The Chi-Square Distributions
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1 MATH 03 The Chi-Square Distributions Dr. Neal, Spring 009 The chi-square distributions can be used in statistics to analyze the standard deviation of a normally distributed measurement and to test the goodness of fit of various population models on a set of data. A chi-square distribution is based on a parameter known as the degrees of freedom n, where n is an integer greater than or equal to 1. Such a random variable is denoted by X ~ (n). The (n) distribution is defined to be the sum of the squares of n independent standard normal distributions. For example, suppose X 1,..., X n are independent normally distributed measurements having mean i and standard deviation i for i = 1,..., n. These measurements could be the heights or IQ scores of various groups of people. By subtracting the mean and then dividing by the standard deviation, we convert each measurement into a standard normal distribution: Z i = X i i ~ N(0, 1), for 1 i n. i So Z 1 ~ N(0, 1) and its distribution graph will be the common bell-shaped curve which is symmetric about the origin. Then Z 1 ~ (1). Its plot will consist of positive values concentrated near the origin, and it will have mean 1 and variance. The standard normal distribution (1) distribution ()distribution (n) distribution By standardizing, squaring, and summing random measurements from the respective normal populations, we obtain a chi-square distribution with n degrees of freedom: X (n) = X X n n = Z 1 + Z Zn. 1 n The distribution graphs for n 3 are skewed bell-shaped curves, defined on [0, ), with increasingly larger values of x as the point at which the graph obtains its maximum. The mean is now n, the variance is n, and the standard deviation is n. For n 3, the maximum (mode) occurs when x = n. X ~ (n) = Z1 + Z Zn Mean = n Variance = n Standard Deviation = n Mode = n (for n 3)
2 Dr. Neal, Spring 009 The theoretical distribution curve is given by f (x) = C n x n/ 1 e x/, for x 0, where C n is a constant that depends on n given by 1 n/ n 1! C n = (n )/ n 1! (n 1)! π for neven for nodd. A chi-square curve can be plotted using the built-in pdf( command from the DISTR menu. For example, to graph the (10) curve, enter pdf( X,10) into the Y= screen. To compute P(a X b) for X ~ (n), enter cdf(a, b, n) or Shade (a, b, n). Example 1. Let X ~ (10). (a) Where does the maximum of the curve occur? (b) Compute P(6 X 10). Is there symmetry at the outer tails; i.e., does P(0 X 6) = P(X 10)? (c) Find the left and right bounds that contain 90% of the distribution. Solution. (a) For X ~ (10), the maximum (mode) occurs when x = n = 8. (b) From the TI output, we see that P(6 X 10) Also, the left-tail is P(0 X 6) , and the right-tail is P(X 10) So the two tails outside of the inner region 6 X 10 are not symmetric. For there to be 90% in the middle of the distribution, we must have 5% at each tail. The values where these occur (chi-square scores) can be found with the table on the next page. In this case, the values are about and
3 Dr. Neal, Spring 009 Left and Right Chi Square Scores for 80%, 90%, 95%, and 98% intervals. (L = Prob. of Left Tail, R = Prob. of Right Tail) d.f. L L L L R R R R
4 Dr. Neal, Spring 009 Theorems I. Let { x 1, x,..., x n } denote the collection of all random samples of size n from n normally distributed measurements having variance. Let S 1 = (x n 1 i x ) be i=1 the distribution of all possible sample variances. Then (n 1)S is a (n 1) distribution. Thus with a normally distributed measurement, we can evaluate P(a S b) by P(a S b) = P(a S b ) (n 1)a (n 1)S (n 1)b = P (n 1)a = P (n 1) (n 1)b provided is known. II. Let S be the sample variance from a random sample of size n of a normally distributed measurement having variance. A confidence interval for, with level of confidence r = 1, is given by (n 1)S R (n 1)S L, where L and R are the left and right bounds of the probability in the middle. A confidence interval for is (n 1) distribution that give r (n 1)S R (n 1)S L. III. To test the null hypothesis H 0 : = M for a normally distributed measurement, we obtain the sample deviation S from a random sample of size n. The test statistic is then (n 1) S (n 1)S x = = M which is compared with the (n 1) distribution. Compute the (left-tail) P -value P (n 1) x (right-tail) P -value P ( ) for the alternative H a : < M, and compute the ( (n 1) x ) for the alternative H a : > M.
5 Dr. Neal, Spring 009 Example. Random samples of size 46 are taken from a measurement that is N(100,15). What is P(13 S 17)? Example 3. From a normally distributed measurement, a sample of size 0 yields S = Find a 98% confidence interval for the true standard deviation. Example 4. From a normally distributed measurement, a sample of size 5 yields a sample deviation of Is there evidence to reject the hypothesis H 0 : = 15? Solutions Example : P(13 S 17) = P(13 S 17 ) (n 1)13 = P = P (n 1)S (n 1) P 33.8 (45) 57.8 (n 1) ( ) (using cdf(33.8, 57.8, 45) ) (n 1)S Example 3: R or (n 1)S L ; hence, , Example 4: For S = 13.96, we use the alternative H a : < 15. The test statistic is x = (n 1) S = 15 = ~ (n 1) = (4) ( ) ( and P (4) cdf(0, , 4). If = 15 were true, then there is still a % chance of obtaining a sample deviation of or lower with a sample of size 5. There is not enough evidence to reject H 0.
6 Dr. Neal, Spring 009 Exercises 1. Let X ~ (15). Find (a) P(13 X 17), (b) P(X < 13) and (c) P(X > 17). Show a graph for each. (d) Find the bounds that contain 95% of the distribution.. Adult heights are found to be normally distributed with mean = 68 inches and standard deviation = 3.5 inches. Suppose various random samples of size n = 6 are collected. Compute P(.8 S 4.). 3. From a normally distributed measurement, a sample of size 5 yields a sample deviation of Find a 95% confidence interval for the true standard deviation. 4. From a normally distributed measurement, a sample of size 16 yields S = 4.6. Is there evidence to reject the hypothesis H 0 : = 3? Answers: 1. (a) (b) (c) (d) L = 6.6 and R = P 3. Use (5) = P 16 (5) 36 ( ) to obtain Test stat = 30.46, P (15) ( ) If = 3 were true, then there is only a 1.1% chance of getting an S of 4.6 or higher with a sample of size 16. Can reject H 0 in favor of H a : > 3.
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