\magnification=\magstep1
\centerline{Mathematics 376 -- Probability and Statistics II}
\centerline{Problem Set 4 Solutions}
\centerline{February 20, 2004}
\bigskip
\noindent
8.17/  With $n = 50$,  $\overline{Y} = 11.5$, and $S = 3.5$, the 2-standard
error bound gives
$$2\sigma_{\overline{Y}} = 2{3.5 \over \sqrt{50}} \doteq .99$$
(That is $\mu = 11.5 \pm .99$.)
\bigskip
\noindent
8.20/ With $n = 1013$, $\widehat{p} = Y/1013 = .54$ 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
$$2 \sigma_{\widehat{p}} \doteq 2\sqrt{(.54)(.46)/1013} \doteq .0313$$
This is the source for the $\pm 3\%$ 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.
\bigskip
\noindent
8.25/ a)  The estimator for the difference in proportions is 
$$\widehat{p_1} - \widehat{p_2} = .59 - .53 = .06$$
The two-standard error bound is (approximately)
$$2\sqrt{(.59)(.41)/1250 + (.53)(.47)/1251} \doteq .0396$$
b)  Since $.06 - .0396 > 0$, the whole interval $.06 \pm .0396$
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).
\bigskip
\noindent
8.26/  Both parts are similar to 8.25.  Avoiding fat:
$$.55 - .51 \pm 2\sqrt{(.55)(.45)/1250 + (.51)(.49)/1251} = .04 \pm .04$$
Avoiding excess salt:
$$.53 - .46 \pm 2\sqrt{(.53)(.47)/1250 + (.46)(.54)/1251} = .07 \pm .04$$
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).
\bigskip
\noindent
8.28/  For this problem we only use the {\it 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
$$S = \sqrt{{1\over 19}\sum_{i=1}^20 (Y_i - 197.1)^2} \doteq \sqrt{8255.04} \doteq 90.85$$
Then the two standard error bound is obtained from:
$$V(500 \overline{Y}) = 500^2 V(\overline{Y}) = 500 S^2/20
\doteq 500^2 (8255.04)/20$$
We take square roots to get
$$\sigma_{500\overline{Y}} = 500 {S\over \sqrt{20}} = {(500)(90.8)\over \sqrt{20}} \doteq 10158$$
We get $500\overline{Y} = 98550 \pm 2\cdot 10158 = 98550 \pm 20316$.
For the individual average account, we have
$$\mu = 197.1 \pm 2 {S\over \sqrt{20}} = 197.1\pm 40.6$$
It is very unlikely that the average account is as large as $250$ (more than
two standard errors above the estimated mean).
\bigskip
\noindent
8.39/a)  By the definition, 
$$\eqalign{F_U(u) &= P(U \le u) \cr
&= P\left( { Y_{(n)}\over \theta } \le u \right)\cr
&= P(Y_{(n)} \le \theta u)\cr
&= F_{(n)}(\theta u) \cr}\leqno(1)$$
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 $[0,\theta]$, so 
$$f_{(n)}(y) = n(F(y))^{n-1}f(y)$$
where $F$ is the uniform cdf and $f$ is the uniform pdf:
$$f(y) = \cases{{1\over \theta} & if $0 \le y\le \theta$\cr
                      0 & otherwise\cr}$$
$$F(y) = \cases{0 & if $y < 0$\cr
                y/\theta & if $0 \le y \le \theta$\cr
                1 & if $y > \theta$ \cr
   }$$ 
So if $0 \le \theta$, then 
$$f_{(n)}(y) = n(y/\theta)^{n-1}(1/\theta) = n y^{n-1}/\theta^n$$
(and equals zero otherwise).
It follows that the cdf for $Y_{(n)}$ is 
$$F_{(n)}(y) = \int n y^{n-1}/\theta^n\ dy = y^n/\theta^n$$
(and $0$ for $y < 0$, $1$ for $y > \theta$).  Hence from (1) 
above:
$$F_U(u) = \cases{ 0 & if $u < 0$\cr
                   (u\theta)^n/\theta^n = u^n & if $0 \le u \le 1$\cr
                   1 & if $u \ge 1$\cr}$$
\bigskip
\noindent
b)  Using this formula, 
$$.95 = P(U \le (.95)^{1/n}) = P\left({Y_{(n)}\over \theta} \le (.95)^{1/n}\right) 
= P\left({Y_{(n)}\over (.95)^{1/n}} \le \theta\right)$$
The desired $95\%$ lower confidence bound is 
${Y_{(n)}\over (.95)^{1/n}}$.
\bigskip
\noindent
8.43/ We estimate the desired proportion $p$ by 
$\widehat{p} = 1912/2374 = .805$.  For the $99\%$ confidence
interval, we want $\alpha = .01$.  From the 
standard normal table, $z_{.005} \doteq 2.575$.
Then the large-sample confidence
interval formula for a proportion gives:
$$p = .805 \pm 2.575 \sqrt{(.805)(.195)/2374} = .805 \pm .021$$
\bigskip
\noindent
8.46/ We are estimating a mean here with $n = 75$, $\overline{Y} = 4.2$,
and $S = 1.5$.  For the $95\%$ confidence interval, $\alpha = .05$ and
$z_{.025} = 1.96$, so:
$$\mu = 4.2 \pm 1.96 {1.5\over \sqrt{75}} = 4.2 \pm .34$$
\bigskip
\noindent
8.51/ We are estimating a difference of proportions here with
$\widehat{p_1} = .19$, $\widehat{p_2} = .70$.  For the $90\%$
confidence interval, we use $\alpha = .1$, and 
$z_{.05} \doteq 1.645$.  So
$$\widehat{p_1} -\widehat{p_2} \doteq 
(.19 - .70) \pm 1.645\sqrt{(.19)(.81)/1250 + (.7)(.3)/1251} = -.51 \pm .028$$
\bigskip
\noindent
8.57/ Here we are estimating a difference of two differences of 
proportions.  The point estimator to use (unbiased) is:
$$\eqalign{(\widehat{p_3} - \widehat{p_1}) - (\widehat{p_4} - \widehat{p_2}) &= 
(Y_1/n_1 - Y_3/n_3) - (Y_4/n_4 - Y_2/n_2)\cr
&= (18/26-20/31) - (7/28 - 13/30) \cr
&\doteq (.692-.645) - (.250-.483)\cr 
&= .230\cr}$$
To derive the confidence interval, we need to know the
{\it 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,
$$V((Y_1/n_1 - Y_3/n_3) - (Y_4/n_4 - Y_2/n_2)) = 
{p_1q_1\over n_1} + {p_2q_2\over n_2}+{p_3q_3\over n_3}+{p_4q_4\over n_4}$$
So for the $95\%$ confidence interval, we take
$$.230 \pm 1.96 \sqrt{(.692)(.308)/26+(.645)(.355)/31+(.25)(.75)/28+(.433)(.567)/30}$$
or 
$$.230 \pm .342$$
Note that the interval contains both positive and negative numbers,
so it is not possible to draw any conclusions about whether
$p_3 - p_1 > p_4 - p_2$(!)
\bye