3,000.I.e. the chances of observing p (our sample proportion) assuming H0 ( = 10%) to be true are so small that we are forced to question this assumption about

n

pzi

)1(

n = 400

x = 22 = 0.1

00.3015.0

045.

40009.0

045.

400)1.01(1.0

1.0055.0

cz

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One tailed test:– (1) H0: = 10%

H1: < 10% (lower tail test)

– (2) = 0.02

– (3) Reject H0 iff P < I.e. if P < 0.02

– (4) Lower tail sig. = P = Prob(z < -3.00) = .001350 since P < 0.02 we can reject H0 knowing that the chances of

incorrect rejection of H0 are less than one in 740

» our cut-off rule for rejecting H0 was no more than a one in 50 chance

» one in 740 is a lot less than one in 50 so we can reject H0 with confidence.

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The macro syntax for one proportion tests is as follows:

H6_L1P n=(400) x=(22) pi=(0.1).

Which comes to the same result.

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5.2 Hypothesis tests about Two population proportions

To test the hypothesis that the population proportions are equal:

H0: 1 = 2

compute the z statistic:

where: SEDp is the pooled standard error =

and

)( 21

DpSE

ppz

2

11)1(

1 nnppSEDp

21

21

nn

xxp

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Example: Two surveys of mortgage payment protection

insurance (MPPI) are carried out, one on single parents with 1 child and one on single parents with 3 children. Amongst the first group, 67 out of a sample of 300 were found to have taken out MPPI, compared with 15 out of a sample of 101 in the second group. Is take-up significantly lower amongst the HHs with three children?

• p1 = 67/300 = 0.2233; • p2 = 15/101 = 0.1485;• p = (300 + 101)/(67+15) = 0.2045;

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– (1) H0: 1 = 2

H1: 1 > 2

– (2) = 0.01 (z* = 2.33)

– (3) Reject H0: if P < 0.01– (4) P = 0.053.

Take-up is not significantly lower amongst HHs with 3 children at the1% sig. level; or even at 5% significance level.

• I.e. we cannot say that the difference in proportions is anything more than the effect of sampling variation.

0464.0100

1

300

1)205.01(205.0

DpSE

6125.10466.0

)1485.02233.0(

)( 21

DpSE

ppz

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H7_L2P n1=(300) n2=(101) x1=(67) x2=(15) .

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Plan: 1. Review of Significance 2. Review of one sample tests on the mean 3. Hypothesis tests about Two population means

• Homogenous variances

• Heterogeneous variances

4. Deciding on whether variances are equal 5. Hypothesis tests about proportions

– One population– Two populations