# X n should be to μ mathematically the central limit

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X n should be to μ . Mathematically, the central limit theorem says that n ( ¯ X n - μ ) d N (0 , σ 2 ). We cannot know the exact value of σ 2 merely by observing our sample X 1 , . . . , X n . But we can estimate it with the sample variance ˆ σ 2 n , defined as ˆ σ 2 n = 1 n n X i =1 ( X i - ¯ X n ) 2 . Some people prefer to define the sample variance with a divisor of n - 1 instead of n , but this makes no difference to us, since we will be looking for limiting results as n → ∞ . With a little algebra, we can see that ˆ σ 2 n = 1 n n X i =1 X 2 i - ¯ X 2 n . 16

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The law of large numbers tells us that 1 n n i =1 X 2 i p E ( X 2 ) as n → ∞ . It also tells us that ¯ X n p E ( X ), and so the continuous mapping theorem implies that - ¯ X 2 n p - E ( X ) 2 . Therefore, applying part 1 of Slutsky’s theorem, we find that ˆ σ 2 n p E ( X 2 ) - E ( X ) 2 = Var( X ) = σ 2 as n → ∞ . In other words, the sample variance ˆ σ 2 n provides a consistent estimate of the population variance σ 2 . Since the square root function is continuous, the continuous mapping theorem implies that ˆ σ n p σ . Combining this result with part 3 of Slutsky’s theorem and the fact that n ( ¯ X n - μ ) d N (0 , σ 2 ), we find that ¯ X n - μ ˆ σ n / n = n ( ¯ X n - μ ) ˆ σ n d N (0 , σ 2 ) σ = N (0 , 1) , provided that σ > 0 (which will be true unless every X i takes the same value). We have shown that the distribution of the ratio ¯ X n - μ ˆ σ n / n approximates the N (0 , 1) distribution in large samples. This fact can be used to test hypotheses about the value of μ , or to form confidence intervals for μ . Suppose we wish to test the hypothesis that μ = μ 0 . Here, μ is the true expected value of X , which we do not know, and μ 0 is a conjectured value of μ . We can calculate the t -statistic t n = ¯ X n - μ 0 ˆ σ n / n from our sample X 1 , . . . , X n . If the hypothesis μ = μ 0 is true, then our earlier argument shows that t n has the N (0 , 1) distribution in large samples. Therefore, it should lie between - 1 . 96 and 1 . 96 with probability approximately equal to 0 . 95. On the other hand, if μ 6 = μ 0 , we have no reason to believe that t n will lie in this range, and in fact it can be shown that it will be greater than 1.96 in absolute value with probability approaching one as n → ∞ . Therefore, if we calculate t n and find that it does not lie between -1.96 and 1.96, we reject the hypothesis that μ = μ 0 . If our hypothesis μ = μ 0 is true, there is an approximately 5% chance that we will incorrectly reject it. This is usually deemed an acceptable rate of error for applied work in the social sciences. The approximation ¯ X n - μ ˆ σ n / n N (0 , 1) can also be used to form large sample confidence intervals for the unknown quantity μ . A standard normal random vari- able lies between -1.96 and 1.96 with probability 0.95. Recalling the definition of convergence in distribution, we can see that lim n →∞ P - 1 . 96 ¯ X n - μ ˆ σ n / n 1 . 96 = 0 . 95 .
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