stat exam2

stat exam2 - With H0: t= 1 2 = 0 vs. H : 1 2 0 , we will...

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With H 0 : 0 2 1 = - μ vs. H a : 0 2 1 - , we will reject H 0 if α < - value p . ( 29 ( 29 ( 29 6 8 . 6 4 5 2 5 240 . 2 6 164 . 2 5 240 . 6 164 . 2 2 2 2 = + + = ν , and the test statistic 17 . 6 1265 . 78 . 95 . 21 73 . 22 5 240 . 6 164 . 2 2 = = + - = t leads to a p-value of 2[ P(t > 6.17)] < 2(.0005) =.001, which is less than most reasonable s ' , so we reject H 0 and conclude that there is a difference in the densities of the two brick types./ We want a 95% confidence interval for 2 1 - . 262 . 2 9 , 025 . = t , so the interval is ( 29 ( 29 80 . 6 , 40 . 20 007 . 3 262 . 2 6 . 13 - - = ± - . Because the interval is so wide, it does not appear that precise information is available./ Let = 1 the true average strength for wire-brushing preparation and let = 2 the average strength for hand-chisel preparation. Since we are concerned about any possible difference between the two means, a two-sided test is appropriate. We test 0 : 2 1 0 = - H vs. 0 : 2 1 - a H . We need the degrees of freedom to find the rejection region: ( 29 ( 29 ( 29 33 . 14 1632 . 0039 . 3964 . 2 11 11 2 5 01 . 4 2 12 58 . 1 2 12 01 . 4 12 58 . 1 2 2 2 2 = + = + + = , which we round down to 14, so we reject H 0 if 145 . 2 14 , 025 . = t t . The test statistic is ( 29 159 . 3 2442 . 1 93 . 3 13 . 23 20 . 19 12 01 . 4 12 58 . 1 2 2 - = - = + - = t , which is 145 . 2 - , so we reject H 0 and conclude that there does appear to be a difference between the two population average strengths. / a. 95% upper confidence bound: x + t .05,65-1 SE = 13.4 + 1.671(2.05) = 16.83 seconds Comment : The above is constructed using a one-sample CI b. Let μ 1 and μ 2 represent the true average time spent by blackbirds at the experimental and natural locations, respectively. We wish to test H 0 : μ 1 μ 2 = 0 v. H a : μ 1 μ 2 > 0. The relevant test statistic is 2 2 76 . 1 05 . 2 7 . 9 4 . 13 + - = t = 1.37, with estimated df = 49 76 . 1 64 05 . 2 ) 76 . 1 05 . 2 ( 4 4 2 2 2 + + ≈ 112.9. Rounding to t = 1.4 and df = 120, the tabulated P - value is very roughly .082. Hence, at the 5% significance level, we fail to reject the null hypothesis. The true average time spent by blackbirds at the experimental location is not statistically significantly higher than at the natural location. c. 95% CI for silvereyes’ average time – blackbirds’ average time at the natural location: (38.4 – 9.7) ± (2.00) 2 2 06 . 5 76 . 1 + = (17.96 sec, 39.44 sec). The t -value 2.00 is based on estimated df = 55. Comment: Notice how standard error of the mean (not standard deviation of individual measurements) was reported in the data and how calculations in the problem were different because of this./ a. Following the usual format for most confidence intervals: statistic ± (critical value)(standard error), a pooled variance confidence interval for the difference between two means is ( 29 n m p n m s t y x 1 1 2 , 2 / + ± - - + . b. The sample means and standard deviations of the two samples are 90 . 13 = x , 225 . 1 1 = s , 20 . 12 = y , 010 . 1 2 = s . The pooled variance estimate is = 2 p s ( 29 ( 29 2 2 2 2 2 1 010 . 1 2 4 4 1 4 225 . 1 2 4 4 1 4 2 1 2 1 - + - + - + - = - + - + - + - s n m n s n m m 260 . 1 = , so 1227 . 1 = p s . With df = m+n-2 = 6 for this interval, 447 . 2 6 , 025 .
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stat exam2 - With H0: t= 1 2 = 0 vs. H : 1 2 0 , we will...

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