Download PSTAT 120B Probability and Statistics - week8

PSTAT 120B Probability and Statistics - week8
Fang-I Chu
University of California, Santa Barbara
November 30, 2012
Fang-I Chu
PSTAT 120B Probability and Statistics
couple notes
Average for Midterm: 39.88/100, median for Midterm:
32/100.
Average for hw#5: 79.69/100.
Average for hw#6: 65.33/100.
About Midterm (2):
Problem 2 requires you to state whether the obtained
probability is exact or approximate.
The answer is approximate because we obtain probability via
using central limit theorem.
Notice though given n = 25 is at border line of being small
sample case, t-table is not available to us. Thus we apply
central limit theorem to obtain approximate probability instead
of exact probability using t distribution.
Please talk to me if you have question about your grade.
Fang-I Chu
PSTAT 120B Probability and Statistics
couple notes
About hw#5 and hw#6:
in hw#5 and hw#6, when looking for MLE, you need to check
2nd order derivative to ensure2 that its maximum but not
L(θ)
< 0.
minimum value. i.e. check d ln
dθ 2
I am aware that the grader penalized you multiple times in
both hw, however, he/she is holding same standard to whole
class. So, REMEMBER to check 2nd order derivative!
in hw#6, problem 2, your final answer should be : MVUE is
X(n)
X(n) n−1
X(n) (n+1)
.
n + 2
n =
2n
hw#7 is the last hw.
For section next week, I will go over the practice final
(previous year finals).
Please email me if there is any particular topic/problem that
you hope to be discussed for next week.
Fang-I Chu
PSTAT 120B Probability and Statistics
Topics for review
Hint for #1 (Exercise 10.5)
Hint for #2 (Exercise 10.18)
Hint for #3 (Exercise 10.30)
Hint for #4 (Exercise 10.39)
Hint for #5 (Exercise 10.64)
Fang-I Chu
PSTAT 120B Probability and Statistics
#1(Exercise 10.5)
10.5
Let Y1 and Y2 be independent and identically distributed with a
uniform distribution over the interval (θ, θ + 1). For testing
H0 : θ = 0 versus Ha : θ > 0, we have two competing tests:
Test 1: Reject H0 if Y1 > .95.
Test 2: Reject H0 if Y1 + Y2 > c.
Find the value of c so that test 2 has the same value for α as test
1.[ Hint: In Example 56.3, we derived the density and distribution
function of the sum of two independent random variables that are
uniformly distributed on the interval (0,1).]
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(Exercise 10.5)
Hint:
Known:
Y1 , Y2 ∼ Unif(θ, θ + 1)
hypothesis: H0 : θ = 0 versus Ha : θ > 0
hypothesis test 1: Reject H0 if Y1 > .95
hypothesis test 2: Reject H0 if Y1 + Y2 > c
Goal: Find c such that test 1 and test 2 give the same α.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(Exercise 10.5)
Hint:
Way to approach:
Under H0 , Y1 , Y2 ∼ Unif(0, 1). From example 6.3, the
distribution
of U = Y1 + Y2 is
u
0≤u≤1
fU (u) =
2−u
1<u≤2
for test 1, we find α = P(Y1 > 0.95) =?
for test 2, we find equation
R2
α = P(Y1 + Y2 > c) = P(U > c) = c fU (u)du, and solve for
c.
Note: be cautious about dealing the range of u.
Fang-I Chu
PSTAT 120B Probability and Statistics
#2(Exercise 10.18)
10.18
The hourly wages in a particular industry are normally distributed
with mean $13.20 and standard deviation $2.50. A company in
this industry employs 40 workers, paying them an average of
$12.20 per hour. Can this company be accused of paying
substandard wages? Use an α = .01 level test.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #2(Exercise 10.18)
Hint:
Known: Denote hourly wages in particular industry as Wi ,
Wi ∼ N (µ, σ 2 )
Goal: Find out whether µW < 13.2 or not, where µW as
mean wage for this particular company. i.e. Hypothesis:
H0 : µW = 13.2 v.s. Ha : µW < 13.2
Way to approach:
2
Note W̄ ∼ N (µW̄ , σW̄
)
The sample mean W̄ is a point estimator of µW , and
2
W̄ ∼ N (µW̄ , σW̄
). i.e. σW̄ = √σn = √2.5
40
Test statistic is Z =
W̄ −µ0
σW̄
=
W̄ −µ0
σ
√
n
.
construct rejection region with α = 0.01.
Make your decision base on obtained test statistic and
rejection region.
Note. look at example 10.5 on page 497.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 10.30)
10.30
A manufacturer claimed that at least 20% of the public preferred
her product. A sample of 100 persons is taken to check her claim.
With α = .05, how small would the sample percentage need to be
before the claim could legitimately be refuted? (Notice that this
would involve a one-tailed test of the hypothesis.)
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 10.30)
Hint:
Known:
X denote the number of observed public preferred such
product, then X is a binomial random variable, with p
denoting the probability that a randomly selected public would
prefer the product. A sample of 100 is taken.
hypothesis: H0 : p = 0.20 v.s. Ha : p < 0.20
Goal: Find how small p̂ needs to be such we will have decision
”reject null hypothesis”.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 10.30)
Hint:
Way to approach:
0
q p̂−p0 (why not using
Note test statistic Z = p̂−p
σp̂ =
p0 (1−p0 )
n
q
p̂(1−p̂)
?)
n
The rejection region should be z < −z0.05 where z0.05 = 1.64
Find the smallest p̂ that satisfy z < −z0.05
Note. Look at example 10.6 on page 498.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #4(Exercise 10.39)
10.39
Refer to Exercise 10.30. Calculate the value of β for the
alternative pa = .15.
Hint:
Known: obtained rejection region from 10.30 :p̂ < 0.134.
Goal: Find the value of β (type II error) when Ha is true. i. e.
true p = 0.15.
Way to approach:
By definition of type II error, we are looking for
P(p̂ not in RR when Ha is true) = P(p̂ > 0.1342|p = 0.15)
Given p = 0.15, we could find P(p̂ > 0.1342) = P(Z > z),
where z = qp̂−p
.
p(1−p)
n
Look up P(Z > z) from table on page 848.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #5(Exercise 10.64)
10.64
A coin-operated soft-drink machine was designed to discharge on
the average 7 ounces of beverage per cup. In a test of the
machine, ten cupfuls of beverage were drawn from the machine
and measured. The mean and standard deviation of the ten
measurements were 7.1 ounces and .12 ounce, respectively. Do
these data present sufficient evidence to indicate that the mean
discharge differs from 7 ounces?
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #5(Exercise 10.64 (a))
10.64(a)
What can be said about the attained significance level for this test
based on the t table in the appendix?
Think:
(1). Look at t-table, what have you observed about the
relation between α and tα for degree of freedom 9?
(2). How does your decision change when given different α?
(3). Draw a graph to explain.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #5(Exercise 10.64 (c))
10.64(c)
What is the appropriate decision if α = .10?
Known:
µ = 7, n = 10, X̄ = 7.1,s = 0.12
Hypothesis: H0 : µ = 7 v.s. Ha : µ 6= 7
Goal: test whether the mean discharge differs from 7 ounces.
Way to approach:
Note test statistic T =
Ȳ −µ0
S
√
n
Construct rejection region based on given α and degrees of
freedom.
Notice this is a two-tailed test. (why?)
Make your decision based rejection region.
Note. Look at example 10.12 on page 521.
Fang-I Chu
PSTAT 120B Probability and Statistics
Remark
For hypothesis testing problem, make sure you understand
what is going to be tested. Your solution should be set up in
five steps:
1. Write down your null hypothesis.
2. Decide whether this should be one-tailed (which direction) or
two-tailed test. Write down your alternative hypothesis.
3. Obtain your test statistic when null hypothesis is true.
4. Determine your rejection region based on given α.
5. Make your decision using test statistic and rejection region.
Fang-I Chu
PSTAT 120B Probability and Statistics