mont4e_sm_ch09_mind

# mont4e_sm_ch09_mind - Mind Expanding Exercises 9-126 The...

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Unformatted text preview: Mind Expanding Exercises 9-126 The parameter of interest is the true,. H0 : = 0 H1 0 a) Reject H0 if z0 < -z- or z0 > z 9-127 X - 0 X - 0 X - 0 X - 0 <- = | = 0 ) + P( > | = 0 ) P / n / n / n / n P( z0 < - z - ) + P( z0 > z ) = (- z - ) + 1 - ( z ) P( = (( - )) + (1 - (1 - )) = b) = P(z X z when 1 = 0 + d ) or = P( - z - < Z 0 < z | 1 = 0 + ) x-0 = P( - z - < < z | 1 = 0 + ) 2 /n = P( - z - - = ( z - 2 /n 2 /n < Z < z - ) ) - ( - z - - 2 /n ) 2 /n 9-128 1) The parameter of interest is the true mean number of open circuits, . 2) H0 : = 2 3) H1 : > 2 4) = 0.05 5) Since n>30 we can use the normal distribution z0 = X - /n 6) Reject H0 if z0 > z where z0.05 =1.65 7) x = 1038/500=2.076 n = 500 9-65 z0 = 2.076 - 2 2 / 500 = 1.202 8) Because 1.202 < 1.65, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true mean number of open circuits is greater than 2 at = 0.01 9-129 a) 1) The parameter of interest is the true standard deviation of the golf ball distance . 2) H0: = 10 3) H1: < 10 4) =0.05 5) Because n > 30 we can use the normal distribution z0 = S -0 2 0 /(2n) 6) Reject H0 if z0 < z where z0.05 =-1.65 7) s = 13.41 n = 100 z0 = 13.41 - 10 10 2 /(200) = 4.82 8) Since 4.82 > -1.65, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true standard deviation is less than 10 at = 0.05 b) 95% percentile: = + 1.645 ^ 95% percentile estimator: = X + 1.645S ^ SE ( ) 2 / n + 1.645 2 2 /(2n) From the independence The statistic S can be used as an estimator for in the standard error formula. c) 1) The parameter of interest is the true 95th percentile of the golf ball distance . 2) H0: = 285 3) H1: < 285 4) = 0.05 5) Since n > 30 we can use the normal distribution z0 = ^ - 0 ^ ^ SE ( ) 282.36 - 285 13.412 / 100 + 1.645 213.412 / 200 6) Reject H0 if z0 < -1.65 7) ^ = 282.36 , s = 13.41, n = 100 z0 = = -1.283 8) Because -1.283 > -1.65, do not reject the null hypothesis. There is not sufficient evidence to indicate that the true is less than 285 at = 0.05 9-130 1) The parameter of interest is the true mean number of open circuits, . 2) H0 : = 0 3) H1 : 0 4) = 0.05 5) test statistic 9-66 2 0 = 2 X i - 0 i =1 n 2 X i i =1 2 0 2 a / 2, 2 n n 6) Reject H0 if 7) compute > n i =1 or 2 0 < 12- a / 2, 2 n 2 X i and plug into 2 0 = 2 X i - 0 i =1 n 2 X i i =1 n 8) make conclusions alternative hypotheses 1) H0 : = 0 H1 : > 0 2 2 0 > a,2n Reject H0 if 2) H0 : = 0 H1 : < 0 Reject H0 if 2 2 0 < a,2n 9-67 ...
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## This note was uploaded on 10/12/2009 for the course IND E 315 taught by Professor Kailashkapur during the Spring '09 term at University of Washington.

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