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M562-HW8-8

# M562-HW8-8 - MTH/STA 562 Exercise 8.68 It is given that Y =...

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MTH/STA 562 Exercise 8.68 . It is given that Y = 26 : 6, S = 7 : 4, and n = 21. For a 95% con°dence level with n ° 1 = 20 degrees of freedom, t 0 : 025 = 2 : 086. Then a 95% con°dence interval for ° is Y ± t 0 : 025 S p n = 26 : 6 ± 2 : 086 7 : 4 p 21 = 26 : 6 ± 3 : 37 or (23 : 33 ; 29 : 97). Exercise 8.69 . 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 = 63 : 5111 : For a 95% con°dence level with n ° 1 = 9 degrees of freedom, t 0 : 025 = 2 : 262. Then a 95% con°dence interval for ° is Y ± t 0 : 025 S p n = 60 : 8 ± 2 : 262 r 63 : 5111 10 = 60 : 8 ± 5 : 701 or (55 : 099 ; 66 : 501). Exercise 8.70 . ( a ) It is given that Y = 419, S = 57, and n = 20. For a 90% con°dence level with n ° 1 = 19 degrees of freedom, t 0 : 05 = 1 : 729. Then a 90% con°dence interval for ° is Y ± t 0 : 05 S p n = 419 ± 1 : 729 57 p 20 = 419 ± 22 : 04 or (396 : 96 ; 441 : 04). ( b ) The con°dence interval does include 422. Thus, 422 is a believable value for ° at the 90% con°dence level. However, numbers such as 397, 410, 441,. for example, are also believable values for ° . ( c ) It is given that Y = 455, S = 69, and n = 20. For a 90% con°dence level with n ° 1 = 19 degrees of freedom, t 0 : 05 = 1 : 729. Then a 90% con°dence interval for ° is Y ± t 0 : 05 S p n = 455 ± 1 : 729 69 p 20 = 455 ± 26 : 67 or (428 : 33 ; 481 : 67). The con°dence interval does include 474. We would conclude, based on our 90% con°dence interval, that the true mean mathematics SAT score

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is not di±erent from 474. Exercise 8.71 . ( a ) It is given that Y 1 = 14 : 5, Y 2 = 11 : 1, S 1 = 3 : 92, S 2 = 3 : 98, and n 1 = n 2 = 10. Then S 2 p = ( n 1 ° 1) S 2 1 + ( n 2 ° 1) S 2 2 n 1 + n 2 ° 2 = 9 (3 : 92) 2 + 9 (3 : 98) 2 18 = 15 : 6034 : For a 95% con°dence level with n 1 + n 2 ° 2 = 18 degrees of freedom, t 0 : 025 = 2 : 101. Then a 95% con°dence interval for ° is ° Y 1 ° Y 2 ± ± t 0 : 025 S p r 1 n 1 + 1 n 2 = (14 : 5 ° 11 : 1) ± 2 : 101 s 15 : 6034 ² 1 10 + 1 10 ³ = 3 : 4 ± 3 : 7 or ( ° 0 : 3 ; 7 : 1). ( b ) It is given that Y 1 = 12 : 2, Y 2 = 11 : 5, S 1 = 3 : 49, S 2 = 4 : 95, and n 1 = n 2 = 10. Then S 2 p = ( n 1 ° 1) S 2 1 + ( n 2 ° 1) S 2 2 n 1 + n 2 ° 2 = 9 (3 : 49) 2 + 9 (4 : 95) 2 18 = 18 : 3413 : For a 90% con°dence level with n 1 + n 2 ° 2 = 18 degrees of freedom, t 0 : 05 = 1 : 734. Then a 90% con°dence interval for ° is ° Y 1 ° Y 2 ± ± t 0 : 05 S p r 1 n 1 + 1 n 2 = (12 : 2 ° 11 : 5) ± 1 : 734 s 18 : 3413 ² 1 10 + 1 10 ³ = 0 : 7 ± 3 : 32 or ( ° 2 : 62 ; 4 : 02).
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