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“An Aggie does not lie, cheat, or steal or tolerate those who do.”
EXAM 3 – FORM A
STAT 211
SPRING 2005
The heat evolved in calories per gram of a cement mixture is approximately normally distributed with the unknown mean 
(true average calories per gram of a cement mixture) and the true standard deviation =2. We collected the random sample
of 10 specimens and computed the sample average calories per gram of a cement mixture as 99. Answer the following 5
questions using this information.
1.
Which of the following is the standard error for the total calories of 10 specimens?
(a) 0.6325
(b) 2
(c) 3.1623
(d) 4.4721
(e) 6.3246
 10 
 10 
SE   xi   Var   xi   10Var ( xi )  10  2 10  6.3246
 i 1 
 i 1 
2.
Which of the following is the standard error for the average calories of 10 specimens?
(a) 0.6325
(b) 2
(c) 3.1623
(d) 4.4721
(e) 6.3246
Var ( xi )

2
_
_
SE x   Var x  


 0.6325
10
10
10
 
 
3.
Which of the following is the 98% confidence interval for ?
(a) (97.70, 100.27)
(b) (97.53 , 100.47)
98% confidence interval for  is
4.
_
x  z / 2
s
n
 99  2.33
(c) (94.34, 103.66)
2
where
10
z / 2  z 0.01  2.33
How many specimens needed to compute the 98% confidence interval with the interval width of 1 calorie per gram?
(a) 6
(b) 21
(c) 22
(d) 86
(e) 87
 2z  
 2(2.33)2 
n=   / 2   
  86.86 87
1


 w 
2
5.
2
I claim that the true average calories per gram of a cement mixture is 100, which of the following is the corresponding
test statistics for decision making?
(a) -5
(b) -1.6
(c) -1
(d) -0.5
_
z
x  0
/ n

99  100
2 / 10
 1.58
Observations on burst strength, X (lb/in2) were obtained for test nozzle closure welds. The data are shown on the following
table
X 5200 7200 6100 7300 7300 8000 7400 7300 7300 8000 6700 8300
The following descriptive statistics are obtained using MINITAB software.
Variable
X
n
12
Mean
7175
Median
7300
StDev
854
Minimum
5200
Maximum
8300
Q1
6825
Q3
7850
95% confidence interval for the true mean of X’s: (6632.39 , 7717.61)
Answer questions 6 to 11 using the information above.
6.
7.
Which of the following is the point estimate for the true median burst strength?
(a) 854
(b) 5200
(c) 7175
(d) 7300
Sample median
Assume X is drawn from a normal distribution, which of the following is the maximum likelihood estimator for the
true standard deviation in the burst strength?
(a) 246.529
(b) 782.833
(c) 817.64
(d) 854
(e) 668539.67
The MLE for  is
8.
(e) 8300
(n  1) s 2
11(854) 2

 668539.6667 =817.64275
n
12
Assume X is drawn from a normal distribution, which of the following is the point estimate for the 97.5th percentile of
burst strength (X)?
(a) 8848.84
(b) 8029
(c) 8007.65
(d) 1.96
“An Aggie does not lie, cheat, or steal or tolerate those who do.”
EXAM 3 – FORM A
STAT 211
SPRING 2005


Let x* be the 97.5th percentile of burst strength then P(Xx*)=0.975= P z 
x *  
 . Using the standard
 
normal table, x*=+1.96. The point estimate for x* is 7175+1.96(854)=8848.84
9.
Which of the following is the point estimate for P(X>6100)?
(a) 0.9167
(b) 0.8962
(c) 0.8333
(d)0.1038
There are 10 data points larger than 6100 then the point estimate is 10/12=0.8333
(e) 0.08333
10. Assume X is drawn from a normal distribution, which of the following is the point estimate for P(X>6100)?
(a) 0.9167
(b) 0.8962
(c) 0.1038
(d) 0.08333


If X is drawn from a normal distribution then P(X>6100)= P z 


P(X>6100) is P z 
6100   
 . The point estimate for


6100  7175 
  P( z  1.26) =0.8962
854

11. If I claimed that the true mean of X’s is 6900, do the data support my claim using the 98% confidence interval for the
true mean of X’s?
(a) Yes
(b) No
6900 falls in the 95% confidence interval for the true mean which implies it will fall in 98% confidence interval
for the true mean.
_
12. Let X1,…,Xn be a random sample from a normal distribution with the unknown mean,  (known x ) and unknown
2
variance, 2 (known s ). What is the confidence level for the interval
_
x  1.697 
s
?
31
(a) 0.95
(b) 0.90
(c) 0.10
(d) 0.05
(e) I do not have the v=31 on the table to determine the confidence level
The degrees of freedom, v=n-1=30 and t / 2;30  1.697 (because small sample and unknown 2 ) then /2=0.05
and 1-=0.90
13. If you increase the standard deviation by a factor of 16 keeping everything else the same, what happens to the
confidence interval for the true mean?
(a) stay the same
(b) becomes wider
(c) becomes narrower
Standard deviation is on the numerator in width computation. If standard deviation is increased then the width
is also increased. It means wider confidence interval.
14. If you decrease the sample mean by a factor of 16 keeping everything else the same, what happens to the width of the
confidence interval for the true mean?
(a) stay the same
(b) increase
(c) decrease
Change in the sample mean does not effect the width.
15. If the two continuous random variables are independent, which of the following is incorrect?
(a) E(XY)=E(X)E(Y)
(b) Var(X-Y)=Var(X)+Var(Y)
(c) E(1/X)=1/E(X)
(d) Var(2X+3Y)=4Var(X)+9Var(Y)
(e) the conditional pdf, f(x|y) is equal to f(x), the pdf of X
16. Keeping all the information on the data the same, if you only change the upper tailed test to two tailed test for testing
the true mean, which of the following is incorrect?
(a) test statistics does not change
(b) critical value does not change
(c) hypothesized value does not change
17. If the P-value = 0.02, which of the following is incorrect?
(a) Reject Ho with =0.05
“An Aggie does not lie, cheat, or steal or tolerate those who do.”
EXAM 3 – FORM A
STAT 211
SPRING 2005
(b) Fail to reject Ho with =0.01
(c) To make a decision, we need to know if we have a one tailed or two tailed test.
18. Let X1,…,Xn be a random sample of n with E(X)=2, which of the following is an unbiased estimator for ?
n
x
_
(a) x
(b)
i 1
n
x
i
(c)
2n
 
  xi
_
 x
E  x   2 then E    E
 2n
 
2

 
_
i 1
_
x
2n
i
(d)
2

 


19. Suppose  has two unbiased estimators,
^
^
 1 and  2 .
^ 
 
^ 
 
If Var   1   5 and Var 2   2 , which of the following
is the minimum variance unbiased estimator for ?
^
^
(a)
1
(b)
2
^
^
(c) Both
1
and
2
^
^
(c) Neither
1
nor
2
The one with smaller variance
20. Suppose we are interested in the average # of violent acts on TV per hour for a specific network. We believe that the
true average number of violent acts on TV per hour,  is now more than 10. It means we are testing, H 0 :   10
_
versus
H a :   10 . Let x is the sample average number of violent acts on TV per hour. Which of the following is
the possible rejection region for the null hypothesis?
_
(a) x > 11
_
(b) x  9
_
_
(c) x  9.8 or x  10.2
21. Which of the following is an incorrect statement?
(a) If the data are normally distributed, the linear combination is also normally distributed.
(b) If the 45 data points are drawn from an unknown distribution, the sample mean is approximately normally
distributed.
(c) If the 45 data points are drawn from an unknown distribution, the reciprocals of data points are
approximately normally distributed.
22. If we are computing the 63.72% large sample confidence interval for the true mean, which of the following is the
critical value?
(a) -0.91
(b) -0.35
(c) 0.35
(d) 0.64
(e) 0.91
1-=0.6372 then z / 2  z 0.1814  0.91
23. Which of the following cannot be a legitimate null hypothesis? Let  be the expected number of cars stop on the
intersection between 10am and 1 pm.
(a)   3
(b)   3
(c)  > 3
24. If the test statistics is z = -1.73 in the lower tailed test of testing the true mean, would you reject H 0 using  =0.05?
(a) Yes
(b) No
In a lower tailed test if z < - z =-1.645 then reject H0. Since -1.73 < -1.645, reject H0.
25. If the test statistics is z = -1.73 in the two tailed test of testing the true mean, would you reject H 0 using  =0.05?
(a) Yes
(b) No
In a two tailed test if |z| > z / 2 =1.96 then reject H0. Since 1.73 < 1.96, fail to reject H0.