Mathematics 376 – Probability and Statistics II Solutions – Final Examination May 13, 2004

I. A) \overline{Y} is normal with mean μ = 4 and variance σ^{2}/5 = 100/5 = 20. See Theorem 7.1 in text.

B) U has a χ^{2} distribution with 3 degrees of freedom, since each (Y_{i} − 4)/10 is standard normal, and they are independent. The sum of the squares has a χ^{2} distribution by Theorem 7.2 in the text.

C) V has a t-distribution with 3 degrees of freedom, since

V = \frac{\sqrt{3}(Y_{4} − 4)}{10\sqrt{U}} = \frac{(Y_{4} − 4)/10}{\sqrt{U/3}}

This is the quotient of a standard normal and the square root of a χ^{2} random variable, divided by its number of degrees of freedom. See Definition 7.2 in the text.

II. A) We have

E(Y_{i}) = ∫_{0}^{1} y · 4y^{3} dy = 4 ∫_{0}^{1} y^{4} dy = 4/5

and

V (Y_{i}) = E(Y_{i}^{2}) − (E(Y_{i}))^{2} = ∫_{0}^{1} y^{2} · 4y^{3} dy − (4/5)^{2} = (4/6) − (4/5)^{2} = 2/75

B) By the Central Limit Theorem,

\sqrt{n} (\frac{\overline{Y} − 4/5}{\sqrt{2/75}}) = \sqrt{n} (\frac{\overline{Y} − .8}{.1633})

converges in distribution to a standard normal as n → ∞. For n = 50, we expect the desired probability to be closely approximated by

P (Z > \sqrt{50} (\frac{.85 − .8}{.1633})) = P (z > 2.16) = .0154

(from the standard normal table).

III. A) For any continuous random variable Y with pdf f(y), the expected value of a function g(Y) of Y is computed by integrating ∫_{−∞}^{∞} g(y)f(y) dy. (Note: the g does not “interact” at all with the density here. For our log-normal Y, we will compute

E(ln(Y)) = ∫_{0}^{∞} ln(y)\frac{1}{y\sqrt{2πσ^{2}}} e^{−(ln(y)−μ)^{2}/2σ^{2}} dy

First, change variables letting u = ln(y). Then du = dy/y and the limits of integration change to −∞, ∞:

\begin{matrix}= ∫_{−∞}^{∞} u · \frac{1}{\sqrt{2πσ^{2}}} e^{−(u−μ)^{2}/2σ^{2}} du \\ = ∫_{−∞}^{∞} uf_{N}(u) du, \end{matrix}

where f_{N} is the pdf for a normally distributed random variable with distribution N(μ, σ^{2}). We are computing the expected value of that normal, which is just μ. Hence

E(ln(Y)) = μ

if Y is log-normal.

B) Given the sample values y_{1}, …, y_{n}, the likelihood function for σ^{2} = 1 is

L(y_{1}, …, y_{n}|μ) = f(y_{1}|μ) ⋯ f(y_{n}|μ)

= ∏_{i=1}^{n} \frac{1}{y_{i}\sqrt{2π}} e^{−(ln(y_{i})−μ)^{2}/2}

= (\frac{1}{2π} )^{\frac{n}{2}} · \frac{1}{∏_{i=1}^{n} y_{i}} · e^{− ∑_{i=1}^{n}(ln(y_{i})−μ)^{2}/2}

Then,

ln(L) = (\frac{−n}{2}) ln(2π) − ln (∏_{i=1}^{n} y_{i}) − ∑_{i=1}^{n}(ln(y_{i}) − μ)^{2}/2

To find the maximum likelihood estimator, we differentiate with respect to μ and set equal to zero:

0 = \frac{d ln(L)}{dμ} = 0 + 0 − ∑_{i=1}^{n}(ln(y_{i}) − μ) = − ∑_{i=1}^{n} ln(y_{i}) + nμ

Hence, solving for μ, we have:

μ = μ̂_{ML} = \frac{1}{n} ∑_{i=1}^{n} ln(Y_{i})

C) This estimator is unbiased since

E(μ̂_{ML}) = E (\frac{1}{n} ∑_{i=1}^{n} ln(Y_{i})) = \frac{1}{n} ∑_{i=1}^{n} E(ln(Y_{i})) = \frac{1}{n} · n · μ = μ

by part A.

IV. A) Since we only have 4, 5 samples in each group, we need to use the small-sample confidence interval for the difference of the means, which uses the assumption that the variances in the samples are equal (see Section 8.8). The best estimate of that common variance is the “pooled estimator”

S_{p}^{2} = \frac{3s_{1}^{2} + 4s_{2}^{2}}{7} = \frac{3(.001) + 4(.002)}{7} = .00157

Then the 95% confidence interval is

μ_{1} − μ_{2} = .25 − .17 ± t_{.025}(7)\sqrt{.00157}\sqrt{\frac{1}{4} + \frac{1}{5}} = .08 ± .0628

B) Since 0 ∉ (.08 − .0628, .08 + .0628) = (.0172, .1428) there is sufficient evidence to see that we should reject the null hypothesis that μ_{1} = μ_{2} with α = .05. (This can also be verified directly by performing a t-test: The test statistic is

t = \frac{.25 − .17}{\sqrt{.00157}\sqrt{\frac{1}{4} + \frac{1}{5}}} = 3.012

The rejection region for the level α = .05 is

RR = {|t| > t_{.025}(7) = 2.365}

We reject the null hypothesis because t = 3.012 is in this RR.

V. A) Since we have n = 50 samples, a lower-tail z-test is appropriate. The rejection region is

RR = {z < −z_{.01}} = {z < −2.33}

The test statistic is

z = \frac{62 − 64}{8/\sqrt{50}} = −1.768

This is not in RR, so there is not sufficient evidence with α = .01 to say that the mean hardness of this type of steel is less than 64 (and the manufacturer breathes a sigh of relief!!)

B) The rejection region in part A is the same as the set of \overline{Y} -values:

RR = \{\overline{y} < 64 − (2.33)\frac{8}{\sqrt{50}}\} = \{\overline{y} < 61.364\}.

To find β, assuming μ = 60, we proceed as follows:

β = P (\overline{y} ∉ RR|μ = 60) = P (\overline{y} ≥ 61.34|μ = 60)

= P (\frac{\overline{y} − 60}{8/\sqrt{50}} ≥ \frac{61.34 − 60}{8/\sqrt{50}})

= P (z ≥ 1.206)

= .1131 VI. A) We use the sample variance S_{2}^{2} = \frac{1}{n−1} ∑_{i=1}^{n}(Y_{i} − \overline{Y})^{2} in each case and obtain

S_{1}^{2} = 15750 S_{2}^{2} = 10920

B) For testing the null hypothesis σ_{1}^{2} = σ_{2}^{2} against the alternative that σ_{1}^{2} ≠ σ_{2}^{2}, we use a 2-tailed F -test (see page 503 in the text). The test statistic is

f = S_{1}^{2}/S_{2}^{2} = 15750/10920 = 1.44

Since F_{.1}(4, 5) = 3.52 and F_{.9}(4, 5) = 1/F_{.1}(5, 4) = .247, the p-value is greater than 2(.1) = .2. This is quite large, which means that there is extremely weak evidence to reject H_{0} here; we would not reject H_{0} for any α < .2.

VII. A) The “highlights” of the computation of the regression line:

\begin{aligned}\overline{x} & = 37.5 \\ \overline{y} & = 26.95 \\ S_{xx} & = 3937.5 \\ S_{xy} & = 2172.38 \\ S_{yy} & = 1202.48 \\ β̂_{1} & = S_{xy}/S_{xx} = 2172.38/3937.5 = .5517 \\ β̂_{0} & = \overline{y} − β̂_{1}\overline{x} = 26.95 − (.5517)(37.5) = 6.26 \\ S & = \sqrt{(S_{yy} − β̂_{1}S_{xy})/4} = \sqrt{3.978/4} = .9972 \end{aligned}

B) For testing H_{0} : β_{1} = .45 against the alternative β_{1} > .45, we use an upper-tail t-test with n − 2 = 4 degrees of freedom. The test statistic is

t = \frac{β̂_{1} − β_{10}}{S/\sqrt{S_{xx}}} = \frac{.5517 − .45}{.9972/\sqrt{3937.5}} = 6.4

We see that t_{.005}(4) = 4.604, so the p-value of this test is p < 2(.005) = .01. Since this is quite small, we have strong evidence to reject the null hypothesis and conclude that β_{1} > .45.