Mathematics 376 – Mathematical Statistics Solutions for Makeup Midterm Exam 2 April 26, 2012

I. Let Y_{1}, …, Y_{n} be a random sample from a population with density function

f(y|θ) = \{ \begin{matrix}\frac{3θ^{3}}{y^{4}} & if 0 < θ ≤ y < ∞ \\ 0 & otherwise. \end{matrix}

A) (10) Determine the method of moments estimator for θ.

Solution: Computing E(Y) from this density, we find:

E(Y) = ∫_{θ}^{∞} y · \frac{3θ^{3}}{y^{4}} dy = \frac{3θ}{2} .

For the method of moments, we set this equal to \overline{Y}, the first sample moment, and solve for θ:

θ̂_{MOM} = \frac{2\overline{Y}}{3} .

B) (10) Determine the maximum likelihood estimator for θ. Solution: The likelihood function in this case is

(1)  L(y_{1}, …, y_{n} | θ) = \frac{3^{n}θ^{3n}}{(∏_{i=1}^{n} y_{i})^{4}}

As a function of θ, neither this nor

ln(L) = n ln(3) + 3n ln(θ) − 4 ∑_{i=1}^{n} ln(y_{i})

has a critical point. Both are strictly increasing as functions of θ. Therefore, we want to choose θ as large as possible, consistent with the sample values, in order to maximize L. This occurs when θ = Y_{(1)}, the sample minimum. Therefore

θ̂_{ML} = Y_{(1)}.

C) (5) Is U = ∑_{i} ln(Y_{i}) a sufficient statistic for θ? Why or why not? Solution: The answer is yes by the factorization criterion. From (1) above we have

L(y_{1}, …, y_{n} | θ) = \frac{3^{n}θ^{3n}}{e^{4U}} = g(U, θ)

We can take the other factor h(y_{1}, …, y_{n}) = 1.

II. In this problem, let S represent the proportion of women who agree that men are basically selfish and egotistical.

A) (15) Suppose that surveys done 20 years ago showed that S was 0.40 at that time, but a new survey finds that Y = 1260 out of a random sample of n = 3000 women said that men are basically selfish and egotistical. Is there sufficient evidence to say the proportion S has increased? Give the null and alternative hypotheses for a suitable test, carry out the analysis, estimate the attained significance level (p-value) and state your conclusion succinctly and clearly.

Solution: The wording of the question is supposed to suggest using H_{0} : S = .4 and H_{a} : S > .4. Since n = 3000 is large, we use a Z-test:

Z = \frac{\frac{1260}{3000} − .4}{\sqrt{\frac{(.4)(.6)}{3000}}} = 2.236.

For the upper tail test, we would reject H_{0} for any α with z_{α} ≤ 2.236. From the standard normal table, this says the p-value is about .0125. There is relatively strong evidence to reject H_{0}.

B) (15) Suppose that the current value of S is actually S = .43. How large would n have to be in a sample like the one from part A to ensure that β (the probability of a Type II error) is .1 or less?

Solution: In terms of \frac{Y}{n}, the rejection region for the test with α = 0.05 (and a general sample size n) is

\{\frac{Y}{n} > .4 + 1.645\sqrt{\frac{(.4)(.6)}{n}}\}

Recall that β is the chance of not rejecting H_{0} when H_{0} is false. If S = .43, we start from

P (\frac{Y}{n} ≤ .4 + 1.645\sqrt{\frac{(.4)(.6)}{n}})

and “restandardize” using S = .43:

β = P (\frac{Y}{n} ≤ .4 + 1.645\sqrt{\frac{(.4)(.6)}{n}})

= P (\frac{\frac{Y}{n} − .43}{\sqrt{\frac{(.43)(.57)}{n}}} ≤ \frac{−.03 + 1.645\sqrt{\frac{(.4)(.6)}{n}}}{\sqrt{\frac{(.43)(.57)}{n}}} )

= P (Z ≤ \frac{−.03 + 1.645\sqrt{\frac{(.4)(.6)}{n}}}{\sqrt{\frac{(.43)(.57)}{n}}} )

To get this to be .1, we want

\frac{−.03 + 1.645\sqrt{\frac{(.4)(.6)}{n}}}{\sqrt{\frac{(.43)(.57)}{n}}} ≐ −1.28

so

\sqrt{n} = \frac{1.645\sqrt{(.4)(.6)} + 1.28\sqrt{(.43)(.57)}}{.03} ≐ 47.98

so n ≥ 2303 will do. (In other words, the test with n = 3000 will already have β this small(!).)

III. Assume that the hardness of iron bars produced by one plant is normally distributed with mean μ and variance σ^{2}. A random sample of n = 7 bars from the output of the plant is measured to determine whether the average hardness meets the desired figure μ_{0} = 172.

A) (20) The following measurements:

167, 174, 179, 164, 163, 160, 168

were obtained. Test H_{0} : μ = 172 versus the alternative H_{a} : μ < 172 using an appropriate test with α = .01.

Solution: Since n < 30, we must use a t-test here. We have \overline{Y} ≐ 167.86 and S^{2} ≐ 43.81. We have

t = \frac{167.82 − 172}{\sqrt{\frac{43.81}{7}}} ≐ −1.66

The rejection region for the test is determined from the t-table. The form of the alternative hypothesis indicates we want

RR = {t | t < −t_{.01}} = {t | t < −3.143}

(for 6 df). There is not sufficient evidence to reject H_{0} here.

B) (10) How is your answer from A related to the 99% lower confidence bound for μ that we learned to compute in the first section of the course?

Solution: The 99% lower confidence bound would be

167.82 − 3.413\sqrt{\frac{43.81}{7}} = 167.82 − 8.54 = 159.28

The value μ_{0} = 172 is one of the numbers in the interval (159.28, ∞), so it is a “believable” value for μ based on the data. Alternatively, the rejection region for the test consists of all numbers less than

172 − 3.413\sqrt{\frac{43.81}{7}} = 159.28

Since the sample \overline{Y} is in this interval we do not reject H_{0}.

IV. (15) Let (x_{i}, y_{i}, z_{i}), i = 1, …, n be a collection of data points. Using the matrix formulation, derive the normal equations for the least squares estimators for the coefficients β_{0}, β_{1}, β_{2}, β_{3} in the model Z = β_{0} + β_{1}x + β_{2}x^{2} + β_{3}y + ε fitting the data. Express in the form Aβ = b where A is a matrix and b is a vector (not matrix products).

Solution: The normal equations are X^{t}Xβ = X^{t}Z where

X = ( \begin{matrix}1 & x_{1} & x_{1}^{2} & y_{1} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ 1 & x_{n} & x_{n}^{2} & y_{n} \end{matrix} ) and Z = ( z_{1} ; ⋮ ; z_{n} )

Hence the normal equations can be written in matrix form as

( \begin{matrix}n & ∑_{i} x_{i} & ∑_{i} x_{i}^{2} & ∑_{i} y_{i} \\ ∑_{i} x_{i} & ∑_{i} x_{i}^{2} & ∑_{i} x_{i}^{3} & ∑_{i} x_{i}y_{i} \\ ∑_{i} x_{i}^{2} & ∑_{i} x_{i}^{3} & ∑ x_{i}^{4} & ∑_{i} x_{i}^{2}y_{i} \\ ∑_{i} y_{i} & ∑_{i} x_{i}y_{i} & ∑_{i} x_{i}^{2}y_{i} & ∑_{i} y_{i}^{2} \end{matrix} ) ( β̂_{0} ; β̂_{1} ; β̂_{2} ; β̂_{3} ) = ( \begin{matrix}∑_{i} z_{i} \\ ∑_{i} x_{i}z_{i} \\ ∑_{i} x_{i}^{2}z_{i} \\ ∑_{i} y_{i}z_{i} \end{matrix} )

Extra Credit “Thought Questions” – Answer in 2 or 3 complete sentences:

A) (5) Exactly why is it the case that a Z-test is not appropriate when the sample size is small and population variance is not known exactly? Explain using the case of a test on a single mean.

Solution: The reason is because of the distribution of the test statistic we can compute from the data. Under the assumption that the data comes from a normal population, we would estimate μ_{0} by \overline{Y} and σ by S. So we would use

t = \frac{\overline{Y} − μ_{0}}{\frac{S}{\sqrt{n}}}

to carry out this sort of test. If the S was the population SD, σ, then (under H_{0} : μ = μ_{0}) this would have a standard normal distribution and the Z-test would be appropriate. However, with the sample SD (S) estimating σ, this statistic actually has a t-distribution with n−1 df, so rejection regions set up using the standard normal would not give enough area in upper and lower tails to account for the variability of t.

B) (5) For the basic small-sample test for equality of means, we need to assume the two samples came from normal populations with equal variances. How would you determine whether applying that test was appropriate?

Solution: Say we observe a test statistic value Z = z_{obs} in a Z-test. The p-value for the 2-tail test would be twice as large as the p-value for either 1-tail test. The reason is that “more extreme” values of in the 2-tail test would be Z > z_{obs} or Z < −z_{obs}.