Download Exam 3

STAT 211
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EXAM 3 - FORM A
The cut off values for the z, t and 2 values are as follows:
: Area on the right
z
FALL 2004
t ;17
0.005
2.575
2.898
0.01
2.33
2.567
0.025
1.96
2.11
0.05
1.645
1.74
0.975
0.99
0.995
Remember that P(z>z)==P(z<- z) or P(t>t;n-1)==P(t<-t;n-1)
Use =0.05 whenever it is not specified but needed.
 2;17
35.716
33.408
30.190
27.857
7.567
6.407
5.697
The health of the bear population in Yellowstone National Park is monitored by periodic measurements taken
from an anesthetized bears. A sample of 54 bears has a mean weight of 182.9lb. Let  be population mean of
all such bear weights and  is 121.8lb and assume that data are normally distributed. Answer questions 1 to 5
using this information.
1. To test the claim that the population mean of all such bear weights is less than 200 lb, which of the
following would be the corresponding null hypothesis?
(a) 200
(b) 200
(c) 200
(d) 200
(e) >200
Claim (<200) is the alternative hypothesis so that the complement is the null hypothesis
2. To test the claim that the population mean of all such bear weights is less than 200 lb, which of the
following is the corresponding test statistics?
(a) -17.1
(b) -1.0317
(c) -0.1404
(d) 0.1404
(e) 1.0317
_
z
x  0
/ n

182.9  200
121.8 / 54
 1.0317
3. If the P-value is 0.1515 for testing the claim of the population mean of all such bear weights is less than 200
lb, would you reject H0 with the significance level 0.05?
(a) Yes
(b) No
P-value>0.05 then fail to reject H0
4. Which of the following is the point estimate for the population mean of all such bear weights?
(a) 54
(b) 121.8
(c) 182.9
(d) do not have enough information
The sample mean is the point estimate for the population mean
5. Which of the following is the standard error for the sample mean of bear weights?
(a) 4.1
(b) 16.6
(c)121.8
(d)182.9
(e) 274.7
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Standard error for the sample mean= Var x    2 / n   / n =121.8/ 54
A package of M&M plain candies is labeled as containing 1361 g. and there are 1498 candies, so the mean
weight of the individual candies should be 1361/1498=0.9085 g. The company wants to produce the M&M’s
with weights that do not cheat consumers, but also they do not want to waste production money by having a
mean that is significantly greater than is necessary. In a test of the claim that ≠0.9085 g, a sample of 100
M&M’s is randomly selected. The following is the MINITAB output to analyze this data. Answer questions 6
to 8 using this information.
Test of
mu=0.9085
vs
mu not = 0.9085
STAT 211
EXAM 3 - FORM A
Variable
M&M
n
100
mean
0.9147
stdev
0.03691
SE mean
0.003691
FALL 2004
95% C.I.
(0.90747, 0.92193)
Z
1.68
P-value
0.093
6. Are consumers being cheated?
(a) Yes
(b) No
mu=0.9085 implies that consumers are not cheated.
significance level is 0.05. We fail to reject H0.
7. Which of the following is the point estimate for /?
(a) 0.003691
(b) 0.03691
(c) 0.9147
P-value=0.0093
and
the
(d) 24.7819
_
The point estimate for / is the x/ s =0.9147/0.03691
8. Which of the following is the point estimate for P(average candy weight < 0.9085)?
(a) P(Z<0.0062)
(b) P(Z<0.9085)
(c) P(Z<-0.1680)
(d) P(Z<-1.6798)
(e) P(Z<-1.6882)
The point estimate for P(average candy weight < 0.9085) is P(z<(0.9085-0.9147)/0.003691)
9. In which of the following case, the large sample hypothesis testing for the proportion (not the confidence
interval) should not be computed?
^
(a) n=100, p0=0.05, p =0.15
n p0<10
^
(b) n=50, p0=0.25, p =0.50
^
(c) n=100, p0=0.15, p =0.05
^
(d) n=50, p0=0.50, p =0.25
10. In a lower tailed test, if the test statistics is z = -1.3 and the significance level is 0.05, would you reject H0?
(a) Yes
(b) No
In a lower tailed test, reject H0 if z<-z=-1.645 with =0.05. -1.3 > -1.645 then fail to reject H0
11. In a two tailed test, if the test statistics is z = -1.3 and the significance level is 0.05, would you reject H0?
(a) Yes
(b) No
In a two tailed test, reject H0 if |z| > z/2 = 1.96 with =0.05. | -1.3|=1.3 < 1.96 then fail to reject H0
^
^
^
Let 1 ,  2 and  3 are the estimators for . The expected values and the variances for those estimators are as
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^
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^

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^

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^
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^
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^
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follows: E  1    , E   2   2   2 , E   3    / 3 , Var  1   3 , Var  2   3 and Var   3   1 .
Use this information to answer questions 12 to 13.
12. Which of the following is a correct statement?
^
(a) Only 1 is the unbiased estimator for 
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(b) Only  2 is the unbiased estimator for 
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(c) Only  3 is the unbiased estimator for 
(d) Exactly two of those three estimators are unbiased estimators for 
(e) All three estimators are unbiased estimators for 
13. Which of the following is the MVUE for ?
STAT 211
EXAM 3 - FORM A
^
^
(a) 1
^
(b)  2
^
^
FALL 2004
^
^
(d) both 1 and 3  3
(c)  3
^
(e) (  2 +2)/2
^
^
1 , 3  3 and (  2 +2)/2 are unbiased estimators for . Among these three, (  2 +2)/2 has the
smallest variance which is 0.75.
14. A journal article reported that the null hypothesis of =100 for the two tailed test was rejected because the
P-value was less than the significance level of 0.01. For that normally distributed random sample, the
sample mean was 103.6 and the sample size was 62. Find the largest possible true standard deviation (in
integers)?
(a) 10
(b) 11
(c) 12
(d) 13
(e) 14
In a two tailed test, reject H0 if |z| > z/2 = 2.575 with =0.01.
103.6  100
 / 62
>2.575 then solve this for 
and see that  < 11.0083.
15. Which of the following is effected by the type of the test (lower tailed, upper tailed, two-tailed) for testing
the population mean?
(a) Test statistics
(b) claimed constant
(c) P-value
16. Let X1,…,Xn be a random sample from a distribution with the expected value, E(X)=2+3. Which of the
following is the MME for ?
_
(a) (X-3)/2
(b) 2X+3
(c) x
_
(d) 2 x +3
_
(e) ( x -3)/2
_
Equate the first population moment to the first sample moment and see that 2+3= x then solve for .
17. Which of the following is incorrect for the confidence interval of the population mean?
(a) Keeping everything constant, if the sample mean is increased, the width of the interval does not change.
(b) Keeping everything constant, if the standard deviation is increased, the width of the interval increases.
(c) Keeping everything constant, if the critical value is increased, the width of the interval decreases.
(d) Keeping everything constant, if the sample size is increased, the width of the interval decreases.
Let’s test the claim that heights of female supermodels vary less than the heights of women in general. The
standard deviation of heights of the population of women is 2.5 in. Listed below are the heights (in inches) of
randomly selected supermodels.
71
71
70
69
69.5
70.5
71
72
70
70
69
69.5
69
70
70
68
70
71
Test of mu = 70 vs mu not = 70
Variable
n
Mean
StDev
heights
18
70.028
0.962
99% C.I for 2: (0.4405 , 2.7615)
SE Mean
0.227
95.0% CI
(69.549,70.506)
t
0.12
P-value
0.904
STAT 211
EXAM 3 - FORM A
FALL 2004
Normal Probability Plot for heights
ML Estimates - 95% CI
99
ML Estimates
95
Mean
70.0278
StDev
0.935002
90
Goodness of Fit
Percent
80
AD*
70
60
50
40
30
1.056
20
10
5
1
67
68
69
70
71
72
73
Data
Answer questions 18 to 25 using this information.
18. If you hypothesize the claim, which of the following that you are testing?
(a) Lower tailed test
(b) Upper tailed test
(c) Two tailed test
Claim: heights of female supermodels vary less than the heights of women in general which is 2.5
Variation is measured by the variance and the population characteristic is the true variance. Then
the less than sign implies that the claim is an alternative hypothesis to be a valid hypothesis.
19. Which of the following is the test statistics for the claim?
(a) 0.9254
(b) 1.5865
(c) 2.5172
(d) 6.3363
(e) 114.8098
When the population characteristic is the true variance and the data is normally distributed, test
statistics is chi-square.  2 
(n  1) s 2
 02

17(0.962) 2
(2.5) 2
20. Do the data support the claim with the 0.01 significance?
(a) Yes
(b) No
In a lower tailed test, if the  2  12 ;n1   02.99;17 =6.407 then reject H0. 2.5172 < 6.407 then
reject H0.
21. If I tell you that the true average heights for female supermodels are 70, do the data support my statement?
(a) Yes
(b) No
My statement is mu=70 is the null hypothesis. If you look at the minitab output, you will see that data
supports the claim because we fail to reject H0 .
22. If the height data (18 observations) is sampled from the 500 supermodels, which of the following is the
estimated total height of all super models?
(a) 0.227
(b) 70.028
(c) 1260.504
(d) 35014
(e) 165862
Estimated total heights of 500 supermodels is 500 times the sample mean of 18 supermodels.
23. Which of the following is the MLE for the 95th percentile of the heights of all female super models?
STAT 211
EXAM 3 - FORM A
FALL 2004
(a)70.028
(b) 70.391
(c) 70.401
(d) 70.611
(e) 71.566
If the data is normally distributed, the MLE for the population mean of the heights of all female
super models is the 70.028 and the MLE for the population variance of the heights of all female super
(n  1) s 2 17(0.962) 2
=0.87403044. Then the 95th percentile of the heights of all female super

n
18
models is 70.028+1.645 0.87403044 .
models is
24. Which of the following is the unbiased estimator for the proportion of all female supermodels taller than 70
inches?
(a) 0.3333
(b) 0.50000
(c) 0.6667
(d) 70.028
(e) none of the listed
There are 6 supermodels taller than 70 inches in the sample then 6/18.
25. If you like to test the proportion of all female supermodels taller than 70 inches, which of the following is
the possible rejection region for the null hypothesis? Note that X is the number of female supermodels taller
than 70 inches in the sample.
(a) X=2
(b) X < 4
(c) X>14
(d) X< 3 or X > 15
IT LOOKS LIKE I MADE A MISTAKE ON THIS ONE AND I WILL OMIT THIS QUESTION BOTH
ON THE EXAM AND ON THE EXTRA CREDIT. IT MEANS FREE POINTS WHO ANSWERED
(A)(B)(D) BECAUSE THE POSTED GRADE ASSUMES (C) CORRECT. CORRECTION WILL BE
MADE NEXT WEEK NOT TODAY (11/12/04).