Mathematics 376 – Mathematical Statistics Final Examination May 16, 2012

Directions

Do all work in the blue exam booklet, and include all work necessary to justify your answers. Do problems I - VI. Be sure to read each question carefully before starting to work. There are 200 possible points, and up to 20 possible Extra Credit points.

General Note: A “random sample” always consists of independent measurements from the indicated distribution.

I. Let X_{1}, X_{2}, X_{3}, X_{4}, X_{5} be a random sample from a normal distribution with mean μ = 6 and variance σ^{2} = 81.

A) (10) What is the distribution (give the type and all relevant parameter values) of

U = \frac{(X_{1} − 6)^{2} + (X_{2} − 6)^{2} + (X_{3} − 6)^{2} + (X_{4} − 6)^{2}}{81} ?

Why? B) (10) What is the distribution (give the type and all relevant parameter values) of

V = \frac{2(X_{5} − 6)}{9\sqrt{U}},

where U is as in part B? Why? C) (10) What is the distribution (give the type and all relevant parameter values) of

W = \frac{(X_{1} − 6)^{2}}{(X_{2} − 6)^{2}} ?

Why?

II. A random variable Y has pdf of the form

f(y) = \{ \begin{matrix}\frac{2y}{α} e^{−y^{2}/α} & if y ≥ 0 \\ 0 & otherwise \end{matrix}

for some unknown α > 0. (This is a particular case of a Weibull distribution; these are often used to model length of life of mechanical components and living organisms.) Let Y_{1}, …, Y_{n} be a random sample from this distribution.

A) (20) Find the maximum-likelihood estimator for α. B) (15) What is the pdf for the sample maximum, Y_{(n)}?

III. Let p be the proportion of letters mailed in Belgium that are delivered the next day.

A) (20) A random sample of n = 400 letters is selected for tracking and 373 of those are delivered the next day. Find an approximate 98% confidence interval for p based on this sample. Is p = .85 a “reasonable” guess based on this data?

B) (15) (“Thought question”) Note that part A says “approximate.” What is the actual distribution of Y = the number of letters delivered the next day (out of a random sample of size n = 400)? Why does the method you used in part A give a reasonable interval estimate for p? Explain.

IV. The fill weights of a random sample of n_{1} = 13 boxes of Twigs and Nuts Extra Crunchy Granola produced at Plant 1 had a mean of 20.5 ounces and standard deviation s = .1 ounce. A similar sample of size n_{2} = 11 produced at Plant 2 had mean fill weight 20.3 ounces and standard deviation s = .083 ounce.

A) (20) Is there sufficient evidence at the α = .01 level to conclude that the variances of

the fill weights at the two plants are different? B) (20) Is there sufficient evidence to conclude that the mean fill weights are different?

Report the results by giving an estimate of the p-value of your test.

V. The following table gives measurements of water depth remaining in an evaporating reservoir as a function of time:

x time (weeks) y depth (meters)
1  19.8
4  16.5
14  12.8
32  8.1
52  7.5

A) (20) Find the least-squares estimators for the coefficients β_{0}, β_{1} in a model Y =

β_{0} + β_{1}x + ε for this data set. B) (20) Is there sufficient evidence to say that β_{1} < −0.1? Test with α = .05.

Note: You may “purchase” a printout of an R session with some results relevant for problem V for 15 points if you wish.

VI. (20) Suppose we have measurements taken from a normal population with unknown μ and known σ^{2}. We test H_{0} : μ = μ_{0} versus H_{a} : μ > μ_{0} using a Z-test and a rejection region designed to produce a Type I error probability α. Show that in order to make the Type II error probability equal a given β when μ is equal to some μ_{a} > μ_{0}, the sample size should be selected as follows:

n = \frac{(z_{α} + z_{β})^{2}σ^{2}}{(μ_{a} − μ_{0})^{2}} .

Extra Credit

A New York Times/CBS poll conducted between April 5 and April 12, 2010 included the question “Do you approve or disapprove of the way your Representative in Congress is handling his or her job?” The poll was carried out in two stages. First a large national simple random sample was asked the question. The results were reported in three categories: 46% of respondents said they approved, 36% said they disapproved, and 18% said they had no opinion. Then the same question was asked of a separate sample of n = 881 people, all of whom had identified themselves as supporters of the “Tea Party” movement. The responses broke down like this in the same three categories:

Approve Disapprove No opinion
Tea Party Sample  352  432  97

(Data is adapted from the New York Times web site.) It seems from the numbers that “Tea Party” supporters may differ from the general population when it comes to their opinions concerning their Congressional Representatives. But is this a real difference, or could it just be a product of chance variation in the sampling process?

We did not discuss this in class, but a standard statistical method in this case would be to use a χ^{2}-test. The test statistic

X^{2} = ∑_{categories} \frac{(observed − expected)^{2}}{expected}

has approximately a χ^{2} distribution with degrees of freedom one less than the number of categories. For a test of H_{0} : the distributions are really the same, versus H_{a} : they are significantly different, we would reject H_{0} if X^{2} fell in an upper tail rejection region.

A) (5) If the “Tea Party” sample did follow the national sample percentages, how many

responses would we expect in each category? B) (15) Carry out the χ^{2} test and interpret the results.

Have a safe, enjoyable, and productive summer!