Answers for HW #1

STAT 554, Spring 2008

Print out this page, put your name on it, fill in the correct answers, and submit it at our 4th class meeting. (Be sure to round answers as indicated below!)

Problem 1

Put your answers for parts (a), (b), (c), (e), (f), (h), (k), and (l) in the answer boxes below. (For some of the parts (which are worth no points), answers are provided. These may serve as useful checks of your work.) Round each test size (parts (b) & (f)) to the nearest thousandth, and round each power (parts (c), (h), (k), & (l)) to the nearest hundredth.

n rejection region size power (p = 0.6) power (p = 0.7)
40 1(a) 1(b) 1(c) 1(d) 0.58
80 1(e) 1(f) 1(g) 0.29 1(h)
120 1(i) {74, 75, ..., 120} 1(j) 0.007 1(k) 1(l)

For part (m), provide a description of the desired randomized test. (For part (n) (which is worth no points), the answer is provided. This may serve as useful check of your work.) Express you answer in the form used for the answer for part (n) below.

(m)

(n)
Reject the null hypothesis if y > 50, and reject it with probability 0.115 if y = 50.

Problem 2

Put your answers for parts (a), (c), and (e) in the answer boxes below, rounding each (possibly approximate) p-value to the nearest thousandth. (Note: I gave an extra significant digit for the answer to part (d) so that it can be determined whether or not the p-value exceeds 0.05.) (For some of the parts (which are worth no points), answers are provided. These may serve as useful checks of your work.)

exact p-value normal approx. w/ c.c. normal approx. w/o c.c.
dealing with outcomes of red 2(a) 2(b) 0.087 2(c)
dealing with outcomes of 13 2(d) 0.0502 2(e) not requested

Problem 3

Put your answers for parts (a) and (c) in the answer boxes below, rounding each probability to the nearest thousandth. (For part (b) (which is worth no points), the answer is provided. This may serve as useful check of your work.)

n probability
10 3(a)
100 3(b) 0.683
1000 3(c)

Comment: A good way to obtain the values above is to note that the sample mean has a normal distribution having mean μ and variance 9/n. If n = 100 (as is the case for part (b)), then the sample mean is a N(μ,0.09) random variable (and so it's standard deviation is 0.3). It follows that the probability that the sample mean is less than or equal to μ + 0.3 is Φ( ( (μ+0.3) - μ )/0.3 ) = Φ( 1.00 ). Also, the probability that the sample mean is less than μ - 0.3 is Φ( ( (μ-0.3) - μ )/0.3 ) = Φ( -1.00 ). So, the probability that the sample mean will take a value between μ - 0.3 and μ + 0.3 is Φ( 1.00 ) - Φ( -1.00 ), which is equal to (about) 0.683.