Inferencing Proportions: Z-Procedures for Single and Two-Sample Hypothesis Testing, Exams of Data Analysis & Statistical Methods

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Chapter 8 Inference for Proportions

This chapter presents the z procedures for one-sample and two-sample inference about popula- tion proportions. The procedures are approximate, based on the large-sample normal approxi- mation.

8.1 Inference for a Single Proportion Example:

(a) What is the current unemployment rate in U.S.?

(b) What is President George W. Bush’s approval rating?

(c) What proportion of students in Math 426/526 will be awarded Grade A?

(d) Does preschool make a difference?

(e) · · ·

Example 8.1 Concern About Mad Cow Disease (Gallup Poll News) On Dec. 23, 2003, the news media reported that the first case of mad cow disease had been found in the United States. After the announcement, many U.S. trading partners immediately banned imports of American beef, causing billions of dollars in losses for the beef industry. A CNN/USA Today/Gallup survey, conducted Jan. 2-5, 2004 found 165 of 1029 adults interviewed express worry that they or their families might become victims of the disease. It is also found that women are somewhat more likely than men to express worry about becoming a victim. 105 of 498 women, compared with 64 of 531 of men, worried that they or someone in their family will become victims of the disease,

(a) Find a 95% confidence interval of proportion of people who worried that they or someone in their family will become victims of the disease.

(b) Are women more likely than men to express worry about becoming a victim?

Solution:

Let p be the population proportion of people who worried that they or someone in their family will become victims of the disease. Then the sample proportion

pˆ =

≈ 0. 16

is a statistic that can be used to estimate the parameter p.

Sampling distribution of a sample proportion: Choose an SRS of size n from a large population with population proportion p. Let X be the count of successes. Recall

• X ∼ B(n, p).
• μX = np, σX =

√

p(1 − p)n.

The sample proportion of successes ˆp = X/n estimates p. Then

• The mean of ˆp is p: μˆp = p
• The standard deviation of ˆp is

σpˆ =

√

p(1 − p) n

.

• The central limit theorem: pˆ − p σpˆ

≈ N (0, 1)

The Wilson Estimate:

p˜ =

X + 2

n + 4 The adjustment is based on the following idea: act as if we have 4 additional observations, 2 of which are successes and 2 of which are failures.

• The standard error of ˜p is

SE˜p =

√

p˜(1 − p˜) n + 4

Confidence interval for a population proportion p: Choose an SRS of size n from a large population with unknown proportion p of successes. An approximate level C confidence interval for p (p. 573) is

p˜ ± z∗SEp˜

The margin of error is m = z∗SEp˜

Use this interval when n ≥ 5 and the confidence level is 90% - 99%.

Example 8.2 Do you approve or disapprove of the way George W. Bush is handling his job as president? A Gallup poll conducted Jan 2-5, 2004 found that 617 of 1029 adults interviewed approved Bush’s job performance. The Gallup poll claims that “For results based on the total sample of adults nationwide, one can say with 95% confidence that the margin of sampling error is no greater than +/ − 3 percentage points.”. Explain.

Solution:

Choosing a sample size The margin of error is

m = z∗

√

p˜(1 − p˜) n + 4 Sample size for desired margin of error: The level C confidence interval for p will have a margin of error approximately equal to a specified value m when the sample size is

n + 4 =

(

z∗ m

) 2

p∗(1 − p∗)

where p∗^ is a guessed value for the sample proportion.

• Conservative sample size:

n =

(

z∗ m

) 2

(1/4) − 4.

The margin of error will be less than or equal to m.

Example 8.3 Find the sample size needed if the margin of error of the 95% confidence interval is

(a) m = 1%

(b) m = 2%

(c) m = 3%

(d) m = 3%, p∗^ =. 3

(e) m = 4%

Solution:

Example 8.4 An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by the local dealer. The customer relations department will survey a random sample of customers and compute a 99% confidence interval for the proportion who are not satisfied.

(a) Past studies suggest that this proportion will be about 0. 2. Find the sample size needed if the margin of error of the confidence interval is to be about 0. 015.

(b) When the sample is actually contacted, 10% of the sample say they are not satisfied. What is the margin of error of the 99% confidence interval? What is the 99% confidence interval?

Solution:

Large-sample significance test for p

Null hypothesis: H 0 : p = p 0 z-test statistic:

z =

pˆ − p 0 √ p 0 (1 − p 0 ) n

Alternative P-value Ha : p > p 0 P (Z ≥ z) Ha : p < p 0 P (Z ≤ z) Ha : p 6 = p 0 2 P (Z ≥ |z|)

Assumptions:

• The data are an SRS
• Population size ≥ 10 × sample size
• n p 0 , n(1 − p 0 ) ≥ 10

Example 8.5 The English mathematician John Kerrich tossed a coin 10,000 times and ob- tained 5067 heads. Is this significant evidence at the 5% level that the coin is not balanced?

Solution:

Example 8.6 Do mathematicians have more girls? Some people think that mathematicians are more likely than other parents to have female children. The Washington State Department of Health lists the parents’ occupations on birth certificates. Between 1980 and 1990, 555 children were born to fathers who were mathematicians. Of these births, 273 were girls. During this period, 48 .8% of all births in Washington State were girls. Is there evidence that the proportion of girls born to mathematicians is higher than the state proportion?

Solution:

8.2 Comparing Two Proportions

Suppose we have two independent samples.

Pop’n Sample Sample Pop’n prop’n size prop’n 1 p 1 n 1 pˆ 1 2 p 2 n 2 pˆ 2

The sampling distribution of ˆp 1 − pˆ 2 :

• The mean of ˆp 1 − pˆ 2 is p 1 − p 2
• The standard deviation of ˆp 1 − pˆ 2 is

σˆp 1 −ˆp 2 =

√

p 1 (1 − p 1 ) n 1

+

p 2 (1 − p 2 ) n 2

The Wilson Estimates:

p˜ 1 =

X 1 + 1

n 1 + 2

, p˜ 2 =

X 2 + 1

n 2 + 2

• The estimate of p 1 − p 2 is D˜ = ˜p 1 − p˜ 2
• The standard error of D˜ is

SE D˜ =

√

p ˜ 1 (1 − p˜ 1 ) n 1 + 2

+

p˜ 2 (1 − p˜ 2 ) n 2 + 2

Confidence intervals for p 1 − p 2 :

p˜ 1 − tildep 2 ± z∗SE (^) D˜

The margin of error is m = z∗SE (^) D˜. Use this method when both sample sizes are at least 10 and the confidence level is 90% - 99%.

Example 8.7 Do drivers reduce excessive speed when they encounter police radar? Researchers studied the behavior of drivers on a rural interstate highway in Oregon where the speed limit was 55 miles per hour. During some time periods, police radar was set up at the measurement location. Here are some of data:

Number of vehicles Number of vehicles over 65 mph No radar 12931 5690 Radar 73285 1051

Give a 95% confidence interval for the effect of radar, as measured by the difference in proportions of vehicles going faster than 65 mph with and without radar.

Solution:

Example 8.8 A study is made to determine if a cold climate results in more students being absent from school during a semester than for a warmer climate. Two groups of students are selected at random, one group from Vermont and the other group from Georgia. Of the 300 students from Vermont, 64 were absent at least 1 day during the semester, and of the 400 students from Georgia, 51 were absent 1 or more days. Find a 90% confidence interval for the difference between the fractions of students who are absent in the two states.

Solution:

Recall: pˆ 1 − pˆ 2 − (p 1 − p 2 ) √ p 1 (1 − p 1 ) n 1

+

p 2 (1 − p 2 ) n 2

≈ N (0, 1)

Significance tests for p 1 − p 2 Null hypothesis: H 0 : p 1 = p 2 z statistic:

z =

pˆ 1 − pˆ 2 SEDp

where the pooled standard error is

SEDp =

√

p ˆ(1 − pˆ)

( 1

n 1

+

n 2

)

and ˆp is the pooled sample proportion

pˆ =

number of successes in both samples number of observations in both samples

=

X 1 + X 2

n 1 + n 2

Alternative P-value Ha : p 1 > p 2 P (Z ≥ z) Ha : p 1 < p 2 P (Z ≤ z) Ha : p 1 6 = p 2 2 P (Z ≥ |z|)

Example 8.9 Does taking aspirin regularly help prevent heart attacks? A double-blind random- ized comparative experiment assigned 11,037 male doctors to take aspirin and another 11, to take a placebo. After 5 years, 104 of the aspirin group and 189 of the control group had died of heart attacks. Is this difference large enough to convince us that aspirin works?

Solution:

Example 8.10 In a Prevention magazine survey related in 2001, Princeton Survey Research Associates examined the weights of children aged 3 to 17. According to this study, 30% of children in this age group were overweight in 1991, and 38% were considered overweight in

2001. Suppose that these percentages are based on random samples of 400 and 600 children in the given age group in 1991 and 2001, respectively. Test at the 1% level of significance if the proportion of overweight children in this age group was less in 1991 than in 2001. Make sure to state the null and alternative hypotheses and report the P -value of the test.

Solution: