Section 9-1Review and

Preview

ReviewIn Chapters 7 and 8 we introduced methods of inferential statistics. In Chapter 7 we presented methods of constructing confidence interval estimates of population parameters. In Chapter 8 we presented methods of testing claims made about population parameters. Chapters 7 and 8 both involved methods for dealing with a sample from a single population.

PreviewThe objective of this chapter is to extend the methods for estimating values of population parameters and the methods for testing hypotheses to situations involving two sets of sample data instead of just one.The following are examples typical of those found in this chapter, which presents methods for using sample data from two populations so that inferences can be made about those populations.

PreviewTest the claim that when college students are

weighed at the beginning and end of their freshman year, the differences show a mean weight gain of 15 pounds (as in the “Freshman 15” belief).

Test the claim that the proportion of children who contract polio is less for children given the Salk vaccine than for children given a placebo.

Test the claim that subjects treated with Lipitor have a mean cholesterol level that is lower than the mean cholesterol level for subjects given a placebo.

Section 9-2 Inferences About Two Proportions

Notation for Two ProportionsFor population 1, we let:

p1 = population proportion

n1 = size of the sample

x1 = number of successes in the sample

The corresponding notations apply to

which come from population 2.

(the sample proportion)11

1

1 1

ˆ

ˆ ˆ1

xp

n

q p

2 2 2 2 2ˆ ˆ, , , ,p n x p q

The pooled sample proportion is denoted by p and is given by:

Pooled Sample Proportion

We denote the complement of p by q, so q = 1 – p

1 2

1 2

x xp

n n

Test Statistic for Two Proportions For H0: p1

= p2

H1: p1

p2 , H1: p1

< p2 , H1: p

1> p

2

1 2 1 2

1 2

ˆ ˆp p p pz

pq pq

n n

where p1 – p

2 = 0 (assumed in the null hypothesis)

1 21 2

1 2

ˆ ˆ and x x

p pn n

1 2

1 2

x xp

n n

1q p

Critical values: Use invNorm(area to the left, 0, 1) *left tailed test, α is in the left tail*right tailed test, α is in the right tail*two tailed test, α is divided equally between the two tails.

 P-values: Use the normalcdf feature on your calculator 

Right-tailed tests: To find this probability in your calculator, type: normalcdf (z test statistic, 99999, 0, 1) 

Left-tailed tests: To find this probability in your calculator, type: normalcdf (–99999, –z test statistic, 0, 1) ***Don’t forget if your test is two-sided, double your P-value.***

Test Statistic for Two Proportions – Cont…

Example 1: The table below lists results from a simple random sample of front-seat occupants involved in car crashes. Use a 0.05 significance level to test the claim that the fatality rate of occupants is lower for those in cars equipped with airbags.

a) State the null hypothesis and the alternative hypothesis.

H0: p1 = p

2

H1: p1 < p

2

b) Find the pooled estimate .

c) Find the critical value(s).

invNorm(0.05, 0, 1) = –1.645

p

41 520.004

11,541 9,853p

d) Find the test statistic.

e) Find the p-value.

normalcdf(–99999, –1.99, 0 , 1) = 0.0233

f) What is the conclusion?

Reject H0. There is enough evidence to suggest that the fatality rate of occupants is lower for those in cars equipped with airbags.

41 520

11,541 9,8531.99

0.004 0.996 0.004 0.996

11,541 9,853

z

Example 2: A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2823 occupants not wearing seat belts, 31 were killed. Among 7765 occupants wearing seat belts, 16 were killed (based on data from “Who Wants Airbags?” by Meyer and Finney, Chance, Vol. 18, No. 2). Use a 0.05 significance level to test the claim that the fatality rate is higher for those not wearing seat belts.

a) State the null hypothesis and the alternative hypothesis.

H0: p1 = p

2

H1: p1 > p

2

b) Find the pooled estimate .

c) Find the critical value(s).

invNorm(1 – 0.05, 0, 1) = 1.645

p

31 160.004

2,823 7,765p

d) Find the test statistic.

e) Find the p-value.

normalcdf(6.43, 99999, 0, 1) ≈ 0

f) What is the conclusion?

Reject H0. There is enough evidence to suggest that the fatality rate is higher for those not wearing seat belts.

31 160

2,823 7,7656.43

0.004 0.996 0.004 0.996

2,823 7,765

z

Example 3: A Pew Center Research poll asked randomly selected subjects if they agreed with the statement that “It is morally wrong for married people to have an affair.” Among the 386 women surveyed, 347 agreed with the statement. Among the 359 men surveyed, 305 agreed with the statement. Use a 0.05 significance level to test the claim that the percentage of women who agree is different from the percentage of men who agree.

a) State the null hypothesis and the alternative hypothesis.

H0: p1 = p

2

H1: p1 ≠ p

2

b) Find the pooled estimate .

c) Find the critical value(s).

invNorm(0.025, 0, 1) = ±1.645

p

347 3050.875

386 359p

d) Find the test statistic.

e) Find the p-value.

2 × normalcdf(2.04, 99999, 0, 1) = 0.0418

f) What is the conclusion?

Reject H0. There is enough evidence to suggest that the percentage of women who agree is different from the percentage of men who agree.

347 3050

386 3592.04

0.875 0.125 0.875 0.125

386 359

z

Confidence Interval Estimate of p1 – p

2

1 2 1 2 1 2ˆ ˆ ˆ ˆp p E p p p p E

1 1 2 2

1 2

ˆ ˆ ˆ ˆwhere | * |

p q p qE z

n n

For population 1, we let:

p1 = population proportion

n1 = size of the sample

x1 = number of successes in the sample

The corresponding notations apply to

which come from population 2.

(the sample proportion)11

1

1 1

ˆ

ˆ ˆ1

xp

n

q p

2 2 2 2 2ˆ ˆ, , , ,p n x p q

Example 4: Use the sample data given in Example 1 to construct a 90% confidence interval estimate of the difference between the two population proportions. (The confidence level of 90% is comparable to the significance level of = 0.05 used in the preceding left-tailed hypothesis test.)

Example 5: Use the sample data given in Example 2 to construct a 90% confidence interval estimate of the difference between the two population proportions.

Example 6: Use the sample data given in Example 3 to construct a 95% confidence interval estimate of the difference between the two population proportions.

Example 7: Lipitor is a drug used to control cholesterol. In clinical trials of Lipitor, 94 subjects were treated with Lipitor and 270 subjects were given a placebo. Among those treated with Lipitor, 7 developed infections. Among those given a placebo, 27 developed infections. Use a 0.05 significance level to test the claim that the rate of infections was the same for those treated with Lipitor and those given a placebo. a) State the null and alternative hypothesis.  

  b) Calculate the pooled estimate.  

  c) Determine the critical value(s).

d) Calculate the test statistic.      e) Determine the P – value.      f) What is the conclusion?

g) Determine

h) Find the margin of error.      i) Construct a 95% confidence interval estimate of the difference between the two population proportions. 

1 2ˆ ˆ .p p