Slide 2 - 25 Copyright © 2008 Pearson Education, Inc. Chapter 11 Inferences for Population...

Copyright © 2008 Pearson Education, Inc. Slide 2 - 25

Chapter 11

Inferences for

Population Proportions


Definition 11.1


Formula 11.1


Key Fact 11.1


Example 11.2

Suppose that 19.1% of all U.S. employees play hooky; that is, that the population proportion is = 0.191. Then, according to Key Fact 11.1, for samples of size 1010, the variable is approximately normally distributed with mean = p = 0.191 and standard deviation

p̂

p̂

p̂ p 1 p n

0.191 1 0.191

10100.012

Use simulation to make that fact plausible.

p̂


Solution Example 11.2

We first simulated 2000 samples of 1010 U.S. employees each. Then, for each of those 2000 samples, we determined the sample proportion, , of those who play hooky. Output 11.1 shows a histogram of those 2000 values of , which is shaped like the superimposed normal curve with parameters 0.191 and 0.012.

p̂

p̂

Output 11.1


Procedure 11.1


Definition 11.2


Formula 11.2


Procedure 11.2


Procedure 11.2 (cont.)


Key Fact 11.2


Procedure 11.3


Procedure 11.3 (cont.)


Example 11.9

Zogby International surveyed 1181 U.S. adults to gauge the demand for vegetarian meals in restaurants. The study, commissioned by the Vegetarian Resource Group and published in the Vegetarian Journal, polled independent random samples of 747 men and 434 women. Of those sampled, 276 men and 195 women said that they sometimes order a dish without meat, fish, or fowl when they eat out. Do the data from the Zogby International poll provide sufficient evidence to conclude that the percentage of U.S. men who sometimes order veg is smaller than the percentage of U.S. women who sometimes order veg? Use a 5% level of significance.



We apply Procedure 11.3, noting first that the assumptions for its use are satisfied.

STEP 1 State the null and alternative hypotheses.Let p1 and p2 denote the proportions of all U.S. men and all U.S. women who sometimes order veg, respectively. The null and alternative hypotheses areH0 : p1 = p2 (percentage for men is not less than that for women)Ha : p1 < p2 (percentage for men is less than that for women).

STEP 2 Decide on the significance level, α.The test is to be performed at the 5% significance level, or α = 0.05.



STEP 3 Compute the value of the test statistic

where

z p̂1 p̂2

p̂p 1 p̂p 1 n1 1 n2

We first obtain , , and . Because 276 of the 747 men sampled and 195 of the 434 women sampled sometimes order veg, x1 = 276, n1 = 747, x2 = 195, and n2 = 434. Therefore,

p̂p x1 x2 n1 n2 .p̂pp̂2p̂1

p̂1 x1

n1

276

7470.369; p̂2

x2

n2

195

4340.449



and

Consequently, the value of the test statistic is

p̂p x1 x2 n1 n2

276 195

747 4340.399.

z p̂1 p̂2

p̂p 1 p̂p 1 n1 1 n2

0.369 0.449

0.399 1 0.399 1 747 1 434 2.71



OR



STEP 6 Interpret the results of the hypothesis test.

Interpretation At the 5% significance level, the data provide sufficient evidence to conclude that, in the United States, the percentage of men who sometimes order veg is smaller than the percentage of women who sometimes order veg.


Procedure 11.4


Formula 11.3


Formula 11.4

Slide 2 - 25 Copyright © 2008 Pearson Education, Inc. Chapter 11 Inferences for Population...

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