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尋找統計高手~~
1.Inasampleof200individuals,120indicatedtheyareDemocrats.Developa95%confidenceintervalfortheproportionofpeopleinthepopulationwhoareDemocrats.2.Arandomsampleof121checkingaccountsatabankshowedanaveragedailybalanceof$300andastandarddeviationof$44.Developa95%...顯示更多1.Inasampleof200individuals,120indicatedtheyareDemocrats.Developa95%confidenceintervalfortheproportionofpeopleinthepopulationwhoareDemocrats.2.Arandomsampleof121checkingaccountsatabankshowedanaveragedailybalanceof$300andastandarddeviationof$44.Developa95%confidenceintervalestimateforthemeanofthepopulation.更新:請問能否詳細解說解題步驟呢?很感謝~
1. Z=[phat-p]/根號{phat*(1-phat)/n)}~N(0,1) phat=120/200=0.6；n=200；alpha=1-0.95=0.05 95% confident interval for the proportion of people in the population who are Democrats is... [phat－Z2/alpha*根號{phat*(1-phat)/n)}, phat＋Z2/alpha*根號{phat*(1-phat)/n)}] =[0.6－Z0.025*根號{0.6*(1-0.6)/200)}, 0.6＋Z0.025*根號{0.6*(1-0.6)/200)}] =[0.6－1.96*根號{0.6*(1-0.6)/200)}, 0.6＋1.96*根號{0.6*(1-0.6)/200)}] =[0.5321, 0.6679] 2. n=121；Xbar=300；Sx=44；alpha=1-0.95=0.05 n夠大，根據中央極限定理Xbar，可用Z分配來找臨界值，則 95% confidence interval estimate for the mean of the population is... [Xbar－Z2/alpha*Sx/根號{n}, Xbar＋Z2/alpha*Sx/根號{n}] =[300－Z0.025*44/根號{121}, 300＋Z0.025*44/根號{121}] =[300－1.96*44/根號{121}, 300＋1.96*44/根號{121}] =[292.16, 307.84]
1.樣本數200人,有120的人說自己是民主黨員.請建立該人口中民主黨員之95%信賴區間.2.某銀行之支票帳戶每日平均存款餘額為300元,標準差為44元,從中隨機選取121個樣本,建立該人口(存款)平均值之95%信賴區間估計值.
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https://tw.answers.yahoo.com/question/index?qid=20070320000010KK01431