Your first and last name:___________________________________ SSN:________________________
STAT 211
SPRING 2002
EXAM 3 - FORM A
1.
Which of the following is incorrect?
a. The criteria to choose the minimum variance unbiased estimator is to find the unbiased statistics
first and then choose the one with the smallest variance among them.
b. The MME for  can be computed maximizing the likelihood function.
c. The large sample confidence interval for estimating the true mean should only be used when the
sample size is greater than 40.
d. The main difference between t and z distributions is the t distribution being more spread out than
the z distribution in small samples.
e. The H0 is rejected when the P-value is at most 
2.
We would like to test the true proportion using hypothesis testing. In which case, the test procedure is
valid using normal distribution?
a. n=100, H 0 : p=0.03
H 0 : p=0.97
c. n=1000, H 0 : p=0.002
d. n=1000, H 0 : p=0.0025
e. n=100, H 0 : p=0.12
b.
n=100,
_
3.
What is the confidence level for the interval
a.
b.
c.
d.
e.
x  1.645 

?
n
0.80
0.85
0.90
0.95
0.99
Consider the following three cases and answer questions 4 to 8 using these cases.
Decision
Population
H
0
Accounting method
Maintenance timing
Garment manufacturing
Decision
Accounting method
Maintenance timing
Price increases
Sugar sack content
Clothing expenditure
Fail to reject
H 0 if
_
x  6% (Keep FIFO)
_
x  9.9 lbs.
  5% (moderate)
  10 lbs. (satisfactory)
 = $200 (on target)
Reject
Ha
 > 5% (excessive)
 < 10 lbs. (underweight)
  $200 (off target)
H 0 if
_
x  6% (Switch to LIFO)
_
x  9.9 lbs. (Do maintenance)
(Continue operations)
Garment manufacturing
_
_
$195  x  $205
x  $195 (Revise plan for lower
(Use present plan)
expenditures) or
_
x  $205 (Revise plan for higher
expenditures)
4.
Which of the following is the type I error for accounting method?
a. Keep FIFO when price increases are moderate
b. Switch to LIFO when price increases are excessive
c. Keep FIFO when price increases are excessive
d. Switch to LIFO when price increases are moderate
e. There is no type I error
5.
Which of the following is the probability of type II error for maintenance timing when =9.85, n=100
and =0.5?
a. 0
b. 0.0228
c. 0.1587
d. 0.8413
e. 0.9772
6.
Which of the following is correct for the garment manufacturing?
_
a.
H 0 is rejected if x  200 and present plan is used
b.
H 0 is not rejected if x  200 and present plan is used
c.
H 0 is rejected if x  200 and revised plan is used
d.
H 0 is not rejected if x  200 and revised plan is used
e.
none of the above
_
_
_
_
7.
If x =9.8, n=100 and =0.5 for maintenance timing, which of the following is the test statistics?
a. -4
b. -2
c. -0.2
d. 2
e. 4
8.
If x =9.8, n=100 and =0.5 for maintenance timing, which of the following is the P-value?
a. 0
b. 0.0228
c. 0.1587
d. 0.8413
e. 0.9772
9.
If I argue that the only difference between the intervals (50,105) and (25,130) for  is the confidence
level and everything else is the same, which of the following is correct?
a. the sample mean is 155
b. The first interval is wider than the second interval
c. The first interval is narrower than the second interval
d. These intervals can only be computed using the t-distribution.
e. These intervals can only be computed using the chisquare distribution.
_
10. Which of the following is not a statistical hypothesis?
a.
 = 10
^
b.
c.
d.
e.
p > 0.5
 > 100
2 = 5
 3
_
11. For the normal distribution, it is proved in the class that the MLE for  and  are x and
 (x
_
i
 x) 2
i
n
,
respectively.
Using
the
normally
distributed
breaking
strength
data
_
^
_
^
  x  384.4 and  
 ( xi  x) 2
i
n
 18.8584 are computed. Estimate the 90th percentile of
the breaking strength.
a. 353.38
b. 360.26
c. 384.40
d. 408.54
e. 415.42
12. Which of the following is correct?
a. If the P-value=0, we should definitely fail to reject H0.
b. We prefer the probability of type I and type II errors to be close to 1.
c. 99% confidence interval is always narrower than the 95% confidence interval.
d. We prefer the power of the test to be close to 1.
e. all of the above
13. Which of the following is the P(2 <18.508 or 2 > 49.802) where 2 is a chi-squared random variable
with degrees of freedom v=35?
a. 0.01
b. 0.05
c. 0.06
d. 0.94
e. 0.96
Let X1,…,Xm be a random sample from normal distribution with the mean
Y1,…,Yn be a random sample
are as follows:
1
2
3
4
X
-3
-2
-1
0
Y
-4
-3
-2
-1
from normal distribution with the mean
5
1
1
6
2
2
7
3
3
8
4
sample mean
0
0
Answer questions 14 and 15 using the information above.
14. Which of the following is the point estimator for
a. -2
b. -1
c. 0
d. 1
e. 2
1 +  2  1?
2
1
and the variance
 12 .
Let
and the variance  2 . The data
sample variance
0.8165
1.0351
2
_
_
15. Which of the following is the point estimate of Var( x - y )?
a. -0.2186
b. -0.0128
c. 0
d. 0.2460
e. 1.8516
A statistician is claiming that exactly half of all MBA's continue their formal education by taking courses
within 10 years of graduation. Using a sample of 200 persons, he found that 111 had taken coursework
since receiving their MBA. Answer questions 16 to 18 using this information.
16. Which of the following is the corresponding alternative hypothesis?
a. H a :   10
b.
H 0 :   10
c.
H a : p  111 / 200
d.
H a : p  0.5
e.
H 0 : X  111
17. Which of the following is the 95% confidence interval for the true proportion of persons that has taken
coursework since receiving their MBA?
a. (0.4861,0.6239)
b. (0.4972,0.6128)
c. (0.5196,0.5904)
d. (0.5216,0.5884)
e. We do not have a large sample to compute the interval.
18. At the significance level of 0.05, should the statistician accept the claim?
a. Since the P-value is 0.0594, claim should be accepted.
b. Since the P-value is 0.0297, claim should be accepted.
c. Since the P-value is 0.1188, claim should be accepted.
d. Since the P-value is 0.0594, claim should be rejected.
e. Since the P-value is 0.1188, claim should be rejected.
19. The confidence interval for  is
_
s _
s 
 x  1.33
, x  2.552
 . Which of the following is the
19
19 

corresponding confidence level?
a. 0.85
b. 0.89
c. 0.90
d. 0.95
e. 0.99
The normally distributed tire mileage data (in thousand of miles) are collected for a new tread design. The
data are based on the 10 tires with the mean 53.87 and the standard deviation 7.76. Answer questions 20 to
23 using this information.
20. Which of the following is the 95% confidence interval for the true average mileage?
a. (36.3169,71.4231)
b. (38.6604,69.0796)
c. (48.3192,59.4208)
d. (49.0603,58.6797)
e.
(49.3719,58.3681)
21. True average mileage for the tires with the new tread design is claimed to be 55 or more. Which of the
following would be the corresponding null hypothesis?
a.   55
b.  > 55
c.  = 55
d.   55
e.  < 55
22. Which of the following would be the test statistics for testing true average mileage for the tires with
the new tread design being 55 or more?
a. z=-0.4605
b. t=-0.4605
c. z=-0.1456
d. t=-0.1456
e. z=-1.13
23. If I claim that the standard deviation is 6, do I have enough evidence to agree with the claim based on
the data using the significance level of 0.05?
a. Yes because 6 fall in the 95% confidence interval for .
b. No because 6 fall in the 95% confidence interval for .
c. Yes because 6 do not fall in the 95% confidence interval for .
d. No because 6 do not fall in the 95% confidence interval for .
e. Yes because 6 do not fall in the 95% confidence interval for 2.
24. You are interested in the true mean income of persons in various professions. Consider surgeons and
teachers. Traditionally, surgeons have the standard deviation, s =$25000 and teachers have the
standard deviation, t =$2000. How many teachers should you consider to find the bound on the error
estimation  500 with the 95% confidence level?
a. 1
b. 62
c. 98
d. 500
e. 9604
25. If the test statistics is z=0.9, which of the following is the P-value for the lower tailed test?
a. 0.1013
b. 0.1841
c. 0.3682
d. 0.8159
e. 0.8987
26. If the test statistics is z=-1.74, which of the following is the P-value for the two-tailed test?
a. 0
b. 0.0205
c. 0.0409
d. 0.0618
e. 0.0818
27. Which of the following does not change the width of a large sample confidence interval for ?
_
a.
b.
c.
x.
The standard deviation of the population.
The confidence level.
d.
e.
The sample size.
All of them except (a)
Answer Key:
1.b
2.e
11.d
12.d
21.d
22.b
3.c
13.c
23.a
4.d
14.b
24.b
5.c
15.d
25.d
6.b
16.d
26.e
7.a
17.a
27.a
8.a
18.c
9.c
19.b
10.b
20.c
Formulas
Random Sample:The random variables X1, X2, ….,Xn are said to form a random sample of size n if the X i's are
independent random variables and every X i's has the same probability distribution.
If X1, X2, ….,Xn are said to form a random sample of size n with the mean  and the variance 2, the sampling
n
_
distribution of
has the mean  and the variance 2/n, the sampling distribution of
x
x
i 1
i
has the mean n and the
variance n2, and so on.
Point estimate of a parameter : single number that can be regarded as the most plausible value of . A point
^
estimator,

^
=  + error of estimation.
^

is an unbiased estimator of  if E(  )=  for every possible value of .
^
Otherwise, it is biased and Bias = E(  )- .
Minimum Variance Unbiased Estimator (MVUE): Among all estimators of  that are unbiased, choose the one that has
^
minimum variance. The resulting
^
The Invariance Principle: Let

^
is MVUE.
^
 1 .,  2 ,...,  m
 1 ,  2 ,...,  m .
be the MLE's of the parameters
^
^
Then the MLE of any
^
function h(  1 ,  2 ,...,  m ) of these parameters is the function h(  1 .,  2 ,..., 
m
) of the MLE's
Confidence Interval for a Population Mean, 
Suppose that the parameter of interest is the population mean,  and that the population distribution is normal and the
value of the population standard deviation  is known. Let X1, X2, ....,Xn be a random sample. Then 100(1-)%
confidence interval for  is
_
 _
 
 x  z / 2
, x  z / 2

n
n

Choosing the sample size: Bound on the error estimation is
z / 2

_
. I mean
n
x
will be within
The sample size required to estimate a population mean  to within an amount B= z / 2
confidence is
 z / 2 
n= 

 B 
 2 z / 2 
n= 

 w 

z / 2
of .
n
with 100(1)%
n
2
.

The same formula can be written using the interval width, w= 2 z / 2
2
.
Large Sample Confidence Interval for 
Suppose that the parameter of interest is the population mean,  and that
a. X1, X2, ...,Xn is a random sample from a population distribution with mean,  and standard deviation, .

n
then
_
For the large sample size n, the CLT implies that x has approximately a normal distribution for any population
distribution.
The value of the population standard deviation  may not be known. Instead, the value of the sample standard
deviation s may be known.
b.
c.
If
n
is
sufficiently
large
(n>40),
100(1-)%
large
sample
confidence
interval
for

is

s
s 
 x  z / 2
, x  z / 2
 .
n
n

_
_
A General Large Sample Confidence Interval
When the estimator satisfies the following properties, the confidence interval can be constructed.
a. The estimator has approximately a normal population distribution
b. It is at least unbiased
c. standard deviation of the estimator is known
Large Sample Confidence Interval for a population proportion, p
^
If n is sufficiently large ( n
^

 p  z / 2


^
^
p  10
p(1  p) ^
, p  z / 2
n
^


n1  p   10 ),


^
^ 
p(1  p) 

n


and
100(1-)% large sample confidence interval for p is
^
Choosing the sample size: Bound on the error estimation is
^
^
I mean
p
will be within
^
p (1  p )
n
z / 2
z / 2
^
p (1  p )
.
n
of p. The sample size required to estimate a population proportion p to within an amount
^
^


z2 / 2 p1  p 
p (1  p )

 . The same formula can be written using
B= z / 2
with 100(1)% confidence is n=
2
n
B
^
^


^
^
4 z2 / 2 p1  p 
p(1  p)

 . The conservative sample size can be found
the interval width, w= 2 z / 2
then n=
2
n
w
^
^
when
^
^
p = 1  p =0.5
Intervals based on a Normal Population Distribution:
When the sample size is small and the population of interest is normal, so that X 1, X2, ...,Xn constitutes a random
sample from a normal distribution with both  and  unknown. 100(1-)% confidence interval for  is
_
s _
s 
 x  t / 2;n 1
, x  t / 2;n 1

n
n

.
Prediction Interval for a Single Future Value:
Let X1, X2, ...,Xn be a random sample from a normal population distribution and we wish to predict the value of X n+1, a
single future observation. 100(1-)% prediction interval for Xn+1 is
_
1 _
1
 x  t / 2;n 1 s 1  , x  t / 2;n 1 s 1  

n
n 

Confidence Intervals for the Variance, 2 and Standard Deviation,  of a Normal Population :
The population of interest is normal, so that X 1, X2, ...,Xn constitutes a random sample from a normal distribution with
parameters  and 2. Then 100(1-)% confidence interval for 2 is
 (n  1)  s 2 (n  1)  s 2

, 2
 2
 1 / 2;n 1

/
2
;
n

1


.


One-sided (One-tailed) test:
Lower tailed:
H0: population characteristics  claimed constant value
(Left-sided)
Ha: population characteristics < claimed constant value
H0: population characteristics  claimed constant value
Ha: population characteristics > claimed constant value
Upper tailed:
(Right-sided)
Two-sided (Two-tailed) test: H0: population characteristics = claimed constant value
Ha: population characteristics  claimed constant value
Hypothesis testing involves two complementary actions or choices, reject H0 and fail to reject H0.
 = P(Type I error) = P(reject H0 when it is true) which is the significance level
 = P(Type II error) = P(fail to reject H0 when it is false)
1- = P(reject H0 when it is false) which is the power of the test.
Population characteristics: Population mean, 
_
0 is the claimed constant,
x
is the sample mean,
 
__
X

and
n
s __ 
X
s
are the population and sample
n
_
standard deviation of
x , respectively.
_
Test statistic :
z
x  0

_
if  is known,
__
x  0
z
s __
X
if  is unknown for a large sample (n >40)
X
_
t
x  0
s __
if  is unknown for a normal population distribution with small sample
X
Population characteristics: Population proportion, p
p0 is the claimed constant,
p 
p0 (1  p0 )
n
is the standard deviation of p.
^
If np0  10 and n(1-p0)  10 is satisfied, test Statistics is
z
p  p0
p
.
Depending on what your test statistics is based on Decision can be made in one of the two ways for the tests above:
a.
Let z* or t* be the computed test statistic values.
When z is used as test statistics
Lower tailed test
P-value = P(z<z*)
Upper tailed test
P-value = P(z>z*)
Two-tailed test
P-value = 2P(z>|z*|)=2P(z<-|z*|)
When t is used as test statistics
P-value = P(t<t*)
P-value = P(t>t*)
P-value = 2P(t > |t*| )=2P(t <- |t*| )
In each case, you reject H0 if P-value   and fail to reject H0 (accept H0) if P-value > 
b.
Rejection region for level  test:
When z is used as test statistics
Lower tailed test
z  -z
Upper tailed test
z  z
Two- tailed test
z  -z/2 or z  z/2
When t is used as test statistics
t  -t;n-1
t  t;n-1
t  -t/2;n-1 or t  t/2;n-1