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Solution ch7

# Solution ch7 - T-Q T—IS 122 n=36 a= b{3 lJ[t——u1...

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Unformatted text preview: T-Q T—IS 122 n=36 a=+b _{3+lJ_ [t——u1+132—1= {—3— l+l]1—1= g _ 2 ﬁ‘ T #—=25—=—III=—III x = x- E 5 7=f—# o‘ﬂn Usingthecentrallirmttheor‘em 1I—2 25—: Pﬂqurtﬂ En“. {E} = HDJ‘HE < 2 < 3.5?42} = Hz < 3.6?42}—P{2 < 0.1343: = l— ELTEEE = {1.2312 EWe%[E(x1}+mare---+E{XTJ]=%oEm)=;on) = n E(é,]=%[EoX.n+E(X¢)+Ecx?)]=%[2n—n+nl=n a) Roth él and it: are unbiased estimates of p. since the expected values ofthese statistics are equivalent to the true mean, |.I.. bJVE-F {WJ 3%? 1H? 2]+"'+F[I?)l= 2 . o- V091): T 1 2 —_ [T'cr )—?ol 1 49 VE,]=V[W]=i1(VEH.J+P(Xs)+F 4))=%{4I’(X1)+I’(X¢)+F(l'q}) Since both estimators are unbiased, the varianees can be compared to decide to select the better estimator. The variance of 191 is smaller than that of '92. Elie the better estimator. a) The average ofthe 26 observations provided can be used as an estimator ofthe meanpull force because we know it is unbiased This value is '?5.615 pounds. b) The median ofttie sample can be used as an estimate ofthe point that divides the population into a “weak" and “strong” half. This estimate is T12 pounds. c) Our estimate ofthe population variance is the sample variance or 2.?38 square pounds. Similarly, our estimate of the population standard deviation is the sample standard deviation or 1. 635 pounds. a) The estimated standard error ofthe meanpull force 15 1. 653215“— — o. 325. This value 15 the standard deviation, not ofthe pull ﬁerce: but ofthe mean pull ﬁerce of the sample. e) Dnlj.r one connector in the sample has a pull force measurement under T3 pounds. Dur point estimate for the proportion requested is then lﬂo = 11.0335 ...
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