STUDENT t DISTRIBUTION

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In the previous chapter, we have used the central limit theorem to calculate con dence intervals and to carry out tests of hypotheses In noting that the ran dom variable ( )/( / n) follows a N(0, 1) distribution, we rely heavily on x the fact that the standard deviation of the population, , is known However, in many practical situations, both population parameters, and , are unknown In 1908, WG Gossett, writing under the pseudonym Student discovered the following: Theorem 102 The ratio of a N(0, 1) random variable divided by the square root of a chi-squared random variable divided by its degrees of freedom (say n) follows a Student t distribution with n degrees of freedom (tn ) 2 Since ( )/( / n) is a N(0, 1) random variable, and (n 1)s2 / 2 is a n 1 x variable, it follows that [( )/( / n)]/ (n 1)s2 / 2 (n 1) follows a tn 1 x

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Student t Distribution

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probability distribution But x / n (n 1)s2 2 (n 1) So we have a random variable that involves alone and values calculated from a sample It is far too simple to say that we simply replace by s in the central limit theorem; much more than that has transpired The sampling can arise from virtually any distribution due to the fact that the central limit theorem is almost immune to the underlying distribution We show two typical Student t distributions in Figure 108 (which was produced by Minitab)

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Distribution plot

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Density

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The curves have 4 and 20 degrees of freedom, respectively, and each appears to be normal-like Here is a table of values of v so that the probability the table value exceeds v is 005 The number of degrees of freedom is n

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Statistical Inference II: Continuous Probability Distributions II

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This is some evidence that the 005 point approaches that of the N(0, 1) distribution, 1645,but the degrees of freedom, depending on the sample size, remain crucial Again a computer is essential in doing calculations as the following example shows

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EXAMPLE 101

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A Sample

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A sample of size 14 showed that x = 298 and s2 = 123 Find the p-value for the test H0 : = 26 and Ha : = 26 / 298 26 Here we nd that t13 = 123/14 = 4797 and P (t13 > 4797) = 17439 10 4 , a rare event indeed

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TESTING THE RATIO OF VARIANCES: THE F DISTRIBUTION

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So far we have considered inferences from single samples, but we will soon turn to comparing two samples, possibly arising from two different populations So we now investigate comparing variances from two different samples We need the following theorem whose proof can be found in texts on mathematical statistics Theroem 103 The ratio of two independent chi-squared random variables, divided by their degrees of freedom, say n in the numerator and m in the denominator, follows the F distribution with n and m degrees of freedom

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2 We know from a previous section that (n 1)s2 / 2 is a n 1 random variable, so if we have two samples, with degrees of freedom n and m, then using the theorem, 2 2 ns1 s1 2 2 1 n 1 = 2 = F [n, m] 2 s2 ms2 2m 2 2 2

Figure 109 shows a graph of a typical F distribution, this with 7 and 9 degrees of freedom The graph also shows the upper and lower 25% points Now notice that the reciprocal of an F random variable is also an F random 1 variable but with the degrees of freedom interchanged, so F [n,m] = F [m, n] This fact can be used in nding critical values Suppose P[F [n, m] < v] = This is equivalent to P[1/F [n, m] > 1/v] or P[F [m, n] > 1/v] = So the reciprocal of the lower point on F [n, m] is the upper point on the F [m, n] distribution