Comparing Two Independent Population Proportions Quiz

PDF page 603: 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: x + x

PDF page 603: 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: x + x

PDF page 603: 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: x + x

PDF page 603: B A The test statistic (z-score) is (p′ − p′ ) − (p − p )

PDF page 604: 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:

PDF page 604: 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 604: = 0.08 1 – pc = 0.92

PDF page 604: P′A – P′B = 0.1 – 0.06 = 0.04.

PDF page 604: 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 604: Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α < p-value.

PDF page 604: Make a decision: Since α < p-value, do not reject H0.

PDF page 605: 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 605: 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: