1. Inferences about two proportions Assumptions 1.We have proportions from two simple random samples that are independent (not paired) 2.For both samples, np ≥ 5 and nq ≥ 5 Possible alternative hypotheses: Null Hypothesis: p 1 = p 2

2. Test Statistic: where p 1 - p 2 = 0 (assumed in null hypothesis) (pooled estimate of p 1 and p 2 ) Critical values and P-values come from the z-score tables

3. Example A drug company wants to determine that their new headache drug is effective. They give 500 people their new drug, and 400 people a placebo. In the experimental group, 350 said their headache went away in 15 minutes. In the placebo group, 235 said their headache went away in 15 minutes. Is the success rate higher in the experimental group?

4. Define hypotheses: Let’s use a 0.01 significance level Because we’re working with a proportion, we use a normal distribution.

5. Value of the test statistic: Since we are working with a 0.01 significance level, and this is a right-tailed test, the critical value is 2.33. Since the test statistic is larger than the critical value, we reject the null hypothesis. The sample data support the claim that a larger proportion of people recovered in 15 minutes using the new drug than using the placebo.

6. P­Value : P(z > 3.516) = 0.0001. Much smaller than our critical value. This would again lead us to reject the null hypothesis. Confidence Intervals for two proportions

7. Example: The 99% confidence interval for our drug test results: We are 99% confident that the difference between the population proportions is between 0.03 and 0.195

8. Inferences about two means Assumptions 1.The two samples are simple random samples, and are independent 2.Either both samples are large (>30), or both samples come from normally distributed populations. Test Statistic: Degrees of freedom is the smaller of n 1 -1 and n 2 -1

9. Example You want to test the theory that talking to plants makes a difference. You put 23 bean plants in one greenhouse and talk to them nicely each day. You put 21 bean plants in another greenhouse and ignore them. After 4 weeks, you find that the mean height of the talked-to plants is 38cm, with a standard deviation of 5cm. The mean height of the ignored plants is 34cm, with a standard deviation of 7cm. Test the claim that the results are different.

10. Define hypotheses: Let’s use a 0.05 significance level Because we’re working with sample means, σ unknown, we use a t distribution. Degrees of freedom = 20

11. Value of the test statistic: This is a two-tailed test with a 0.05 significance level. From our critical t table, our critical values are -2.086 and 2.086 Since the test statistic is larger than the critical value, we reject the null hypothesis. The sample data support the claim that plants that are talked to grow differently than plants that are ignored.

12. Confidence Intervals for two means P-Value: Since it is a two-tailed test with the test statistic to the right of center, we want to find twice the area to the right of the test statistic: 2. P(t > 2.163) = 2. (0.020) = 0.04 (using technology) Since this is less than our significance level, we reject the null hypothesis

13. Example: The 95% confidence interval for our plant results above: We are 95% confident that the difference between the populations is between 0.14cm and 7.86cm. In other words, we are 95% confident that the plants that are talked to grow between 0.14cm and 7.86cm higher than plants that are ignored.

14. Matched Pairs When comparing two populations where the samples are not independent, we must use the Matched Pairs test (8.4) Example of matched pairs: People are given a test while listening to music, and another test in silence. Claim: the mean test score while listening to music is higher than the mean test score in silence

15. Homework 8.2: 5, 7, 11, 21 8.3: 1, 3, 5, 7, 15