Two Proportion z-Interval and z-Tests

Suppose that A follows a Normal distribution with mean 24 and standard deviation of 5. B follows a Normal distribution with mean 42 and standard deviation of 8.5. What is the probability that a randomly selected A is greater than a randomly selected B? Two Proportion z-Interval and z-Tests Small differences. Recall from Random Variables y If X and Y are two independent Random Variables from Normal distributions, we can combine them to get a new Random Variable that is also from a Normal distribution. Mean: E ( X − Y ) = E ( X ) − E (Y ) Standard Deviation: SD( X − Y ) = SD 2 ( X ) + SD 2 (Y ) Differences between two proportions y We can create a confidence interval for the difference between the two proportions. y Conditions: 1. Randomization 2. Independence: Independence between the two groups 3. 10% Condition: check for both groups 4. Success/Failure: check for both groups separately, use p̂1 and p̂2 All conditions have been met to use a Normal model for a two-proportion z-interval. Two-proportion z-interval mean and standard deviation(error). y Mean: pˆ1 − pˆ 2 ˆ ˆ ˆ ˆ y Standard Error: p1q1 + p2q2 Conclusion: Three possibilities for the interval: 1. Both Negative 2. Both Positive 3. One Negative; One Positive pˆ1qˆ1 pˆ 2qˆ2 y CI: ( pˆ1 − pˆ 2 ) ± z * + n1 n2 We are ____% confident that the true proportion of (Group 1) is between ___% to ___% (lower/higher) than (Group 2) y Since the Normal model is being used we need a n1 n2 Example: Gender and Binge Drinking y College students were randomly selected and asked about how much alcohol they consumed on a weekly basis. Over a certain amount of alcohol consumed was considered binge drinking and those individuals were labeled frequent binge drinkers. The following table shows the number of men and women who were frequent binge drinkers. What is the difference in the proportions of male and female binge drinkers? Male Female Frequent Binge Total Drinker (x) Questioned(n) 1392 5348 1748 8471 p1 – proportion of males p2 – proportion of females Conditions: 1. Randomization: Stated as random samples 2. Independence: It is safe to assume that men and women are independent of each other. 3. 10% Condition: 5348 male college students is less than 10% of all male college students 8471 female college students is less than 10% of all female college students 4. Success/Failure: n1pˆ1 = 1392 ≥ 10 n2 pˆ 2 = 1748 ≥ 10 n1qˆ1 = 3956 ≥ 10 n2qˆ2 = 6723 ≥ 10 All conditions have been met to use a Normal model for a two-proportion z-interval. Two-proportion z-interval at 95% confidence CI:( pˆ1 − pˆ 2 ) ± z * pˆ1qˆ1 pˆ 2qˆ2 + = n1 n2 ⎛ 1392 ⎞ ⎛ 3956 ⎞ ⎛ 1748 ⎞⎛ 6723 ⎞ ⎜ 5348 ⎟ ⎜ 5348 ⎟ ⎜ 8471 ⎟⎜ 8471 ⎟ ⎛ 1392 1748 ⎞ ⎝ ⎠⎝ ⎠+⎝ ⎠⎝ ⎠ − ± 1.96 ⎜ 5348 8471 ⎟ 5348 8471 ⎝ ⎠ =(0.039, 0.069) We are 95% confident that the true proportion of men who are frequent binge drinkers is between 3.9% to 6.9% higher than the proportion of women who are frequent binge drinkers. Mechanics: p. 520 #9, 11 9. Arthritis p1 = proportion of males suffering from arthritis p2 = proportion of females suffering from arthritis Conditions: Independence: It is safe to assume that males and females would be independent of each other Randomization: stated as a random sample 10% Condition: 1012 males is less than 10% of all males over 65, and 1062 females is less than 10% of all females over 65 $ = 411 ≥ 10 n p $ Success/Failure: n1 p 1 2 2 = 535 ≥ 10 n1q$1 = 601 ≥ 10 n2 q$ 2 = 527 ≥ 10 All conditions have been met to use the Normal model for a 2proportion z-interval. p1 − dogs fromhomes whereherbicidesareused; $ = 411 p $ = 535 p 1 2 1012 1062 11. Pets ⎛ 411 ⎞ ⎛ 601 ⎞ ⎛ 535 ⎞ ⎛ 527 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎛ 411 535 ⎞ ⎝ 1012 ⎠ ⎝ 1012 ⎠ + ⎝ 1062 ⎠ ⎝ 1062 ⎠ CI : ⎜ − ± 1 . 960 ⎟ 1012 1062 ⎝ 1012 1062 ⎠ Conditions: 1. Random: Not random so we will assume they are representative samples 2. Independence: It is safe to assume the two groups would be independent of each other 3. 10% Condition: 827 dogs is less than 10% of all dogs from homes that use the herbicides; 130 dogs is less than 10% of all dogs from homes that don’t use the herbicides. 4. Success/Failure: n1pˆ1 = 473 ≥ 10; n2 pˆ 2 = 19 ≥ 10 ( −0.140, −0.055 ) Conclusion: We are 95% confident that the proportion of men over 65 who suffer from arthritis is between 5.5% to 14.0% lower than the proportion of women over 65 who suffer from arthritis. Since the interval for the difference in proportions of arthritis sufferers does not contain 0, there is strong evidence that arthritis is more likely to afflict women than men. p2 − dogs fromhomes whereherbicidesarenot used n1qˆ1 = 354 ≥ 10; n2qˆ2 = 111 ≥ 10 All conditions have been met to use the Normal model for a 2-proportion z-interval Mechanics: p$1 = 473 = 0.5719 p$ 2 = 19 = 0.1462 827 130 ⎛ 473 ⎞ ⎛ 354 ⎞ ⎛ 19 ⎞ ⎛ 111 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎛ 473 19 ⎞ ⎝ 827 ⎠ ⎝ 827 ⎠ + ⎝ 130 ⎠ ⎝ 130 ⎠ − CI : ⎜ ⎟ ± 1.960 827 130 ⎝ 827 130 ⎠ ( 0.5719 − 0.1462 ) ± 1.960 ( 0.5719 )( 0.4281) + ( 0.1462 )( 0.8538 ) 827 130 ( 0.3563,0.4953 ) Conclusion: We are 95% confident that the true proportion of dogs from homes that use herbicides that develop malignant lymphoma is between 35.6% to 49.5% higher than the proportion of dogs from homes that don’t use herbicides that develop malignant lymphoma. Hypotheses y The null should be that there is no difference between the two proportions or that they are equal. H 0 : p1 = p2 y But we have no probability model for this situation. y Since both proportions come from independent(we hope) Normal distributions, the difference of the two would be a new Normal distribution. H 0 : p1 – p2 = 0 Comparing 2 Group Proportions y Three possibilities: 1. The two proportions are the same. 2. The two proportions are different. 3. One proportion is larger than the other. The null hypothesis would be that they are the same. We’ll then run a test to see if they are different(twotail) or one is larger than the other(one-tail) Alternative Hypothesis y The alternative hypotheses are then: y p1 < p2 or p1 – p2 < 0 The proportion of group 1 is lower than group 2 y p1 > p2 or p1 – p2 > 0 The proportion of group 1 is higher than group 2 y p1 ≠ p2 or p1 – p2 ≠ 0 The proportion of group 1 is not the same as group 2 Mechanics Conditions y We need to calculate a pooled proportion from 1. Randomization 2. Independence: Independence between the two groups 3. 10% Condition: check for both groups 4. Success/Failure: check for both groups separately the two we are given. y The assumption is that they are equal, so creating a combined proportion would give us a better estimate of the true proportion. So given: pˆ1 = x1 x ; pˆ 2 = 2 n1 n2 y The pooled proportion is: All conditions have been met to use a Normal model for a two proportion z-test pˆ pooled = x1 + x2 n1 + n2 y This will is used to calculate a standard deviation for the two proportions. y Pooled Standard Deviation: SE ( pˆ pooled ) = pˆ pooled qˆ pooled n1 + Conclusion pˆ pooled qˆ pooled n2 ( pˆ1 − pˆ2 ) − 0 = SE ( pˆ pooled ) y Talks about difference between the two proportions only, not the values of the individual proportions. y Test Statistic: z= y Compare P-Value to stated alpha. pˆ1 − pˆ 2 pˆ pooled qˆ pooled pˆ pooled qˆ pooled + n1 n2 y P-Value = P(z <,> ___) Example: Binge Drinkers Is there evidence that male college students have a higher proportion of frequent binge drinking than female college students? Hypothesis: p1 – proportion of males H 0 : p1 = p2 p2 – proportion of females H A: p1 > p2 Conditions: 1. Randomization: Stated as random samples 2. Independence: It is safe to assume that men and women are independent of each other. 3. 10% Condition: 5348 male college students is less than 10% of all male college students 8471 female college students is less than 10% of all female college students 4. Success/Failure: n1pˆ1 = 1392 ≥ 10 n1qˆ1 = 3956 ≥ 10 n2 pˆ 2 = 1748 ≥ 10 n2qˆ2 = 6723 ≥ 10 All conditions have been met to use a Normal model for a 2 proportion z-test. Mechanics: 1392 1748 α = 0.05 = 0.2603 pˆ 2 = = 0.2064 5348 8471 x + x2 1392 + 1748 3140 pˆ pooled = 1 = = = 0.2272 n1 + n2 5348 + 8471 13819 pˆ1 = z= z= 1392 1748 − 5348 8471 = 7.369 ⎛ 3140 ⎞ ⎛ 10679 ⎞ ⎛ 3140 ⎞ ⎛ 10679 ⎞ ⎜ 13819 ⎟ ⎜ 13819 ⎟ ⎜ 13819 ⎟ ⎜ 13819 ⎟ ⎝ ⎠⎝ ⎠+⎝ ⎠⎝ ⎠ 5348 8471 0.2603 − 0.2064 ( 0.2272 )( 0.7728 ) + ( 0.2272 )( 0.7728 ) 5348 8471 = 7.369 P − Value = P ( z > 7.369) = 8.66 × 10 −14 ≈ 0 Conclusion: Since the P-Value is so much smaller than alpha (8.66 x 10-14 < 0.05), we reject the null hypothesis. There is statistically significant evidence that the proportion of male college students who are frequent binge drinkers is more than female college students. 18. p1 – proportion of cardiac patients without depression who died within 4 years p2 – proportion of cardiac patients with depression who died within 4 years Hypotheses: H0: p1 – p2 = 0 n1q$1 = 294 ≥ 10 HA : p1 – p2 < 0 67 + 26 93 $ = 67 p $ = 26 p $ p = ≈ 0.207 α = 0.05 1 2 Poooled = 361 89 361 + 89 450 ⎛ 67 ⎞ ⎛ 26 ⎞ ⎜ ⎟−⎜ ⎟ ⎝ 361 ⎠ ⎝ 89 ⎠ z= = −2.223 ⎛ 93 ⎞ ⎛ 357 ⎞ ⎛ 93 ⎞ ⎛ 357 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ 450 ⎠ ⎝ 450 ⎠ + ⎝ 450 ⎠ ⎝ 450 ⎠ 361 89 or ( 0.186 ) − ( 0.292 ) ( 0.207 )( 0.793 ) + ( 0.207 )( 0.793 ) 361 n2 q$ 2 = 63 ≥ 10 All conditions have been met to use the Normal model for a 2-proportion z-test. Mechanics: z= Conditions: Independence: It is safe to assume that one group would have an influence on the deaths of another group. Randomization: Not random, we will assume that the samples are representative of the populations. 10% Condition: 361 patients is less than 10% of all patients without depression. 89 patients is less than 10% of all patients with depression $ = 26 ≥ 10 Success/Failure: n1 p$1 = 67 ≥ 10 n2 p 2 = −2.223 89 P-Value = P(z < -2.223) = 0.013 Conclusion: Since the P-Value is less than alpha (0.013 < 0.05), we reject the null hypothesis. There is evidence that the proportion of cardiac patients without depression that died is less than for cardiac patients with depression.