Download math-111 test 3 answers

MATH-111 DUPRE’ TEST 3 (F2009)
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Suppose that Y is a normal random variable and that E[Y] = 200
and that SD(Y)= 28. Suppose that X is the average of 16 independent random observations of
Y. Calculate the probability that
1.
X is between 192 and 203
__normalcdf(192,203,200,28/4)=.5393333918___
2.
X is = 201, exactly
____ZERO___________________
3.
|X -201| is less than 5
_normalcdf(201-5,201+5,200,28/4)=.5204625387_
We have a box containing 50 blocks of unknown color. We ASSUME as a NULL
HYPOTHESIS, H, that exactly 20 are red. We draw 12 blocks, at random.
4.
What is the expected number of red blocks given H?__12*(20/50)=4.8_____
5.
What is the standard deviation in the number of red blocks DRAWN, given H and also
given that we draw WITH REPLACEMENT?
__sqrt(4.8*3/5)=1.697056275__
6.
What is the standard deviation in the number of red blocks DRAWN, given H and also
given that we draw WITHOUT REPLACEMENT?
__sqrt(4.8*3/5)*sqrt(38/49)=1.494479638____
7.
Given H, what is the probability that when we draw 12 blocks, we get no more than 1
red block, assuming we draw without replacement?
(20 nCr 0)(30 nCr 12)/(50 nCr 12)+(20 nCr 1)(30 nCr 11)/(50 nCr 12)=.0097120479
Suppose a population of fish has normally distributed weight with standard deviation
9 pounds, and we have a sample of size 25 with mean 203 pounds.
8.
What is the MARGIN OF ERROR of the 95 percent confidence interval for the true
population mean weight?
ME=9*invNorm(.975,0,1)/sqrt(25)=3.527935175 or use ZInterval in TEST menu
9.
What is the 95 percent CONFIDENCE INTERVAL for the true population mean
weight?
203 plus or minus 3.528 or (199.472, 206.528)
10.
What is the P-VALUE of the sample data as evidence that the true mean weight of this
population of fish exceeds 200 pounds?
USE Z-Test from TEST menu
P-VALUE=.0477903304
Suppose that we assume that fish weight is normally distributed and that we do not
know the population standard deviation of the fish weight but we have a sample of 25 fish
with mean 203 pounds and standard deviation 12 pounds.
11.
What is the MARGIN OF ERROR of the 95 percent confidence interval for the true
population mean weight?
Use TInterval from TEST menu
ME=4.9534
12.
What is the 95 percent CONFIDENCE INTERVAL for the true population mean
weight?
Use TInterval from TEST menu
203 plus or minus 4.9534 or (198.05, 207.95)
13.
What is the P-VALUE of this sample data as evidence that the true mean weight of
fish in this population exceeds 200 pounds?
Use T-Test from TEST menu
P-VALUE=.1116757391
14.
What is the P-VALUE of this sample data as evidence that the true mean weight of
fish in this population is NOT EQUAL to 200 pounds?
Use T-Test from TEST menu
P-VALUE=.223514781
Notice that the answer here is simply double that of the previous problem.
15.
Senator Snort is concerned he is going to lose the upcoming election. He asked 10
voters chosen at random and only 4 said they will vote for him, the other 6 are going to vote
for his opponents. At the level of significance .15 should he be concerned?
P-VALUE=binomcdf(10, .5, 4)=.376953125 >.15, so NO he should NOT be concerned.
The 1-PropZTest is not really accurate for such a small sample,
but it gives the P-VALUE=.2635445651 > 1.5, so if you had used the 1-PropZTest, you would
have reached the same result.
BUT BEWARE THAT ON THE FINAL EXAM YOU MUST USE THE binomcdf FOR ALL
PROPORTION PROBLEMS INVOLVING HYPOTHESIS TESTING. The 1-PropZTest is
just not accurate enough unless the sample size is enormous.