Small differences. Two Proportion z-Interval and z-Tests.

Small differences.

Two Proportion z-Interval and z-Tests

Recall from Random VariablesIf X and Y are two independent Random

Variables from Normal distributions, we can combine them to get a new Random Variable that is also from a Normal distribution.

Mean:

Standard Deviation:2 2( ) ( ) ( )SD X Y SD X SD Y

( ) ( ) ( )E X Y E X E Y

Differences between two proportionsWe can create a confidence interval for the

difference between the two proportions.Conditions:

1.Randomization

2. Independence: Independence between the two groups

3.10% Condition: check for both groups

4.Success/Failure: check for both groups separately, use and

All conditions have been met to use a Normal model for the differences in two proportions.

1p̂ 2p̂

Two-proportion z-intervalSince the Normal model is being used we need a

mean and standard deviation(error).Mean: Standard Error:

CI:

Be very careful with the conclusion.

1 2ˆ ˆp p

1 1 2 2

1 2

ˆ ˆ ˆ ˆp q p q

n n

1 1 2 21 2

1 2

ˆ ˆ ˆ ˆˆ ˆ( ) *

p q p qp p z

n n

Example: Gender and Binge DrinkingCollege students were randomly selected and asked

about how much alcohol they consumed on a weekly basis. Over a certain amount of alcohol consumed was considered binge drinking and those individuals were labeled frequent binge drinkers. The following table shows the number of men and women who were frequent binge drinkers. What is the difference in the proportions of male and female binge drinkers?

 Frequent Binge Drinker (x)

Total Questioned(n)

Male 1392 5348Female 1748 8471

Conditions:

1.Randomization: Stated as random samples

2. Independence: It is safe to assume that men and women are independent of each other.

3.10% Condition: 5348 male college students is less than 10% of all male college students

8471 female college students is less than 10% of all female college students

4.Success/Failure:

All conditions have been met to use a Normal model for the differences of two proportions.

1 1

1 1

ˆ 1392 10

ˆ 3956 10

n p

n q

2 2

2 2

ˆ 1748 10

ˆ 6723 10

n p

n q

Two-proportion z-interval at 95% confidence

CI:

=(0.039, 0.069)

We are 95% confident that the true proportion of men who are frequent binge drinkers is between 3.9% to 6.9% higher than the proportion of women who are frequent binge drinkers.

1 1 2 21 2

1 2

ˆ ˆ ˆ ˆˆ ˆ( ) *

p q p qp p z

n n

1392 3956 1748 67231392 1748 5348 5348 8471 8471

1.965348 8471 5348 8471

Comparing 2 Group ProportionsThree possibilities:

1.The two proportions are the same.

2.The two proportions are different.

3.One proportion is larger than the other.

The null hypothesis would be that they are the same.

We’ll then run a test to see if there are different(two-tail) or one is larger than the other(one-tail)

HypothesesThe null should be that there is no difference

between the two proportions or that they are equal.

H0: p1 = p2

But we have no probability model for this situation.Since both proportions come from independent(we

hope) Normal distributions, the difference of the two would be a new Normal distribution.

H0: p1 – p2 = 0We can test against this null hypothesis now.

Alternative HypothesisThe alternative hypotheses are then:

p1 – p2 < 0

The proportion of group 1 is lower than group 2 p1 – p2 > 0

The proportion of group 1 is higher than group 2 p1 – p2 ≠ 0

The proportion of group 1 is not the same as group 2

Conditions1. Randomization

2. Independence: Independence between the two groups

3. 10% Condition: check for both groups

4. Success/Failure: check for both groups separately

All conditions have been met to use a Normal model for the difference of two proportions

MechanicsWe need to calculate a pooled proportion from

the two we are given. The assumption is that they are equal, so given:

The pooled proportion is:

This will is used to calculate a standard deviation for the two proprotions.

1 21 2

1 2

ˆ ˆ;x x

p pn n

1 2

1 2

ˆpooled

x xp

n n

Pooled Standard Deviation:

Test Statistic:

P-Value = P(z <,> ___)

1 2

ˆ ˆ ˆ ˆˆ( ) pooled pooled pooled pooled

pooled

p q p qSE p

n n

1 2 1 2

1 2

ˆ ˆ 0 ˆ ˆ

ˆ( ) ˆ ˆ ˆ ˆpooled pooled pooled pooled pooled

p p p pz

SE p p q p q

n n

ConclusionCompare P-Value to stated alpha. Talks about difference between the two

proportions only, not the values of the individual proportions.

Example: Binge DrinkersIs there evidence that male college students have a

higher proportion of frequent binge drinking than female college students?

Hypothesis:

H0: p1 – p2 = 0 The proportion of male and female college students that are frequent binge drinkers is the same.

HA: p1 – p2 > 0 The proportion of male college students that are frequent binge drinkers is more than female college students.

Conditions:

1. Randomization: Stated as random samples

2. Independence: It is safe to assume that men and women are independent of each other.

3. 10% Condition: 5348 male college students is less than 10% of all male college students

8471 female college students is less than 10% of all female college students

4. Success/Failure:

All conditions have been met to use a Normal model for the differences of two proportions.

1 1

1 1

ˆ 1392 10

ˆ 3956 10

n p

n q

2 2

2 2

ˆ 1748 10

ˆ 6723 10

n p

n q

Mechanics:1

11

1392ˆ5348

xp

n 2

22

1748ˆ8471

xp

n 0.05

1 2

1 2

1392 1748 3140ˆ5348 8471 13819pooled

x xp

n n

1392 17485348 8471 7.369

3140 10679 3140 1067913819 13819 13819 13819

5348 8471

1 2ˆ ˆ 0

ˆ( )pooled

p pz

SE p

14( 7.369) 8.66 10 0P Value P z

Conclusion:

Since the P-Value is so much smaller than alpha (8.66 x 10-14 < 0.05), we reject the null hypothesis. There is statistically significant evidence that the proportion of male college students who are frequent binge drinkers is more than female college students.