CHAPTER 5 Sampling Theory

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2, s2 10.8 [from Problem 5.1(b)]. For s2 1 ns2 1 s2 (2)(7.2) 10.8 1.33 7.2, we have

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(a) We have n

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According to Theorem 5-6, x2 nS2 >s2 2S2 >10.8 has the chi-square distribution with 1 degree of freedom. From the table in Appendix E it follows that P(S2 s2) 1 P(x2 1.33) 0.25

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Therefore, we would expect about 25% of the samples, or 6, to have variances greater than 7.2. (b) From Problem 5.18 we find by counting that there are actually 6 variances greater than 7.2, so that there is agreement.

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Case where population variance is unknown 5.24. Prove Theorem 5-7, page 159.

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Let Y nS2 , n n 1. Then since the Xj are normally distributed with mean m and s2 s> !n # variance s2, we know (Problem 5.21) that X is normally distributed with mean m and variance s2 >n, so that y is normally distributed with mean 0 and variance 1. Also, from Theorem 5-6, page 158, or Problem 5.22, Z is chi square distributed with n n 1 degrees of freedom. Furthermore, it can be shown that Y and Z are independent. It follows from Theorem 4-6, page 116, that , Z T has the t distribution with n Y !Z>n # X S> !n m 1 # X

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S > !n

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1 degrees of freedom.

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5.25. According to the table of Student s t distribution for 1 degree of freedom (Appendix D), we have P( 1.376 T 1.376) 0.60. Check whether this is confirmed by the results obtained in Problem 5.1.

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# From the values of X in (1) on page 155, and the values of S2 in Problem 5.18(a), we obtain the following # values for T (X m)>(S> !1): ` 7.0 1.0 0.33 0.11 7.0 ` 1.0 0.20 0.25 1.0 1.0 c 1.0 1.0 0.33 0.20 1.0 ` 2.33 0.11 0.25 1.0 2.33 `

There are actually 16 values for which 1.376 T 1.376 whereas we would expect (0.60) (25) 15. This is not too bad considering the small amount of data involved. This method of sampling was in fact the way Student originally obtained the t distribution.

Sampling distribution of ratios of variances 5.26. Prove Theorem 5-8, page 159.

Denote the samples of sizes m and n by X1, . . . , Xm and Y1, . . . ,Yn, respectively. Then the sample variances are given by S2 1 # # where X, Y are the sample means. 1 m a (Xj

j 1 m

# X)2,

S2 2

1 n a (Yj

# Y)2

CHAPTER 5 Sampling Theory

Now from Theorem 5-6, page 158, we know that mS2 >s2 and nS2 >s2 are chi-square distributed with m 1 1 2 2 and n 1 degrees of freedom, respectively. Therefore, from Theorem 4-7, page 117, it follows that mS2 >(m 1 nS2 >(n 2 1)s2 1 1)s2 2 S2 >s2 1 1

S2 >s2 2 2

has the F distribution with m

1, n

1 degrees of freedom.

5.27. Two samples of sizes 8 and 10 are drawn from two normally distributed populations having variances 20 and 36, respectively. Find the probability that the variance of the first sample is more than twice the variance of the second.

We have m 8, n 10, s2 1 20, s2 2 36. Therefore, 8S2 >(7)(20) 1 10S2 >(9)(36) 2 S2 1 S2 2

The number of degrees of freedom for numerator and denominator are n1 m 1 7, n2 n 1 9. Now if S2 is more than twice S2, i.e., S2 2S2, then F 3.70. Referring to the tables in Appendix F, we see that the 1 2 1 2 probability is less than 0.05 but more than 0.01. For exact values we need a more extensive tabulation of the F distribution.

Frequency distributions 5.28. In Table 5-4 the weights of 40 male students at State University are recorded to the nearest pound. Construct a frequency distribution.

Table 5-4 138 146 168 146 161 164 158 126 173 145 150 140 138 142 135 132 147 176 147 142 144 136 163 135 150 125 148 119 153 156 149 152 154 140 145 157 144 165 135 128

The largest weight is 176 lb, and the smallest weight is 119 lb, so that the range is 176 If 5 class intervals are used, the class interval size is 57>5 If 20 class intervals are used, the class interval size is 57>20 11 approximately. 3 approximately.