Statistics Midterm Exam Solutions: Hypothesis Testing and Confidence Intervals, Exams of Statistics

Download Statistics Midterm Exam Solutions: Hypothesis Testing and Confidence Intervals and more Exams Statistics in PDF only on Docsity! Midterm Exam Solution 1.(i) 24:,24:0 ≠= µµ aHH 186.1 90/32. 2496.23 / −= − = − = ns Xt µ , so we can not reject the null hypothesis at significance level 0.05. 987.1|| 89,025. =< tt (ii) The 99% CI for µ is )0487.24,8713.23(089.96.2390/32.*63.296.23/89,005. =±=±=± nstX (iii) ))11(,0(~ 2σ n NXX +− , an estimate of the proportion P(X < 23.76) is 2671.)6216.( ) 90/1132. 96.2376.23() /11 96.2376.23()96.2376.23()76.23(ˆ =−<= + − <= + − <=−<−=< TP TP ns TPXXPXP 2. (a) FMaFM HH µµµµ ≠= :,:0 Since we assume equal variance, we need to calculate the pooled standard error. 64.83 279145 9294.82*7802.84*144 2 )1()1( 22 21 2 2 2 1 = −+ + = −+ −+− = nn SnSn s YXp 97.1 .9968.3 64.83 0253.5657724.611 222,025. 79 1 145 111 21 => = + − = + − = tt s YXt nnp So is rejected at significance level 0.05. 0H (b) 10:,10:0 >−=− FMaFM HH µµµµ 34.2 .1418.3 64.83 100253.5657724.61110 222,01. 79 1 145 111 21 => = + −− = + −− = tt s YXt nn So is rejected at significance level 0.01. 0H (c) The overall sample average is ),(~ 21 2 21 21 nn N nn YnXn ++ + σµ The formula for 95% CI of the overall population mean is therefore 21 222,025. 21 21 1 nn st nn YnXn p + ± + + From the particular sample, the 95% CI of the overall population mean is )606.2949,584.2765( 0092.112857.595 79145/64.83*97.1 79145 0253.565*797724.611*145 1 21 222,025. 21 21 = ±= +± + + = + ± + + nn st nn YnXn p (d) 594:,594:0 >= µµ aHH Under , 0H 222 21 0 21 21 ~ 1 T nn s nn YnXn p + − + + µ So the test statistics is 65.1 085. 224/64.83 5942857.595 )/(1 222,05. 21 0 21 21 =< = − = + − + + = tt nns nn YnXn t p µ So we cannot reject the null hypothesis at significance level 0.05.