mont4e_sm_ch08_supplemental

# mont4e_sm_ch08_supplemental - Supplemental Exercises 8-77...

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Supplemental Exercises 8-77 Where α = + 2 1 . Let 05 . 0 = Interval for 025 . 0 2 / 2 1 = = = The confidence level for n x n x / 96 . 1 / 96 . 1 σ μσ + is determined by the by the value of z 0 which is 1.96. From Table III, we find Φ (1.96) = P(Z<1.96) = 0.975 and the confidence level is 95%. Interval for 04 . 0 , 01 . 0 2 1 = = The confidence interval is n x n x / 75 . 1 / 33 . 2 + , the confidence level is the same since 05 . 0 = . The symmetric interval does not affect the level of significance; however, it does affect the length. The symmetric interval is shorter in length. 8-78 μ = 50 σ unknown a) n = 16 x = 52 s = 1.5 1 16 / 8 50 52 = = o t The P -value for t 0 = 1, degrees of freedom = 15, is between 0.1 and 0.25. Thus we would conclude that the results are not very unusual. b) n = 30 37 . 1 30 / 8 50 52 = = o t The P-value for t 0 = 1.37, degrees of freedom = 29, is between 0.05 and 0.1. Thus we conclude that the results are somewhat unusual. c) n = 100 (with n > 30, the standard normal table can be used for this problem) 5 . 2 100 / 8 50 52 = = o z The P-value for z 0 = 2.5, is 0.00621. Thus we conclude that the results are very unusual. d) For constant values of x and s, increasing only the sample size, we see that the standard error of X decreases and consequently a sample mean value of 52 when the true mean is 50 is more unusual for the larger sample sizes. 8-79 5 , 50 2 = = σμ a) For find or 16 = n ) 44 . 7 ( 2 s P ) 56 . 2 ( 2 s P () 10 . 0 32 . 22 05 . 0 5 ) 44 . 7 ( 15 ) 44 . 7 ( 2 15 2 2 15 2 = = χ P P S P Using Minitab =0.0997 ) 44 . 7 ( 2 S P = = 68 . 7 05 . 0 5 ) 56 . 2 ( 15 ) 56 . 2 ( 2 15 2 15 2 P P S P 0.10 Using Minitab =0.064 ) 56 . 2 ( 2 S P 8-27

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b) For find or 30 = n ) 44 . 7 ( 2 S P ) 56 . 2 ( 2 S P () 05 . 0 15 . 43 025 . 0 5 ) 44 . 7 ( 29 ) 44 . 7 ( 2 29 2 29 2 = = χ P P S P Using Minitab = 0.044 ) 44 . 7 ( 2 S P 025 . 0 85 . 14 01 . 0 5 ) 56 . 2 ( 29 ) 56 . 2 ( 2 29 2 29 2 = = P P S P Using Minitab = 0.014. ) 56 . 2 ( 2 S P c) For or 71 = n ) 44 . 7 ( 2 s P ) 56 . 2 ( 2 s P 01 . 0 16 . 104 005 . 0 5 ) 44 . 7 ( 70 ) 44 . 7 ( 2 70 2 70 2 = = P P S P Using Minitab =0.0051 ) 44 . 7 ( 2 S P 005 . 0 84 . 35 5 ) 56 . 2 ( 70 ) 56 . 2 ( 2 70 2 70 2 = = P P S P Using Minitab < 0.001 ) 56 . 2 ( 2 S P d) The probabilities get smaller as n increases. As n increases, the sample variance should approach the population variance; therefore, the likelihood of obtaining a sample variance much larger than the population variance will decrease. e) The probabilities get smaller as n increases. As n increases, the sample variance should approach the population variance; therefore, the likelihood of obtaining a sample variance much smaller than the population variance will decrease. 8-80 a) The data appear to follow a normal distribution based on the normal probability plot since the data fall along a straight line. b) It is important to check for normality of the distribution underlying the sample data since the confidence intervals to be constructed should have the assumption of normality for the results to be reliable (especially since the sample size is less than 30 and the central limit theorem does not apply).
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mont4e_sm_ch08_supplemental - Supplemental Exercises 8-77...

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