Download Week 10 - UCSB Statistics - University of California, Santa Barbara

PSTAT 120B Probability and Statistics - Week 10
Fang-I Chu
University of California, Santa Barbara
June 7, 2013
Fang-I Chu
PSTAT 120B Probability and Statistics
Topic for review
Exercise #10.55
Exercise #10.82
Exercise #10.88
Exercise #10.92
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.55
10.55
A check-cashing service found that approximately 5% of all checks
submitted to the service were bad. After instituting a
check-verification system to reduce its losses, the service found
that only 45 checks were bad in a random sample of 1124 that
were cashed. Does sufficient evidence exist to affirm that the
check-verification system reduced the proportion of bad checks?
What attained significance level is associated with the test? What
would you conclude at the α= .01 level?
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.55
Solution:
Known:
approximately 5% of all checks submitted to the service were
bad.
45 checks were bad in a random sample of 1124 after
instituting a check-verification system.
Goal: decide whether check-verification system reduced the
proportion of bad checks. i.e. Hypothesis: H0 : p = .05 versus
Ha : p < .05
Way to approach:
45
Compute p̂ = 1124
= 0.040
Formula for test statistic: Z =
q p̂−p0
p0 (1−p0 )
n
∼ N (0, 1)
0.04−0.05
= −1.538
Under H0 , test statistic is Z = √
0.05×0.95
1124
The p-value= P(Z < −1.538) = 0.0616.
Fail to reject H0 with α = .01
There is not enough evidence to conclude that the proportion
of bad checks has decreased from 5%.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.82
10.82
Exercises 8.83 and 10.73 presented some data collected in a 1993
study by Susan Beckham and her colleagues. In this study,
measurements of anterior compartment pressure (in millimeters of
mercury) were taken for ten healthy runners and ten healthy
cyclists. The researchers also obtained pressure measurements for
the runners and cyclists at maximal O2 consumption. The data
summary is given in the accompanying table.
condition
Rest
80%maximal O2
consumption
Maximal O2 consumption
Fang-I Chu
Runners
Cyclists
Mean
s
Mean
s
14.5
3.92
11.1
3.98
12.2
3.49
11.5
4.95
19.1
16.9
12.2
PSTAT 120B Probability and Statistics
4.67
Exercise 10.82
10.82 (a)
(a)Is there sufficient evidence to support a claim that the
variability of compartment pressure differs for runners and cyclists
who are resting? Use α = .05.
Solution:
Known:
Let σ12 , σ22 denote the population variances for compartment
pressure for resting runners and cyclists, respectively.
Goal: test whether the variability of compartment pressure
differs for runners and cyclists. i.e. Hypothesis:
H0 : σ12 = σ22 , Ha : σ12 6= σ22
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.82
10.82 (a)
(a)Is there sufficient evidence to support a claim that the
variability of compartment pressure differs for runners and cyclists
who are resting? Use α = .05.
Solution:
Way to approach:
Note s12 = 3.982 , s22 = 3.922 , as sample variance for runners
and cyclists, respectively.
S2
Formula for F- statistic: F = S12 ∼ Fn1 −1,n2 −1, α2
2
2
Compute F-statistics: F = 3.98
3.922 = 1.03
With α = .05, F9,9,.025 = 4.03 (look up from table)
1.03 < 4.03, we fail to reject H0 with α = .05
There is no evidence to conclude that the variability of
compartment pressure is different for runners and cyclists.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.82
10.82 (b)
(b) i. What can be said about the attained significance level using
a table in the appendix?
(b) ii. Applet Exercise What can be said about the attained
significance level using the appropriate applet?
Solution:
(i) From Table 7, p-value > .1.
(ii) Using the Applet, 2P(F > 1.03) = 2(.4828) = .9656
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.82
10.82 (c)
(c)Is there sufficient evidence to support a claim that the
variability in compartment pressure between runners and cyclists
differs at maximal O2 consumption? Use α = .05.
Known:
Let σ12 , σ22 denote the population variances for compartment
pressure for 80% maximal O2 consumption for runners and
cyclists, respectively.
Goal: test whether the variability of compartment pressure
differs for runners and cyclists. i.e. Hypothesis:
H0 : σ12 = σ22 , Ha : σ12 6= σ22
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.82
10.82 (c)
(c)Is there sufficient evidence to support a claim that the
variability in compartment pressure between runners and cyclists
differs at maximal O2 consumption? Use α = .05.
Solution:
Way to approach:
Note s12 = 16.92 , s22 = 4.672 , as sample variance for runners
and cyclists, respectively.
S2
Formula for F- statistic: F = S12 ∼ Fn1 −1,n2 −1, α2
2
2
Compute F-statistics: F = 16.9
4.672 = 13.096
With α = .05, F9,9,.025 = 4.03 (look up from table)
13.096 > 4.03, we reject H0 with α = .05
There is evidence to conclude that the variability of
compartment pressure between runners and cyclists exits.
Fang-I Chu
PSTAT 120B Probability and Statistics
#10.82
10.82 (d)
(d)i. What can be said about the attained significance level using
a table in the appendix?
(d)ii. Applet Exercise What can be said about the attained
significance level using the appropriate applet?
Solution:
(i) From Table 7, p-value< 0.005.
(ii) Using the Applet, 2P(F > 13.096) = 2(.00036) = .00072
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.88
10.88
Refer to Exercise 10.2. Find the power of the test for each
alternative in (a)-(d).
(a) p = .4
(b) p = .5
(c) p = .6
(d) p = .7
(e) Sketch a graph of the power function.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.2
10.2
An experimenter has prepared a drug dosage level that she claims
will induce sleep for 80% of people suffering from insomnia. After
examining the dosage, we feel that her claims regarding the
effectiveness of the dosage are inflated. In an attempt to disprove
her claim, we administer her prescribed dosage to 20 insomniacs
and we observe Y , the number for whom the drug dose induces
sleep. We wish to test the hypothesis H0 : p = .8 versus the
alternative, Ha : p < .8. Assume that the rejection region y ≤ 12 is
used.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.88
10.88
Refer to Exercise 10.2. Find the power of the test for each
alternative in (a)-(d).
Known:
Y is binomial with parameter n = 20 and p
Hypothesis: H0 : p = .8 versus Ha : p < .8
Rejection Region: reject H0 if y ≤ 12
definition of power for a test: P( reject H0 |Ha true)
Goal: Find the power of the test for alternative p = .4
Note: Power of a test =1- type II error
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.88
Solution
(a) p = .4
Way to approach:
P( reject H0 |Ha true) = P(Y ≤ 12|p = .4)
12 X
20
=
0.4y 0.620−y
y
y =0
= 0.979
(b) p = .5
Way to approach:
P( reject H0 |Ha true) = P(Y ≤ 12|p = .5)
12 X
20
=
0.5y 0.520−y
y
y =0
= 0.86
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.88
Solution
(c) p = .6
Way to approach:
P( reject H0 |Ha true) = P(Y ≤ 12|p = .6)
12 X
20
=
0.6y 0.420−y
y
y =0
= 0.584
(d) p = .7
Way to approach:
P( reject H0 |Ha true) = P(Y ≤ 12|p = .7)
12 X
20
=
0.7y 0.320−y
y
y =0
= 0.228
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.88
(e) Sketch a graph of the power function.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.92
10.92
Consider the situation described in Exercise 10.91. What is the
smallest sample size such that an α = .05-level test has power at
least .80 when µ = 8?
10.91
Let Y1 , Y2 , . . . , Y20 be a random sample of size n = 20 from a
normal distribution with unknown mean µ and known variance
σ 2 = 5. We wish to test H0 : µ = 7 versus Ha : µ > 7.
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.92
Solution
Known:
Yi follows normal distribution with unknown mean µ and
known σ 2 = 5, sample size is 20.
Hypothesis:H0 : µ = 7 versus Ha : µ > 7
Goal: Find the smallest sample size such that an α = .05-level
test has power at least .80 when µ = 8 i.e. Ha : µ = 8
Fang-I Chu
PSTAT 120B Probability and Statistics
Exercise 10.92
Solution
Way to approach:
from 10.91(a), we obtain rejection region as reject H0 if
Ȳ > 7..82. (Check this!)
Using the definition of power, we write
P(reject H0 |Ha true) = P(Ȳ > 7.82|µ = 8)
7.82 − 8
= P(Z > q
) = .80
5
n
√ 5 = −.84
From table, we have z.80 = −.84. So we have 7.82−8
n
Solve this equation, we obtain n = 108.89.
After rounding up, at least 109 observations must be taken
such that an α = .05-level test has power at least .80 when
µ = 8.
Fang-I Chu
PSTAT 120B Probability and Statistics