Lesson #22 Inference for Two Independent Means. Two independent samples: Again interested in (  1 –  2 ) n1n1 n2n2 Use to estimate (  1 –  2 )

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Presentation transcript:

Lesson #22 Inference for Two Independent Means

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

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

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

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 -

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

= (76)(0.026) 2 +(160)(0.025) =  s p =

= = 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.

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