Download PSTAT 120B Probability and Statistics - Week 9

PSTAT 120B Probability and Statistics - Week 9
Fang-I Chu
University of California, Santa Barbara
May 29, 2013
Fang-I Chu
PSTAT 120B Probability and Statistics
Announcement
Office hour: Tuesday 11:00AM-12:00PM
Please make use of office hour or email if you have question
about hw problem.
Midterm 2 section average: 46.86
Fang-I Chu
PSTAT 120B Probability and Statistics
Topic for review
Exercise #10.19
Exercise #10.37
Exercise #10.46
Exercise #10.48
Exercise #10.40
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.19
10.19
The output voltage for an electric circuit is specified to be 130. A
sample of 40 independent readings on the voltage for this circuit
gave a sample mean 128.6 and standard deviation 2.1. Test the
hypothesis that the average output voltage is 130 against the
alternative that it is less than 130. Use a test with level .05.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.19
Solution:
Known: denote sample mean as Ȳ = 128.6 and standard
deviation as s = 2.1, sample size n = 40
Goal: Test the hypothesis that the average output voltage is
130 against the alternative that it is less than 130. Use a test
with level .05. i.e. Hypothesis: H0 : µ = 130 v.s Ha : µ < 130
Way to approach:
Formula for test statistic: Z =
Under H0 , test statistic is Z =
Ȳ −µy
s
√
∼ N (0, 1)
n
128.6−130
2.1
√
40
= −4.22
Look up from z-table, with α = 0.05, we have −z.05 = −1.645
−4.22 < −1.645 (Note. in form of z < −z0.05 ), we reject H0 .
we conclude that, there is evidence that the mean output
voltage is less than 130.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.37
10.37
Refer to Exercise 10.19. If the voltage falls as low as 128, serious
consequences may result. For testing H0 : µ = 130 versus
Ha : µ = 128, find the probability of a type II error,β, for the
rejection region used in Exercise 10.19.
Solution:
Known: from 10.19, we obtain the rejection region as
z = ȳ −130
< −z.05 where z.05 = 1.645
σ
√
n
Goal: find the probability of a type II error,β, for the rejection
region used in Exercise 10.19.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.37
Solution:
Way to approach:
Simplify z =
ȳ −130
σ
√
n
< −z.05 to obtain ȳ < 130 −
1.645×2.1
√
.
40
i.e.
rejection region:Ȳ < 129.45
Formula for type II error, β:
β = P( fail to reject H0 |Ha is true)
Fail to reject H0 implies that test statistics fall into acceptance
region: z > −z05
β = P(Ȳ > 129.45|µ = 128)
129.45 − 128
= P(Z >
)
2.1
√
40
= P(Z > 4.37)
= 0.000031
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.46
10.46
A large-sample α -level test of hypothesis for H0 : θ = θ0 versus
Ha : θ > θ0 rejects the null hypothesis if
θ̂ − θ0
> zα .
σθ̂
Show that this is equivalent to rejecting H0 if θ0 is less than the
large-sample 100(1 − α)% lower confidence bound for θ.
Fang-I Chu
PSTAT 120B Probability and Statistics
#10.46
Proof:
1. Information:
Hypothesis: H0 : θ = θ0 versus Ha : θ > θ0
0
Obtained the rejection region, θ̂−θ
> zα
σ
θ̂
0
2. Goal: Show that the rejection region, θ̂−θ
σθ̂ > zα , is equivalent
to rejecting H0 if θ0 is less than the large-sample 100(1 − α)%
lower confidence bound for θ.
3. Bridge:
the rejection region,
θ̂−θ0
σθ̂
> zα , can be rewritten as
θ0 < θ̂ − zα σ̂θ̂
θ̂ − zα σ̂θ̂ is the 100(1 − α)% lower confidence bound for θ.
4. Fine tune: we’ve got our proof.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.48
10.48
A large-sample α -level test of hypothesis for H0 : θ = θ0 versus
Ha : θ < θ0 rejects the null hypothesis if
θ̂ − θ0
< −zα .
σθ̂
Show that this is equivalent to rejecting H0 if θ0 is greater than
the large-sample 100(1 − α)% upper confidence bound for θ.
Fang-I Chu
PSTAT 120B Probability and Statistics
#10.48
Proof:
1. Information:
Hypothesis: H0 : θ = θ0 versus Ha : θ < θ0
0
< −zα
Obtained the rejection region, θ̂−θ
σ
θ̂
0
2. Goal: Show that the rejection region, θ̂−θ
σθ̂ < −zα , is
equivalent to rejecting H0 if θ0 is less than the large-sample
100(1 − α)% lower confidence bound for θ.
3. Bridge:
the rejection region,
θ̂−θ0
σθ̂
< −zα , can be rewritten as
θ0 > θ̂ + zα σ̂θ̂
θ̂ + zα σ̂θ̂ is the 100(1 − α)% upper confidence bound for θ.
4. Fine tune: we’ve got our proof!
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.40
10.40
Refer to Exercise 10.33. The political researcher should have
designed a test for which β is tolerably low when p1 exceeds p2 by
a meaningful amount. For example, find a common sample size n
for a test with α = .05 and β ≤ .20 when in fact p1 exceeds p2 by
.1. [Hint: The maximum value of p(1 − p) is .25.]
Solution:
Known:
Hypothesis: H0 : p1 − p2 = 0 v.s Ha : p1 − p2 > 0
define α = P( reject H0 |H0 is true) = .05
define β = P( fail to reject H0 |Ha is true) ≤ .20
Goal: find a common sample size n for a test with α = .05
and β ≤ .20 when in fact p1 exceeds p2 by .1
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.40
Solution:
Way to approach:
1. For α = .05, we use the test statistic Z = √p̂p11−q1p̂2 −0
p2 q2 , such
n
+
n
that the rejection region as Z ≥ z.05 = 1.645
2. Rewrite the rejection
p region in 1., we have reject H0 if
:p̂1 − p̂2 ≥ 1.645 p1nq1 + p2nq2
3. For β ≤ .20, we fix it at the largest acceptable value so
P(p̂1 − p̂2 ≤ c|p1 − p2 = .1) = .20 for some c.
4. Simplify the equation in 3. the acceptance region is restated
p̂2 −.1
as: fail to reject H0 if √p̂1p−
p2 q2 = z.20 where z.20 = −.84
1 q1
n
Fang-I Chu
+
n
PSTAT 120B Probability and Statistics
Exercise 10.40
Solution:
Way to approach:
p
5. Let p̂1 − p̂2 ≥ 1.645 p1nq1 + p2nq2 , substitute this in 4. After
simplifying, We obtain 2.485 = √ p1 q1.1 p2 q2
n
+
n
6. Using the hint, set p1 = p2 = .5 as a worse case scenario.
7. For the scenario p1 = p2 = .5, we have 2.485 = r .1
.
.5×.5
n
1
1
n+n
Solve this equation to obtain n = 308.76
8. Round up n, the common sample size for the researcher’s test
should be n = 309.
Fang-I Chu
PSTAT 120B Probability and Statistics