Mathematics 376 – Probability and Statistics II Problem Set 4 Solutions February 20, 2004

8.17/ With n = 50, \overline{Y} = 11.5, and S = 3.5, the 2-standard error bound gives

2σ\overline{Y} = 2\frac{3.5}{\sqrt{50}} ≐ .99

(That is μ = 11.5 ± .99.)

8.20/ With n = 1013, p̂ = Y/1013 = .54 is the point estimate for p (the fraction of the population saying that religion is an important part of their lives). The 2-standard error bound is approximated by

2σ_{p} ≐ 2\sqrt{(.54)(.46)/1013} ≐ .0313

This is the source for the ±3% error bound for the poll quoted in the question. From this, we can conclude that it is probably true that a majority (> .5 of the population) answer the same way.

8.25/ a) The estimator for the difference in proportions is

p̂_{1} − p̂_{2} = .59 − .53 = .06

The two-standard error bound is (approximately)

2\sqrt{(.59)(.41)/1250 + (.53)(.47)/1251} ≐ .0396

b) Since .06 − .0396 > 0, the whole interval .06 ± .0396 consists of positive numbers, and it is likely that the actual proportion of people who watched their fiber intake in 1983 was higher than in 1992 (a demonstrable decrease).

8.26/ Both parts are similar to 8.25. Avoiding fat:

.55 − .51 ± 2\sqrt{(.55)(.45)/1250 + (.51)(.49)/1251} = .04 ± .04

Avoiding excess salt:

.53 − .46 ± 2\sqrt{(.53)(.47)/1250 + (.46)(.54)/1251} = .07 ± .04

Whether or not this is “extremely disturbing” is a subjective question. What we can say is that people’s attention to these issues also probably declined from 1983 to 1992 (demonstrable decreases, with the one for avoiding fat close to the borderline).

8.28/ For this problem we only use the amounts column in the table of data provided. The sample mean of these 20 account amounts is \overline{Y} = 197.1. So our estimate for the total accounts receivable is 500\overline{Y} = 98550. To find a bound on the error of estimation, we need to compute the sample standard deviation

S = \sqrt{\frac{1}{19} ∑_{i=1}^{2} 0(Y_{i} − 197.1)^{2}} ≐ \sqrt{8255.04} ≐ 90.85

Then the two standard error bound is obtained from:

V (500\overline{Y}) = 500^{2}V (\overline{Y}) = 500S^{2}/20 ≐ 500^{2}(8255.04)/20

We take square roots to get

σ_{500\overline{Y}} = 500\frac{S}{\sqrt{20}} = \frac{(500)(90.8)}{\sqrt{20}} ≐ 10158

We get 500\overline{Y} = 98550 ± 2 · 10158 = 98550 ± 20316. For the individual average account, we have

μ = 197.1 ± 2\frac{S}{\sqrt{20}} = 197.1 ± 40.6

It is very unlikely that the average account is as large as 250 (more than two standard errors above the estimated mean).

8.39/a) By the definition,

(1)  \begin{aligned}F_{U}(u) & = P (U ≤ u) \\ & = P (\frac{Y_{(n)}}{θ} ≤ u) \\ & = P (Y_{(n)} ≤ θu) \\ & = F_{(n)}(θu) \end{aligned}

where F_{(n)} is the cdf for the sample maximum. We can compute that function from the information given in the problem as follows. The samples come from the uniform distribution on [0, θ], so

f_{(n)}(y) = n(F (y))^{n−1}f(y)

where F is the uniform cdf and f is the uniform pdf:

f(y) = \{ \begin{matrix}\frac{1}{θ} & if 0 ≤ y ≤ θ \\ 0 & otherwise \end{matrix}

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

So if 0 ≤ θ, then

f_{(n)}(y) = n(y/θ)^{n−1}(1/θ) = ny^{n−1}/θ^{n}

(and equals zero otherwise). It follows that the cdf for Y_{(n)} is

F_{(n)}(y) = ∫ ny^{n−1}/θ^{n} dy = y^{n}/θ^{n}

(and 0 for y < 0, 1 for y > θ). Hence from (1) above:

F_{U}(u) = \{ \begin{matrix}0 & if u < 0 \\ (uθ)^{n}/θ^{n} = u^{n} & if 0 ≤ u ≤ 1 \\ 1 & if u ≥ 1 \end{matrix}

b) Using this formula,

.95 = P (U ≤ (.95)^{1/n}) = P (\frac{Y_{(n)}}{θ} ≤ (.95)^{1/n}) = P (\frac{Y_{(n)}}{(.95)^{1/n}} ≤ θ)

The desired 95% lower confidence bound is \frac{Y_{(n)}}{(.95)^{1/n}}.

8.43/ We estimate the desired proportion p by p̂ = 1912/2374 = .805. For the 99% confidence interval, we want α = .01. From the standard normal table, z_{.005} ≐ 2.575. Then the large-sample confidence interval formula for a proportion gives:

p = .805 ± 2.575\sqrt{(.805)(.195)/2374} = .805 ± .021

8.46/ We are estimating a mean here with n = 75, \overline{Y} = 4.2, and S = 1.5. For the 95% confidence interval, α = .05 and z_{.025} = 1.96, so:

μ = 4.2 ± 1.96\frac{1.5}{\sqrt{75}} = 4.2 ± .34

8.51/ We are estimating a difference of proportions here with p̂_{1} = .19, p̂_{2} = .70. For the 90% confidence interval, we use α = .1, and z_{.05} ≐ 1.645. So

p̂_{1} − p̂_{2} ≐ (.19 − .70) ± 1.645\sqrt{(.19)(.81)/1250 + (.7)(.3)/1251} = −.51 ± .028

8.57/ Here we are estimating a difference of two differences of proportions. The point estimator to use (unbiased) is:

(p̂_{3} − p̂_{1}) − (p̂_{4} − p̂_{2}) = (Y_{1}/n_{1} − Y_{3}/n_{3}) − (Y_{4}/n_{4} − Y_{2}/n_{2})
= (18/26 − 20/31) − (7/28 − 13/30)
≐ (.692 − .645) − (.250 − .483)
= .230

To derive the confidence interval, we need to know the standard error of this estimator – the square root of its standard deviation. This is similar to what we did before for differences of proportions (see Problem 7.58 b from Problem Set 3, for instance). Since the Y_{i} are independent binomial random variables,

V ((Y_{1}/n_{1} − Y_{3}/n_{3}) − (Y_{4}/n_{4} − Y_{2}/n_{2})) = \frac{p_{1}q_{1}}{n_{1}} + \frac{p_{2}q_{2}}{n_{2}} + \frac{p_{3}q_{3}}{n_{3}} + \frac{p_{4}q_{4}}{n_{4}}

So for the 95% confidence interval, we take

.230 ± 1.96\sqrt{(.692)(.308)/26 + (.645)(.355)/31 + (.25)(.75)/28 + (.433)(.567)/30}

or

.230 ± .342

Note that the interval contains both positive and negative numbers, so it is not possible to draw any conclusions about whether p_{3} − p_{1} > p_{4} − p_{2}(!)