Lecture 39 Section 11.5

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1 on Lecture 39 Section 11.5 Hampden-Sydney College Mon, Nov 10, 2008

2 Outline 1 on 2 3 on 4

3 on Exercise 11.27, page 715. A researcher was interested in comparing body weights for two strains of laboratory mice. She observed the following weights (in grams) of adult mice: Strain 1: Strain 2:

4 on Exercise 11.27, page 715. The researcher was interested in seeing whether these data provide evidence of a difference in mean body weight of the two strains. Assume that the data are the observed values of independent random samples that is, the two distributions are normal with common population standard deviations.

5 on Exercise 11.27, page 715. (a) State the appropriate null and alternative hypotheses H 0 and H 1. (b) Calculate the value of a test statistic for testing H 0 and H 1 and give the p-value. (c) State the results for a test with significance level Give your decision and a conclusion in terms of the mean body weight for the two strains.

6 on Solution (a) Let µ 1 = the average body weight in Strain 1. Let µ 2 = the average body weight in Strain 2. The hypotheses are H 0 : µ 1 = µ 2. H 1 : µ 1 µ 2.

7 on Solution (b) Under these circumstances, the test statistic is t = (x 1 x 2 ) 0 s p 1 n n 2 with 13 degrees of freedom, where (n 1 1)s 2 1 s p = + (n 2 1)s 2 2. n 1 + n 2 2

8 on Solution (b) Use to get the stats: Strain 1 Strain 2 x 1 = x 2 = 42.5 s 1 = s 2 = n 1 = 7 n 2 = 8 Then compute the value of s p : s p = =

9 on Solution (b) Now compute the value of t: t = = =

10 on Solution (b) Finally, compute the p-value: p-value = 2 tcdf(-e99,-2.578,13) = =

11 on Solution (c) At the 5% level of significance, we reject H 0. We conclude that the mean body weights of the two strains of mice are different.

12 on intervals for use the same theory as do the other confidence intervals. The point estimate is x 1 x 2. The standard deviation of x 1 x 2 is approximately s p 1 n n 2.

13 The confidence interval is either on or or (x 1 x 2 ) ± z α/2 σ 2 1 n 1 + σ2 2 n 2, (x 1 x 2 ) ± z α/2 s p 1 n n 2, (x 1 x 2 ) ± t α/2,df s p 1 n n 2, depending on the circumstances.

14 on The choice depends on Whether σ 1 and σ 2 are known and/or are assumed to be equal. Whether the populations are normal. Whether the sample sizes are large.

15 Example on Example ( interval for ) Find a 95% confidence interval for difference in the average time to recovery for the old and new drugs. New Drug (# 1) Old Drug (# 2) n 1 = 40 n 2 = 60 x 1 = 5.4 x 2 = 6.8 s 1 = 1.8 s 2 = 1.3

16 Example on Example ( interval for ) The formula is (x 1 x 2 ) ± z s 2 1 n 1 + s2 2 n 2. (We could use a pooled estimate for σ, but it is not necessary when using z.) We get 95% C.I. = ( ) ± = 1.4 ± (1.96)(0.3304) = 1.4 ±

17 Example on Example ( interval for ) Now find a 95% confidence interval for difference in the average time to recovery for the old and new drugs for smaller samples. New Drug (#1) Old Drug (#2) x 1 = 5.3 x 2 = 6.4 s 1 = 1.4 s 2 = 2.0 n 1 = 8 n 2 = 16 The researchers made QQ plots, which indicated that both populations are normal.

18 Example on Example ( interval for ) The formula is (x 1 x 2 ) ± t 0.025,22 s p 1 n n 2. We first compute the pooled estimate for σ. (n 1 1)s 2 1 s p = + (n 2 1)s 2 2 = n 1 + n 2 2 (We must use the pooled estimate for σ when using t.)

19 Example on Example ( interval for ) We get 95% C.I. = 1 ( ) ± t 0.025,22 (1.831) = 1.1 ± (2.074)(1.831)(0.4330) = 1.1 ±

20 The TI-83 - of Samples on TI-83 intervals for Press STAT > TESTS. Choose either 2-SampZInt or 2-SampTInt, depending on circumstances. Choose Data or Stats. Provide the information that is called for. 2-SampTTest will ask whether to use a pooled estimate of σ. Answer yes." Press Calculate.

21 Example on Example (TI-83 interval for ) Use to find a 95% confidence intervals for in each of the two previous examples.

22 on Read Section 11.4, pages (confidence intervals). Exercises 21, 23(omit c), 25(omit d), 26(c), 27(d)(i), page 713.

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