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ch7_slides - INFERENCES ABOUT POPULATION VARIANCES...

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INFERENCES ABOUT POPULATION VARIANCES Estimation and Tests for a Single Population Variance Recall that the sample variance s 2 = ( y - ¯ y ) 2 / ( n - 1) is the point estimate of σ 2 . For tests and confidence intervals about σ 2 we use the fact that the sampling random variable ( n - 1) S 2 2 = χ 2 has the Chi-square Distribution with n - 1 degrees of freedom or df , when the sample is from a normal population. 1
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The chi-square distribution is nonsymmetric. Like the Student’s t distribution, there is a different curve for each sample size n i.e., for each value of df . Percentiles of the chi-square distribution are given in Table 7. Plots of the chi-square distribution for df = 5 , 15 , and 30 are shown in Fig. 7.3 The distribution appear to be more skewed for smaller values of df and become more symmetric as df increases. Because the chi-square distribution is nonsymmetric, the percentiles for probabilities at both ends needs to be tabulated. 2
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A 100(1- α ) % Confidence Interval for σ 2 This has the form: ( n - 1) s 2 χ 2 U < σ 2 < ( n - 1) s 2 χ 2 L where Since df = n - 1 look up χ 2 n - 1 percentiles. χ 2 L is the lower-tail value with area α/ 2 to the left. χ 2 U is the upper-tail value with area α/ 2 to the right. The confidence interval for the standard deviation σ is found by taking square roots of both end points of above. 4
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Example 7.1 The normal probability plot of the data (see text book) appear to show that the sample is from a normal distribution. From the data n = 30 , ¯ y = 500 . 453 , s = 3 . 433 were calculated. A 99% confidence interval for σ 2 is computed as follows: Since α = . 1 , α/ 2 = . 005 and 1 - α/ 2 = . 995 , we compute χ 2 L = χ 2 0 . 995 , 29 = 13 . 12 χ 2 U = χ 2 0 . 005 , 29 = 52 . 34 Thus the endpoints for the required C.I. are, respectively,: ( n - 1) s 2 χ 2 U = 29(3 . 433 2 ) 52 . 34
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