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Hypothesis Testing: Two Population Means and Two Population Proportions: Comparing Two Independent Population Proportions

Module by: Dr. Barbara Illowsky, Susan Dean

  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.

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 ( P' A − P'B ) ( P' A − P'B ) reflects a difference in the populations.

The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, H o : p A = p B H o : p A = p B . To conduct the test, we use a pooled proportion, p c p c .

The pooled proportion is calculated as follows:

p c = X A + X B n A + n B p c = X A + X B n A + n B (1)

The distribution for the differences is:

P ' A − P ' B ~ N [ 0 , p c · ( 1 − p c ) · ( 1 n A + 1 n B ) ] P ' A − P ' B ~N[0, p c · ( 1 − p c ) · ( 1 n A + 1 n B ) ](2)

The test statistic (z-score) is:

z = ( p ' A − p ' B ) − ( p A − p B ) p c · ( 1 − p c ) · ( 1 n A + 1 n B ) z= ( p ' A − p ' B ) − ( p A − p B ) p c · ( 1 − p c ) · ( 1 n A + 1 n B ) (3)

Example 1: Two population proportions

Two types of medication for hives are being tested to determine if there is a difference in the percentage 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% level of significance.

Determining the solution

This is a test of 2 population proportions.

Exercise 1

How do you know?

Solution 1

The problem asks for a difference in percentages.

Let AA and BB be the subscripts for medication A and medication B. Then p A p A and p B p B are the desired population proportions.

Random Variable:

P' A − P'B = P' A − P'B = difference in the percentages of adult patients who did not react after 30 minutes to medication A and medication B.

H o : p A = p B p A - p B = 0 H o : p A = p B p A - p B =0

H a : p A ≠ p B p A - p B ≠ 0 H a : p A ≠ p B p A - p B ≠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:

p c = X A + X B n A + n B = 20 + 12 200 + 200 = 0.08 1 − p c = 0.92 p c = X A + X B n A + n B = 20 + 12 200 + 200 =0.081− p c =0.92

Therefore, P' A − P'B ~ N [ 0 , ( 0.08 ) ⋅ ( 0.92 ) ⋅ ( 1 200 + 1 200 ) ] P' A − P'B ~N[0, ( 0.08 ) ⋅ ( 0.92 ) ⋅ ( 1 200 + 1 200 ) ]

P' A − P'B P' A − P'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 = X A n A = 20 200 = 0.1 p' A = X A n A = 20 200 =0.1

Estimated proportion for group B: p' B = X B n B = 12 200 = 0.06 p' B = X B n B = 12 200 =0.06

Graph:

Figure 1
Normal distribution curve of the difference in the percentages of adult patients who don't react to medication A and B after 30 minutes. The x-axis has values of -0.04 and 0.04. Two vertical upward lines extend from these values to the curve. 1/2(p-values) point to the areas on either side of these values.

P' A − P'B = 0.1 - 0.06 = 0.04 P' A − P'B =0.1-0.06=0.04.

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

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

Make a decision: Since α < p-value α<p-value, you cannot reject H o H o .

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

TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 20 for x1x1, 200 for n1n1, 12 for x2x2, and 200 for n2n2. Arrow down to p1: and arrow to does not equal p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1404p =0.1404 and the test statistic is 1.47. Do the procedure again but instead of Calculate do Draw.

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