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Mth562-HW8-9

# Mth562-HW8-9 - MTH/STA 562 n X i=1 n X i=1 Exercise 8.81...

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MTH/STA 562 Exercise 8.81 . Calculate n X i =1 Y i = 514 : 4 and n X i =1 Y 2 i = 44 ; 103 : 74 with n = 6. Then Y = 1 n n X i =1 Y i = 514 : 4 6 = 85 : 73 and S 2 = 1 n ° 1 n X i =1 Y 2 i ° n Y 2 ! = 44 ; 103 : 74 ° 6 (85 : 73) 2 5 = 0 : 502667 : With con°dence coe±cient 1 ° ° = 0 : 90 and n ° 1 = 5 degrees of freedom, we read from Table 5 that ± 2 1 ° °= 2 = ± 2 0 : 95 = 1 : 145476 and ± 2 °= 2 = ± 2 0 : 05 = 11 : 0705. Hence, the 90% con°dence interval for ² 2 is 5 (0 : 502667) 11 : 0705 < ² 2 < 5 (0 : 502667) 1 : 145476 or 0 : 227 < ² 2 < 2 : 194. Exercise 8.82 . Calculate n X i =1 Y i = 608 and n X i =1 Y 2 i = 37 ; 538 with n = 10. Then Y = 1 n n X i =1 Y i = 608 10 = 60 : 8 and S 2 = 1 n ° 1 n X i =1 Y 2 i ° n Y 2 ! = 37 ; 538 ° 10 (60 : 8) 2 9 = 571 : 6 : With con°dence coe±cient 1 ° ° = 0 : 90 and n ° 1 = 9 degrees of freedom, we read from Table 5 that ± 2 1 ° °= 2 = ± 2 0 : 95 = 3 : 32511 and ± 2 °= 2 = ± 2 0 : 05 = 16 : 9190. Hence, the 90% con°dence interval for ² 2 is 9 (571 : 6) 11 : 0705 < ² 2 < 9 (571 : 6) 16 : 9190 or 33 : 785 < ² 2 < 171 : 90. Exercise 8.83 . ( a ) Find a number ± 2 ° with n ° 1 degrees of freedom such that P ( ( n ° 1) S 2 ² 2 ± ± 2 ° ) = 1 ° °: Then P ( ( n ° 1) S 2 ± 2 ° ± ² 2 ) = 1 ° °:

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Hence, °1 ; ( n ° 1) S 2 ± 2 ° # is a 100 (1 ° °
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Mth562-HW8-9 - MTH/STA 562 n X i=1 n X i=1 Exercise 8.81...

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