45-733 PROBABILITY AND STATISTICS I Topic #7C
22 February 2000, 24 February 2000
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P[|Xn - m| < c] = 1 - a, or
_ _
P[Xn - c < m < Xn + c] = 1 - a, where
a = .01 or .05 or .10 typically.
Note that the random variable here is a line length. The proper
interpretation of it is that, in the long run, 1 - a
_
percent of the time the interval around Xn will contain m.
_
Also note that, once we insert a value for Xn we no
longer have a random variable -- we have a defined interval.
Hence, we say that we are "1 - a confident
that the true mean is in the interval."
To find the value of c: _ _
P[|Xn - m| < c] = P[-c < Xn - m < c] =
_
P[-cn1/2/s < (Xn - m)/s/n1/2 < cn1/2/s] =
P[-cn1/2/s < Z < cn1/2/s] = F(cn1/2/s) - F(-cn1/2/s) =
F(za/2) - F(-za/2) = 1 - a
Hence: za/2 = c/s/n1/2 and c = za/2s/n1/2
Which gives us our confidence interval:
_ _
P[Xn - za/2s/n1/2 < m < Xn + za/2s/n1/2] = 1 - a
Example: Suppose we take a random sample of 25 from
N(m, 1).
Construct a 95% confidence interval for
m._ _ P[Xn - za/2s/n1/2 < m < Xn + za/2s/n1/2] = 1 - aSuppose we have a large order of bolts delivered to our factory. We are concerned about the precision with which these bolts have been machined. In particular, we want to construct 95% confidence limits for the true mean length of the bolts. Assume that the length of the bolts is normally distributed.
_ We are given: n = 500, Xn = 6.1cm, s = .1cm, 1 - a = .95, a = .05, a/2 = .025, Hence, z.025 = 1.96So the confidence limits are: 6.1 ± (1.96*.1)/5001/2 or
Ù Ù Ù p ~ N[p, p(1 - p)/n] so that the confidence interval is:Problem 8.47 p.346
Ù Ù Ù Ù Ù Ù P{p - za/2[p(1 - p)/n]1/2 < p < p + za/2[p(1 - p)/n]1/2} = 1 - a
Ù We are given: n = 1506, p = .73, 1 - a = .95, a = .05, a/2 = .025. Hence, z.025 = 1.96So the confidence limits are: .73 ± 1.96[(.73*.27)/1506]1/2
_ (Xn - m)/s/n1/2 ~ tn-1 So that we can write the confidence interval as: _ _ P[Xn - ta/2s/n1/2 < m < Xn + ta/2s/n1/2] = 1 - aProblem 8.68 p.358
_ _
Xn ~ N[mx, sx2/n] and Ym ~ N[my, sy2/m]
_ _
Then: Xn - Ym ~ N[mx - my, sx2/n + sy2/m]
And the confidence interval is:
_ _ _ _
P{Xn - Ym - za/2[sx2/n + sy2/m]1/2 < mx - my < Xn - Ym + za/2[sx2/n + sy2/m]1/2}
= 1 - a
Confidence Interval for the Difference Between Two Means when
sampling from two separate, independent, Normal Distributions with
unknown variances but large (n > 30 and m > 30) sample
sizes.
_ _ _ _
P{Xn - Ym - za/2[sx2/n + sy2/m]1/2 < mx - my < Xn - Ym + za/2[sx2/n + sy2/m]1/2}
= 1 - a
_ _ We are given: n = 252, m = 307, Xn = 11.48, Ym = 13.21, sx = 5.69, sy = 5.31, 1 - a = .95, a = .05, a/2 = .025. Hence, z.025 = 1.96Our confidence limits are:
_ _ We are given: n = 252, m = 307, Xn = 22.05, Ym = 25.96, sx = 5.12, sy = 5.07, 1 - a = .90, a = .10, a/2 = .05. Hence, z.05 = 1.645Our confidence limits are:
Ù Ù Ù Ù Ù Ù p1 - p2 ~ N[p1 - p2, p1(1 - p1)/n + p2(1 - p2)/m] And the confidence limits can be computed from: Ù Ù Ù Ù Ù Ù p1 - p2 ± za/2[p1(1 - p1)/n + p2(1 - p2)/m]1/2Problem 8.49 p.347
Ù Ù We are given: p1 = .19, p2 = .70, n = 1250, m = 1251, 1 - a = .9, a = .1, a/2 = .05. Hence, z.05 = 1.645.The confidence limits are:
Ù Ù We are given: p1 = .67, p2 = .90, n = 1250, m = 1251, 1 - a = .98, a = .02, a/2 = .01. Hence, z.01 = 2.33.The confidence limits are:
_ _ _ _
P{Xn - Ym - ta/2s(1/n + 1/m)1/2 < mx - my < Xn - Ym + ta/2s(1/n + 1/m)1/2} =
1 - a
Problem 8.71 p.359_ _ We are given: Xn = 11, Ym = 20, n = 16, m = 20, sx = 6, sy = 8, 1 - a = .95, a = .05, a/2 = .025. Hence, t.025, 34df = 1.96Pooling the sample sums of squares: