1. Lesson #22 Inference for Two Independent Means

2. Two independent samples: Again interested in (  1 –  2 ) n1n1 n2n2 Use to estimate (  1 –  2 )

3. Two approaches: Equal variances - - assuming equal variances, - unequal variances, Calculate the pooled estimate of variance,

4. A 100(1-  )% confidence interval for (  1 –  2 ) is: To test H 0 : (  1 –  2 ) = 0, use the pooled t-test. The test statistic is: under H 0

5. A 100(1-  )% confidence interval for (  1 –  2 ) is: To test H 0 : (  1 –  2 ) = 0, use the test statistic is: under H 0 Unequal variances -

6. Reject H 0 if | t 0 | > t (236),.975 H 0 : (  1 –  2 ) = 0H 1 : (  1 –  2 )  0 n 1 + n 2 – 2 = 77 + 161 – 2 = 236 n 1 = 77= 0.098 s 1 = 0.026 n 2 = 161= 0.095 s 2 = 0.025  1.980

7. = (76)(0.026) 2 +(160)(0.025) 2 236 =.0006414  s p = 0.0253

8. = 0.098- 0.095.0006414 1 77 1 161 + 0.00351 = 0.855 Do not reject H 0 Not enough evidence to detect a difference in mean mineral content between babies born to smoking mothers and non-smoking mothers.

9.  (-0.004, 0.010) (0.098 – 0.095)  (1.980) (0.00351)