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Small differences. Two Proportion z-Interval and z-Tests
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Recall from Random Variables 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: Standard Deviation:
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Differences between two proportions We can create a confidence interval for the difference between the two proportions. 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 and All conditions have been met to use a Normal model for the differences in two proportions.
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Two-proportion z-interval Since the Normal model is being used we need a mean and standard deviation(error). Mean: Standard Error: CI: Be very careful with the conclusion.
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Example: Gender and Binge Drinking 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? Frequent Binge Drinker (x) Total Questioned(n) Male13925348 Female17488471
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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: All conditions have been met to use a Normal model for the differences of two proportions.
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Two-proportion z-interval at 95% confidence CI: =(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.
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Comparing 2 Group Proportions 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 there are different(two-tail) or one is larger than the other(one-tail)
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Hypotheses The null should be that there is no difference between the two proportions or that they are equal. H 0 : p 1 = p 2 But we have no probability model for this situation. Since both proportions come from independent(we hope) Normal distributions, the difference of the two would be a new Normal distribution. H 0 : p 1 – p 2 = 0 We can test against this null hypothesis now.
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Alternative Hypothesis The alternative hypotheses are then: p 1 – p 2 < 0 The proportion of group 1 is lower than group 2 p 1 – p 2 > 0 The proportion of group 1 is higher than group 2 p 1 – p 2 ≠ 0 The proportion of group 1 is not the same as group 2
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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 All conditions have been met to use a Normal model for the difference of two proportions
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Mechanics We need to calculate a pooled proportion from the two we are given. The assumption is that they are equal, so given: The pooled proportion is: This will is used to calculate a standard deviation for the two proprotions.
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Pooled Standard Deviation: Test Statistic: P-Value = P(z ___)
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Conclusion Compare P-Value to stated alpha. Talks about difference between the two proportions only, not the values of the individual proportions.
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Example: Binge Drinkers Is there evidence that male college students have a higher proportion of frequent binge drinking than female college students? Hypothesis: H 0 : p 1 – p 2 = 0 The proportion of male and female college students that are frequent binge drinkers is the same. H A : p 1 – p 2 > 0 The proportion of male college students that are frequent binge drinkers is more than female college students.
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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: All conditions have been met to use a Normal model for the differences of two proportions.
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Mechanics:
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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.
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