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# Introduction

### Introduction

When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present:

1. The two independent samples are simple random samples that are independent.
2. The number of successes is at least five, and the number of failures is at least five, for each of the samples.
3. Growing literature states that the population must be at least 10 or 20 times the size of the sample. This keeps each population from being over-sampled and causing incorrect results.

Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions.

The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, H0: pA = pB. To conduct the test, we use a pooled proportion, pc.

The pooled proportion is calculated as follows:

$pc=xA+xBnA+nB.pc=xA+xBnA+nB.$

The distribution for the differences is

$P′A−P′B~N[0,pc(1−pc)(1nA+1nB)].P′A−P′B~N[0,pc(1−pc)(1nA+1nB)]$

The test statistic (z-score) is

$z=(p′A−p′B)−(pA−pB)pc(1−pc)(1nA+1nB).z=(p′A−p′B)−(pA−pB)pc(1−pc)(1nA+1nB)$

### Example 10.8

Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given Medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given Medication B still had hives 30 minutes after taking the medication. Test at a 1 percent level of significance.

Solution 10.8

The problem asks for a difference in proportions, making it a test of two proportions.

Let A and B be the subscripts for Medication A and Medication B, respectively. Then, pA and pB are the desired population proportions.

Random Variable: P′AP′B = difference in the proportions of adult patients who did not react after 30 minutes to Medication A and to Medication B.

H0: pA = pB

pApB = 0

Ha: pApB

pApB ≠ 0

The words is a difference tell you the test is two-tailed.

Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal:

$pc=xA+xBnA+nB=20+12200+200=0.08 1–pc=0.92pc=xA+xBnA+nB=20+12200+200=0.08 1–pc=0.92$

$P′A–P′B~N[0,(0.08)(0.92)(1200+1200)]P′A–P′B~N[0,(0.08)(0.92)(1200+1200)]$

P′AP′B follows an approximate normal distribution.

Calculate the p-value using the normal distribution: p-value = 0.1404.

Estimated proportion for group A: $p′A=xAnA=20200=0.1p′A=xAnA=20200=0.1$

Estimated proportion for group B: $p′B=xBnB=12200=0.06p′B=xBnB=12200=0.06$

Graph:

Figure 10.7

P′AP′B = 0.1 – 0.06 = 0.04.

Half the p-value is below –0.04, and half is above 0.04.

Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α p-value.

Make a decision: Since α p-value, do not reject H0.

Conclusion: At a 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to Medication A and Medication B.

### Using the TI-83, 83+, 84, 84+ Calculator

Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 20 for x1, 200 for n1, 12 for x2, and 200 for n2. Arrow down to p1: and arrow to not equal p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1404, and the test statistic is 1.47. Do the procedure again, but instead of Calculate do Draw.

Try It 10.8

Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5 percent level of significance.

### Example 10.9

A research study was conducted about gender differences in texting. The researcher believed that the proportion of girls involved in texting is less than the proportion of boys involved. The data collected in spring 2010 among a random sample of middle and high school students in a large school district in the southern United States is summarized in Table 10.10. Is the proportion of girls sending texts less than the proportion of boys texting? Test at a 1 percent level of significance.

Males Females
Sent texts 183 156
Total number surveyed 2231 2169
Table 10.10
Solution 10.9

This is a test of two population proportions. Let M and F be the subscripts for males and females. Then, pM and pF are the desired population proportions.

Random variable: p′Fp′M = difference in the proportions of males and females who sent texts.

H0: pF = pMH0: pFpM = 0

Ha: pF pMHa: pFpM

The words less than tell you the test is left-tailed.

Distribution for the test: Since this is a test of two population proportions, the distribution is normal:

$pc=xF+xMnF+nM=156+1832169+2231=0.077pc=xF+xMnF+nM=156+1832169+2231=0.077$

$1−pc=0.9231−pc=0.923$

Therefore,

$p′F–p′M∼N(0,(0.077)(0.923)(12169+12231))p′F–p′M∼N(0,(0.077)(0.923)(12169+12231))$

p′Fp′M follows an approximate normal distribution.

Calculate the p-value using the normal distribution:

p-value = 0.1045

Estimated proportion for females: 0.0719

Estimated proportion for males: 0.082

Graph:

Figure 10.8

Decision: Since α p-value, do not reject H0.

Conclusion: At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of girls sending texts is less than the proportion of boys sending texts.

### Using the TI-83, 83+, 84, 84+ Calculator

Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1045 and the test statistic is z = –1.256.

### Example 10.10

Researchers conducted a study of smartphone use (Phone A versus Phone B) among adults. A cell phone company claimed that Phone B smartphones are more popular with whites (non-Hispanic) than with African Americans. The results of the survey indicate that of the 232 African American cell phone owners randomly sampled, 5 percent own Phone B. Of the 1,343 white cell phone owners randomly sampled, 10 percent own Phone B. Test at the 5 percent level of significance. Is the proportion of white Phone B owners greater than the proportion of African American Phone B owners?

Solution 10.10

This is a test of two population proportions. Let W and A be the subscripts for the whites and African Americans. Then, pW and pA are the desired population proportions.

Random variable:p′Wp′A = difference in the proportions of Phone A and Phone B users.

H0: pW = pAH0: pWpA = 0

Ha: pW > pAHa: pWpA > 0

The words more popular indicate that the test is right-tailed.

Distribution for the test: The distribution is approximately normal:

$1−pc=0.90731−pc=0.9073$

Therefore,

$p′W–p′A∽N(0,(0.0927)(0.9073)(11343+1232))p′W–p′A∽N(0,(0.0927)(0.9073)(11343+1232))$

$p′W–p′Ap′W–p′A$ follows an approximate normal distribution.

Calculate the p-value using the normal distribution:

p-value = 0.0077
Estimated proportion for group A: 0.10
Estimated proportion for group B: 0.05

Graph:

Figure 10.9

Decision: Since α > p-value, reject the H0.

Conclusion: At the 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of white cell phone owners use Phone B than African Americans.

### Using the TI-83, 83+, 84, 84+ Calculator

TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 135 for x1, 1343 for n1, 12 for x2, and 232 for n2. Arrow down to p1: and arrow to greater than p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.0092, and the test statistic is z = 2.33.

Try It 10.10

A group of citizens wanted to know if the proportion of homeowners in their small city was different in 2011 than in 2010. Their research showed that of the 113,231 available homes in their city in 2010, 7,622 of them were owned by the families who live there. In 2011, 7,439 of the 104,873 of the available homes were owned by city residents. Test at a 5 percent significance level. Answer the following questions:

a. Is this a test of two means or two proportions?

b. Which distribution do you use to perform the test?

c. What is the random variable?

d. What are the null and alternative hypotheses? Write the null and alternative hypotheses in symbols.

e. Is this test right-, left-, or two-tailed?

f. What is the p-value?

g. Do you reject or not reject the null hypothesis?

h. At the ______ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.