STATS 13 Study Guide - Final Guide: Confidence Interval, Cheddar Cheese, Analysis Of Variance
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E(m + n) = e(m) + e(n) = 31 + 35. 5 = 66. 5 minutes sd(m + n) = E(m n) = e(m) e(n) = 31 35. 5 = 4. 5 minutes sd(m n) = Normal ( = 4. 5, = 4. 60977) pr(m n > 0) = 1 pr(m n < 0) = 1 0. 8355 = 0. 1645. = 4. 60977, assuming m and n are independent. = 4. 60977, assuming n and m are independent. E(n m) = e(n) e(m) = 35. 5 31 = 4. 5 minutes sd(n m) = E(tm) = e(m1 + m2 + m3 + m4 + m5) = 5e(m) = 5 x 31 = 155 minutes sd(tm) = sd(m1 + m2 + m3 + m4 + m5) = assuming independence of the morning travel times. Normal ( = 155min, = 6. 7082min) (ii) let tn = n1 + n2 + n3 + n4 + n5, where ni ~ approx. 35 = 6. 7082 minutes, (iii) let t = tm + tn.
