Lecture 12

Comparing Two Proportions

Let and be the two population proportions of successes. Use to estimate .

Population Population proportion Sample size Count of successes Sample proportion

1

2

Large-Sample CI for Comparing Two Proportions:

Choose an SRS of size from a large population having proportion of successes and

an independent SRS of size from another population having proportion of

successes. The estimate of the difference in the population proportions is

A ( ) CI for is

( ) √ ( )

( )

where is the value for the standard Normal density curve with area between

and .

The standard error of is √ ( )

( )

and the margin of error for confidence level is

Use this method for 90%, 95%, or 99% confidence when the number of successes and the

number of failures in each sample are at least 10.

Example: Let’s find a 95% CI for the difference between proportion of men and of

women who are frequent binge drinkers.

Population n X

1 (men) 5,348 1,392 0.260

2 (women) 8,471 1,748 0.206

Total 13,819 3,140 0.227

Report: Men are 5.4% more likely to be binge drinkers with ME = 1.5%.

Plus Four CI for a Difference in Proportions

We add 1 success and 1 failure to each sample:

Choose an SRS of size from a large population having proportion of successes and

an independent SRS of size from another population having proportion of

successes. The plus four estimate of the difference in proportions is

Where

and

.

A ( ) CI for is

( ) √ ( )

( )

where is the value for the standard Normal density curve with area between

and .

The standard error of is √ ( )

( )

and the margin of error for confidence level is .

Use this method for 90%, 95%, or 99% confidence when both sample sizes are at least 5.

Example: In studies that look for a difference between genders, a major concern is

whether or not apparent differences are due to other variables that are associated with

gender. Because boys mature more slowly than girls, a study of adolescents that

compares boys and girls of the same age may confuse a gender effect with an effect of

sexual maturity.

The “Tanner score” is a commonly used measure of sexual maturity.

Subjects are asked to determine their score by placing a mark next to a rough drawing of

an individual at their level of sexual maturity.

There are five different drawings, so the score is an integer between 1 and 5.

A pilot study included 12 girls and 12 boys from a population that will be used for a large

experiment. Four of the boys and three of the girls had Tanner scores of 4 or 5, a high

level of sexual maturity.

Let’s find a 95% CI for the difference between the proportions of boys and girls who

have high (4 or 5) Tanner scores in this population.

Report: The difference in proportions of boys and girls with high scores is 7.1% with

ME=34.5%.

Significance Test for a Difference in Proportions:

We want to test

To test the hypothesis or

Compute the z statistic

where the pooled standard error is

√ ( ) (

)

and where

.

In terms of a standard Normal random variable Z, the P-value for a test of against

is ( )

is ( )

is ( )

Example: Are men and women college students equally likely to be frequent binge

drinkers?

Population n X

1 (men) 5,348 1,392 0.260

2 (women) 8,471 1,748 0.206

Total 13,819 3,140 0.227

Report: 26% of men and 20.6% of women are binge drinkers, the difference is significant

(z = 7.37, p < 0.001)

Paired Samples

Example: Speed-skating races are run in pairs.

Two skaters start at the same time, one on the inner lane and one on the outer lane.

Halfway through the race, they cross over, switching lanes so that each will skate the

same distance in each lane.

Even though this seems fair, at the 2006 Olympics some fans thought there might have

been an advantage to starting on the outside.

Inner Lane | Outer Lane

Name Time Name Time

Zhang 125.75 Nemoto 122.34

Abramova 121.63 Lamb 122.12

Remfel 122.24 Noh 123.35

Lee 120.85 Timmer 120.45

Rokita 122.19 Marra 123.07

Yakshina 122.15 Opitz 122.75

Bjelkevik 122.16 Haugli 121.22

… … … …

Here are the partial data for the women’s 1500-m race:

Skating pair Inner Time Outer Time Inner - Outer

1 125.75 122.34 3.41

2 121.63 122.12 -0.49

3 122.24 123.35 -1.11

4 120.85 120.45 0.40

5 122.19 123.07 -0.88

6 122.15 122.75 -0.60

7 122.16 121.22 0.94

… … … …

This is an example of paired data.

What test shall we use to see if there is a difference in two lanes?

Paired t-Test

This is just the one-sample t-test applied to a single sample of differences.

To test whether the mean of paired differences is significantly different from zero, we test

the hypothesis , where d’s are the pairwise differences.

We use the t statistic

where is the mean of the pairwise differences, n is the number of pairs, and

√ .

When the conditions are met, under the null hypothesis we can model the sampling

distribution of this statistic with a distribution to find a P-value.

Example: Was there a difference in speeds between the inner and outer speed-skating

lanes at the 2006 Winter Olympics?

No difference between lanes

So we can use the paired t-test:

CIs for Matched Pairs

Example: The table below represents ages of 170 married couples. How much older, on

average, are husbands?

Wife’s Age Husband’s Age Difference

43 49 6

28 25 -3

30 40 10

57 52 -5

52 58 6

27 32 5

52 43 -9

43 47 4

23 31 8

25 26 1

… … …

Paired t CI:

A ( ) CI for the difference is given by

where is the value for density curve with area between and ,

and

√ .

Conclusion: We are 95% confident that husbands are, on average, 1.6 to 2.8 years older

than their wives.