Comparing Two Independent Population Proportions Quiz

PDF page 588: 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: x + x

PDF page 588: 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: x + x

PDF page 588: 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% level of significance.

PDF page 588: P′A – P′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 pA – pB = 0 Ha: pA ≠ pB pA – pB ≠ 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: x + x 20 + 12

PDF page 588: Estimated proportion for group A: p′ = = = 0.1

PDF page 588: B A ⎣ ⎦ 200 200 P′A – P′B follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.1404. x 20

PDF page 589: P′A – P′B = 0.1 – 0.06 = 0.04.

PDF page 589: Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α < p-value.

PDF page 589: Make a decision: Since α < p-value, do not reject H0. 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 proportions of adult patients who did not react after 30 minutes to medication A and medication B.

PDF page 589: 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.

PDF page 590: Ha: pF < pM Ha: pF – pM < 0 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: x + x 156 + 183

PDF page 590: = = 0.077 pc = n + n 2169 + 2231

PDF page 590: F M ⎝ ⎝ ⎠⎠ 2169 2231 p′F – p′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: