L21 - This following material covers Sec- tion 4.4:...

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Unformatted text preview: This following material covers Sec- tion 4.4: confidence intervals for proportions and variances. You are not responsible for the deriva- tion (shown in purple ) of these confidence intervals on a test or final exam. I have shown the details for those who are inter- ested. You should know how to use the tables, how to construct and interpret the confidence in- tervals, and the meanings of stan- dard error, margin of error. 1 Estimating a Variance Given . N ( , 2 ) population, random sample { X 1 ,...,X n } . Find C.I. on 2 Use fact that W = ( n- 1) S 2 2 has a chi-square distribution on n- 1 degrees of freedom. Upper percentage points 2 ( A,n- 1) satisfy upper tail area condition P ( W > 2 ( A,n- 1)) = A Given by Table C5 p 570. For confidence in- tervals A will be / 2 where typically = 0 . 05 but 2 ( A,n- 1) has the same meaning as an upper percentile of W for all values of A. 2 3 Deriving 100(1- ) C.I. on 2 . Start with statement P ( 2 (1- / 2 ,n- 1) W 2 ( / 2 ,n- 1)) = 1- Substitute W = ( n- 1) S 2 2 and rearrange. Get P ( n- 1) S 2 2 ( / 2 ,n- 1) 2 ( n- 1) S 2 2 (1- / 2 ,n- 1) ! = 1- 100(1- )% CI: ( n- 1) S 2 2 ( / 2 ,n- 1) , ( n- 1) S 2 2 (1- / 2 ,n- 1) ! To get C.I. on take square roots of end points. 4 Example 4.4-1 . n = 16 , x = 4 . 3 ,s = 0 . 6 . 95% CI on 2 is ( n- 1) S 2 2 ( / 2 ,n- 1) , ( n- 1) S 2 2 (1- / 2 ,n- 1) ! = 15(0 . 6) 2 2 (0 . 025 , 15) , 15(0 . 6) 2 2 (1- . 025 , 15) ! = 15(0 . 6) 2 2 (0 . 025 , 15) , 15(0 . 6) 2 2 (1- . 025 , 15) ! = 15(0 . 6) 2 27 . 488 , 15(0 . 6) 2 6 . 262 ! =(0 . 20 , . 86) 95% CI on 2 is ( . 20 , . 86) = (0 . 0447 , . 929) 5 Comparing Two Variances Independent samples of sizes n 1 ,n 2 from two normal populations, variances 1 , 2 2 respectively.respectively....
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L21 - This following material covers Sec- tion 4.4:...

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