Mathematics 376 – Probability and Statistics II Solutions – Problem Set 3 February 13, 2004

7.48/ a) Y = number of lawyers in sample is binomial with p = 1/410. So

P (Y ≥ 1) = 1 − P (Y = 0) = 1 − (409/410)^{1500} = .9743

Note we could also approximate this using the normal approximation to the binomial distribution. With “continuity correction” we replace 1 by .5

\begin{aligned}P (Y ≥ .5) & = P (\frac{Y/1500 − 1/410}{\sqrt{(1/410)(409/410)(1/1500)}} ≥ \frac{.5/1500 − 1/410}{\sqrt{(1/410)(409/410)(1/1500)}}) \\ & ≐ P (Z ≥ −1.6528) = .9505 \end{aligned}

(reasonably close to the exact value).

b) Again using the “continuity correction” to replace 30 by 30.5

\begin{aligned}P (Y ≥ 30) & ≐ P (\frac{Y/1500 − 1/64}{\sqrt{(1/64)(63/64)(1/1500)}} ≥ \frac{30.5/1500 − 1/64}{\sqrt{(1/64)(63/64)(1/1500)}}) \\ & = P (Z ≥ 1.47) = .0708 \end{aligned}

c) Yes, part b says the chance of getting 30 or more lawyers in a sample of n = 1500 is very small (7%), so with a sample of size n = 1000, the chances are even smaller. In fact, with p = 1/64, the expected number of lawyers in a sample of size 1000 is E(Y) = np = 1000/64 = 15.625, and the standard deviation is σ = \sqrt{npq} = \sqrt{1000(1/64)(63/64)} = 3.922. The value Y = 30 is approximately 3.67 standard deviations above the mean, so very unlikely.

7.49/ We take the true population fraction favoring the bond issue as p = .2. Then for Y the number of people in the sample who favor the bond issue, we want

P (|Y/64 − .2| ≤ .06) = P (−.06 ≤ Y/64 − .2 ≤ .06)

= P (\frac{−.06}{\sqrt{(.2)(.8)/64}} ≤ \frac{Y/64 − .2}{\sqrt{(.2)(.8)/64}} ≤ \frac{.06}{\sqrt{(.2)(.8)/64}})

= P (−1.2 ≤ Z ≤ 1.2) = .7698

from the standard normal table.

7.50/ a) The variance of Y/n is V (Y/n) = npq/n^{2} = pq/n = (p − p^{2})/n. Call this function f(p) (n is fixed but arbitrary). Then to maximize f(p) we find f′(p) = (1 − 2p)/n = 0 if and only if p = 1/2. This is a maximum by the 2nd derivative test. b) We want to determine n that guarantees P (|Y/n − p| ≤ .1) = .95 (at least). This is equivalent to

P (\frac{−.1}{\sqrt{pq/n}} ≤ \frac{Y/n − p}{\sqrt{pq/n}} ≤ \frac{.1}{\sqrt{pq/n}}) = .95

Note that we do not know a value for p here. However, part a says that the variance is maximized when p = 1/2. Hence the interval here is minimized when p = 1/2 (since we’re dividing by the square root of the variance). If we substitute p = 1/2, then from the standard normal table, we want

\frac{.1}{\sqrt{pq/n}} = 2\sqrt{n}(.1) = 1.96

to get probability .95. Hence n = 4(19.5)^{2} = 96.04, so take n = 97.

7.51/The idea is the same as 50 part b. Use the value p = 1/2 because it maximizes the variance:

P (|Y/n − p| ≤ .15) = .98

when 2\sqrt{n}(.15) = 2.33 (standard normal table), so n = 61.

7.57/ a) The probability that a car turns right is p = 1/3. Let Y be the number of cars that turn right. With continuity correction

P (Y ≤ 150) ≐ P (\frac{Y/500 − 1/3}{\sqrt{(1/3)(2/3)(1/500)}} ≤ \frac{150.5/500 − 1/3}{\sqrt{(1/3)(2/3)(1/500)}})

This is approximately P (Z ≤ −1.53) = P (Z ≥ 1.53) = .0630.

b) Now let Y be the number that turn (either left or right). This makes p = 1/3+1/3 = 2/3.

P (Y ≥ 350) ≐ P (\frac{Y/500 − 2/3}{\sqrt{(2/3)(1/3)(1/500)}} ≥ \frac{349.5/500 − 2/3}{\sqrt{(2/3)(1/3)(1/500)}})

This is approximately P (Z ≥ 1.53) = .0630 also.

7.58/ a) E(Y_{1}/n_{1} − Y_{2}/n_{2}) = (1/n_{1})E(Y_{1}) − (1/n_{2})E(Y_{2}) = p_{1} − p_{2}.

b) By independence,

V (Y_{1}/n_{1} − Y_{2}/n_{2}) = (1/n_{1}^{2})V (Y_{1}) + (1/n_{2}^{2})V (Y_{2}) = p_{1}q_{1}/n_{1} + p_{2}q_{2}/n_{2}

8.2/a) We have since θ̂_{1} and θ̂_{2} are unbiased estimators for θ,

E(θ̂_{3}) = E(aθ̂_{1} + (1 − a)θ̂_{2}) = aθ + (1 − a)θ = θ

b) By independence, we have

V (θ̂_{3}) = V (aθ̂_{1} + (1 − a)θ̂_{2}) = a^{2}σ_{1}^{2} + (1 − a)^{2}σ_{2}^{2}

To minimize this as a function of a, as in 7.50a above, we differentiate with respect to a and set equal to 0:

2aσ_{1}^{2} − 2(1 − a)σ_{2}^{2} = 0 ⇒ a = \frac{σ_{1}^{2}}{σ_{1}^{2} + σ_{2}^{2}}

This seen to be a minimum since the second derivative is 2(σ_{1}^{2} + σ_{2}^{2}) > 0.

8.3/ If θ̂_{1} and θ̂_{1} are not independent, then

V (θ̂_{3}) = V (aθ̂_{1} + (1 − a)θ̂_{2}) = a^{2}σ_{1}^{2} + (1 − a)^{2}σ_{2}^{2} − 2a(1 − a)c

where c = Cov(θ̂_{1}, θ̂_{2}). As in 2 b, we differentiate with respect to a and set equal to zero to minimize:

2(σ_{1}^{2} + σ_{2}^{2} − 2c)a + 2c − 2σ_{2}^{2} = 0

so

a = \frac{σ_{2}^{2} − c}{σ_{1}^{2} + σ_{2}^{2} − 2c}

As before, this is a minimum by the second derivative test.

8.5/ The distribution of each Y_{i} is exponential with parameter β = θ +1. Hence E(Y_{i}) = θ for all i, and E(\overline{Y}) = θ + 1. Hence θ̂ = \overline{Y} − 1 is an unbiased estimator for θ since E(\overline{Y} − 1) = θ “on the nose”.

8.7/The mean is μ = 3, so

E((Y − 3)^{3}) = E(Y^{3}) − 9E(Y^{2}) + 81 − 27 = E(Y^{3}) − 9E(Y^{2}) + 54

Hence by the given information

θ̂ = θ̂_{3} − 9θ̂_{2} + 54

is an unbiased estimator for the third central moment. E((Y − 3)^{3})

8.9/a) Y is binomial so the exact variance is V (Y) = npq = np(1 − p). So it is natural, thinking of the fraction Y/n as an estimator for p, to use n(Y/n)(1 − Y/n) to estimate V (Y). But

E(n(Y/n)(1 − Y/n)) = E(Y − Y^{2}/n)

= E(Y) − (1/n)E(Y^{2})

= np − (1/n)(npq + n^{2}p^{2}) = (n − 1)pq Since E(n(Y/n)(1 − Y/n)) ≠ npq, this is a biased estimator for V (Y). But if we “adjust” to

θ̂ = \frac{n}{n − 1} · n(Y/n)(1 − Y/n)

we get E(θ̂) = npq = V (Y) “on the nose”, so θ̂ is an unbiased estimator for V (Y). 8.10/a) Let F (y) be the cdf for Y, then

F (y) = \{ \begin{matrix}0 & if y < 0 \\ y^{α}/θ^{α} & if 0 ≤ y ≤ θ \\ 1 & if y > θ \end{matrix}

Hence by our formula from class for the pdf of the sample maximum:

f_{(n)}(y) = n(F (y))^{n−1}f(y)

we get for θ̂:

\begin{aligned}E(θ̂) & = ∫_{0}^{θ} nα (\frac{y^{nα}}{θ^{nα}} ) dy \\ & = \frac{nα}{nα + 1}θ \end{aligned}

b) As in 9 above, we can adjust by a constant multiple to get an unbiased estimator for θ:

θ̂_{2} = \frac{nα + 1}{nα}θ̂ = \frac{nα + 1}{nα}Y_{(n)} = \frac{nα + 1}{nα} max(Y_{1}, …, Y_{n})

c)

MSE(θ̂) = V (θ̂) + B(θ̂)^{2}

= E(θ̂^{2}) − (E(θ̂))^{2} + B(θ̂)^{2}

= ∫_{0}^{θ} nα (\frac{y^{nα+1}}{θ^{nα}} ) dy − (\frac{nα}{nα + 1}θ)^{2} + (\frac{1}{nα + 1}θ)^{2}

= \frac{2θ^{2}}{(nα + 1)(nα + 2)}

(after a lot of algebra!) 8.13/a) Since Y is binomial and p̂_{2} = (Y + 1)/(n + 2),

B(p̂_{2}) = (np + 1)/(n + 2) − p = \frac{1 − 2p}{n + 2}.

b) p̂_{1} is unbiased, so

MSE(p̂_{1}) = V (p̂_{1}) = pq/n

On the other hand after some algebra,

MSE(p̂_{2}) = \frac{npq + (2p − 1)^{2}}{(n + 2)^{2}}

Note: The answer in the text is unfortunately incorrect (they reversed the inequality in deriving their solution). The idea is you want to rewrite the inequality

\frac{p(1 − p)}{n} < \frac{np(1 − p) + (2p − 1)^{2}}{(n + 2)^{2}}

by clearing denominators and then putting all terms on the same side:

(8n + 4)p^{2} − (8n + 4)p + n > 0

This is true if and only if p is smaller than the smaller root of this quadratic equation or larger than the larger root:

p < \frac{1}{2} − \frac{\sqrt{(8n + 4)^{2} − 4n(8n + 4)}}{2(8n + 4)}

or

p > \frac{1}{2} + \frac{\sqrt{(8n + 4)^{2} − 4n(8n + 4)}}{2(8n + 4)}