1. Independent Samples: Comparing Means Lecture 39 Section 11.4 Fri, Apr 1, 2005

2. Independent Samples In a paired study, two observations are made on each subject, producing one sample of bivariate data. In a paired study, two observations are made on each subject, producing one sample of bivariate data. Or we could think of it as two samples of paired data. Or we could think of it as two samples of paired data. Often these are “before” and “after” observations. Often these are “before” and “after” observations. By comparing the “before” mean to the “after” mean, we can determine whether the intervening treatment had an effect. By comparing the “before” mean to the “after” mean, we can determine whether the intervening treatment had an effect.

3. Independent Samples On the other hand, with independent samples, there is no logical way to “pair” the data. On the other hand, with independent samples, there is no logical way to “pair” the data. One sample might be from a population of males and the other from a population of females. One sample might be from a population of males and the other from a population of females. Or one might be the treatment group and the other the control group. Or one might be the treatment group and the other the control group. The samples could be of different sizes. The samples could be of different sizes.

4. Independent Samples We wish to compare population means  1 and  2. We wish to compare population means  1 and  2. We do so by comparing sample means  x 1 and  x 2. We do so by comparing sample means  x 1 and  x 2. More specifically, we will use  x 1 –  x 2 as an estimator of  1 –  2. More specifically, we will use  x 1 –  x 2 as an estimator of  1 –  2. If we want to know whether  1 =  2, we test to see whether  1 –  2 = 0 by computing  x 1 –  x 2. If we want to know whether  1 =  2, we test to see whether  1 –  2 = 0 by computing  x 1 –  x 2.

5. The Distributions of  x 1 and  x 2 Let n 1 and n 2 be the sample sizes. Let n 1 and n 2 be the sample sizes. If the samples are large, then  x 1 and  x 2 have (approx.) normal distributions. If the samples are large, then  x 1 and  x 2 have (approx.) normal distributions. However, if either sample is small, then we will need an additional assumption. However, if either sample is small, then we will need an additional assumption. The populations are normal. The populations are normal.

6. Further Assumption We will also assume that the two populations have the same standard deviation. We will also assume that the two populations have the same standard deviation. Call it . Call it . If this assumption is not supported by the evidence, then it should not be made. If this assumption is not supported by the evidence, then it should not be made. This assumption is often not warranted, but without it, the formulas become much more complicated. See p. 658. This assumption is often not warranted, but without it, the formulas become much more complicated. See p. 658.

7. The t Distribution Let s 1 and s 2 be the sample standard deviations. Let s 1 and s 2 be the sample standard deviations. Whenever we use s 1 and s 2 instead of , then we will have to use the t distribution instead of the standard normal distribution, unless the sample sizes are large. Whenever we use s 1 and s 2 instead of , then we will have to use the t distribution instead of the standard normal distribution, unless the sample sizes are large.

8. The Distribution of  x 1 –  x 2 Suppose that  x 1 and  x 2 have normal distributions with means  1 and  2 and standard deviations  1 /  n 1 and  2 /  n 2 (according to the Central Limit Theorem, p. 500). Suppose that  x 1 and  x 2 have normal distributions with means  1 and  2 and standard deviations  1 /  n 1 and  2 /  n 2 (according to the Central Limit Theorem, p. 500). Then  x 1 –  x 2 is a normal random variable with the following properties: Then  x 1 –  x 2 is a normal random variable with the following properties: The mean is  1 –  2. The mean is  1 –  2. The standard deviation is  (  1 2 /n 1 +  2 2 /n 2 ). The standard deviation is  (  1 2 /n 1 +  2 2 /n 2 ).

9. The Distribution of  x 1 –  x 2 If we assume that  1 =  2, then the standard deviation may be simplified to If we assume that  1 =  2, then the standard deviation may be simplified to That is, That is,

10. The Distribution of  x 1 –  x 2 11 0

11. 22 11 0

12. 22 11 0  1 –  2

13. The Distribution of  x 1 –  x 2 If If then it follows that

14. Estimating  Individually, s 1 and s 2 estimate . Individually, s 1 and s 2 estimate . However, we can get a better estimate than either one if we “pool” them together. However, we can get a better estimate than either one if we “pool” them together. The pooled estimate is The pooled estimate is

15.  x 1 –  x 2 and the t Distribution If we use s p instead of , and the sample sizes are small, then we should use t instead of Z. If we use s p instead of , and the sample sizes are small, then we should use t instead of Z. The number of degrees of freedom is The number of degrees of freedom is df = df 1 + df 2 = n 1 + n 2 – 2. That is That is

16. Hypothesis Testing See Example 11.4, p. 647 – Comparing Two Headache Treatments. See Example 11.4, p. 647 – Comparing Two Headache Treatments. State the hypothesis. State the hypothesis. H 0 :  1 =  2 H 0 :  1 =  2 H 1 :  1 >  2 H 1 :  1 >  2 State the level of significance. State the level of significance.  = 0.05.  = 0.05.

17. The t Statistic, a.k.a. Our Second Bad Formula Compute the value of the test statistic. Compute the value of the test statistic. The test statistic is The test statistic is with df = n 1 + n 2 – 2.

18. Computations

19. Hypothesis Testing Calculate the p-value. Calculate the p-value. The number of degrees of freedom is The number of degrees of freedom is df = df 1 + df 2 = 18. p-value = P(t > 1.416) p-value = P(t > 1.416) = tcdf(1.416, E99, 18) = 0.0869.

20. Hypothesis Testing State the conclusion. State the conclusion. Since p-value > , we conclude that, Since p-value > , we conclude that, At the 5% level of significance, the data do not support the claim that Treatment 1 is more effective than Treatment 2.

21. Confidence Intervals Confidence intervals for  1 –  2 use the same theory. Confidence intervals for  1 –  2 use the same theory. The point estimate is  x 1 –  x 2. The point estimate is  x 1 –  x 2. The standard deviation of  x 1 –  x 2 is approximately s p. The standard deviation of  x 1 –  x 2 is approximately s p.

22. Confidence Intervals The confidence interval is The confidence interval isoror (  known, large samples) (  unknown, large samples) (  known, normal pops., small samples)

23. Confidence Intervals The choice depends on The choice depends on Whether  is known. Whether  is known. Whether the populations are normal. Whether the populations are normal. Whether the sample sizes are large. Whether the sample sizes are large.

24. Example Find a 95% confidence interval for  1 –  2 in Example 11.4. Find a 95% confidence interval for  1 –  2 in Example 11.4.  x 1 –  x 2 = 3.2.  x 1 –  x 2 = 3.2. s p = 5.052. s p = 5.052. Use t = 2.101. Use t = 2.101. The confidence interval is The confidence interval is 3.2  (2.101)(2.259) = 3.2  4.75.