Exercise 5. Assume that we have observed the following values from a normal distribution with known variance o2 = 1 and unknown mean u. 1.23 -0.67 1.16 1.67 -0.24 2.99 0.02 1.17 0.27 1.21. Test the hypothesis Ho:μ = 0 against the alternative H₁: μ0 at significance level a = 5%. The table gives Þ(x) = X 1 = [² e-4²2²0 dt √2π and this corresponds to the shaded area in the figure to the right. Þ(x) is the probability that a random variable, normally distributed with zero mean and unit variance, will be less than or equal to x. Þ(x) X Þ(x) X Þ(x) 0.00 0.5000 0.50 0.6915 1.00 0.05 0.5199 0.55 0.7088 1.05 0.10 0.5398 0.60 0.7257 1.10 0.15 0.5596 0.65 0.7422 1.15 0.20 0.5793 0.70 0.7580 1.20 0.25 0.5987 0.75 0.30 0.6179 0.80 0.35 0.6368 0.85 0.40 0.6554 0.90 0.45 0.6736 0.95 0.7734 1.25 0.7881 1.30 0.8023 1.35 0.8159 1.40 0.8289 1.45 0.50 0.6915 1.00 0.8413 Þ(x) X Þ(x) X Þ(x) 0.8413 1.50 0.9332 2.00 0.9772 2.50 0.8531 1.55 0.9394 2.05 0.9798 2.55 0.8643 1.60 0.9452 2.10 0.9821 2.60 0.8749 1.65 0.9505 2.15 0.9842 2.65 0.8849 1.70 0.9554 2.20 0.9861 2.70 X 0.8944 1.75 0.9599 2.25 0.9032 1.80 0.9641 2.30 0.9115 1.85 0.9678 2.35 0.9192 1.90 0.9713 2.40 0.9265 1.95 0.9744 2.45 X Þ(x) 0.9938 0.9946 0.9953 0.9960 0.9965 0.9878 2.75 0.9970 0.9893 2.80 0.9974 0.9906 2.85 0.9978 0.9918 2.90 0.9981 0.9929 2.95 0.9984 1.50 0.9332 2.00 0.9772 2.50 0.9938 3.00 0.9987
Exercise 5. Assume that we have observed the following values from a normal distribution with known variance o2 = 1 and unknown mean u. 1.23 -0.67 1.16 1.67 -0.24 2.99 0.02 1.17 0.27 1.21. Test the hypothesis Ho:μ = 0 against the alternative H₁: μ0 at significance level a = 5%. The table gives Þ(x) = X 1 = [² e-4²2²0 dt √2π and this corresponds to the shaded area in the figure to the right. Þ(x) is the probability that a random variable, normally distributed with zero mean and unit variance, will be less than or equal to x. Þ(x) X Þ(x) X Þ(x) 0.00 0.5000 0.50 0.6915 1.00 0.05 0.5199 0.55 0.7088 1.05 0.10 0.5398 0.60 0.7257 1.10 0.15 0.5596 0.65 0.7422 1.15 0.20 0.5793 0.70 0.7580 1.20 0.25 0.5987 0.75 0.30 0.6179 0.80 0.35 0.6368 0.85 0.40 0.6554 0.90 0.45 0.6736 0.95 0.7734 1.25 0.7881 1.30 0.8023 1.35 0.8159 1.40 0.8289 1.45 0.50 0.6915 1.00 0.8413 Þ(x) X Þ(x) X Þ(x) 0.8413 1.50 0.9332 2.00 0.9772 2.50 0.8531 1.55 0.9394 2.05 0.9798 2.55 0.8643 1.60 0.9452 2.10 0.9821 2.60 0.8749 1.65 0.9505 2.15 0.9842 2.65 0.8849 1.70 0.9554 2.20 0.9861 2.70 X 0.8944 1.75 0.9599 2.25 0.9032 1.80 0.9641 2.30 0.9115 1.85 0.9678 2.35 0.9192 1.90 0.9713 2.40 0.9265 1.95 0.9744 2.45 X Þ(x) 0.9938 0.9946 0.9953 0.9960 0.9965 0.9878 2.75 0.9970 0.9893 2.80 0.9974 0.9906 2.85 0.9978 0.9918 2.90 0.9981 0.9929 2.95 0.9984 1.50 0.9332 2.00 0.9772 2.50 0.9938 3.00 0.9987
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![Exercise 5. Assume that we have observed the following values from a normal distribution with
known variance o2 = 1 and unknown mean u.
1.23 -0.67 1.16 1.67 -0.24 2.99 0.02 1.17
0.27 1.21.
Test the hypothesis Ho:μ = 0 against the alternative H₁: μ0 at significance level a = 5%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb7f27c73-756c-4a47-8ea7-79b047cfb8fe%2F0b5f31c9-7afa-4288-8937-6b0a704864cd%2Fke09z3m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Exercise 5. Assume that we have observed the following values from a normal distribution with
known variance o2 = 1 and unknown mean u.
1.23 -0.67 1.16 1.67 -0.24 2.99 0.02 1.17
0.27 1.21.
Test the hypothesis Ho:μ = 0 against the alternative H₁: μ0 at significance level a = 5%.
![The table gives
Þ(x) =
X
1
= [² e-4²2²0
dt
√2π
and this corresponds to the shaded area in the
figure to the right. Þ(x) is the probability that
a random variable, normally distributed with
zero mean and unit variance, will be less than
or equal to x.
Þ(x) X Þ(x) X Þ(x)
0.00 0.5000 0.50 0.6915 1.00
0.05 0.5199 0.55 0.7088 1.05
0.10 0.5398 0.60 0.7257 1.10
0.15 0.5596 0.65 0.7422 1.15
0.20 0.5793 0.70 0.7580 1.20
0.25 0.5987 0.75
0.30 0.6179 0.80
0.35 0.6368 0.85
0.40 0.6554 0.90
0.45 0.6736 0.95
0.7734 1.25
0.7881 1.30
0.8023 1.35
0.8159 1.40
0.8289 1.45
0.50 0.6915 1.00 0.8413
Þ(x)
X
Þ(x) X
Þ(x)
0.8413 1.50 0.9332 2.00 0.9772 2.50
0.8531 1.55 0.9394 2.05 0.9798 2.55
0.8643 1.60 0.9452 2.10 0.9821 2.60
0.8749 1.65 0.9505 2.15 0.9842 2.65
0.8849 1.70 0.9554 2.20 0.9861 2.70
X
0.8944 1.75 0.9599 2.25
0.9032 1.80 0.9641 2.30
0.9115 1.85 0.9678 2.35
0.9192 1.90 0.9713 2.40
0.9265 1.95
0.9744 2.45
X
Þ(x)
0.9938
0.9946
0.9953
0.9960
0.9965
0.9878 2.75 0.9970
0.9893 2.80 0.9974
0.9906 2.85 0.9978
0.9918 2.90 0.9981
0.9929 2.95
0.9984
1.50 0.9332 2.00 0.9772 2.50 0.9938 3.00
0.9987](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb7f27c73-756c-4a47-8ea7-79b047cfb8fe%2F0b5f31c9-7afa-4288-8937-6b0a704864cd%2Fbosprt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The table gives
Þ(x) =
X
1
= [² e-4²2²0
dt
√2π
and this corresponds to the shaded area in the
figure to the right. Þ(x) is the probability that
a random variable, normally distributed with
zero mean and unit variance, will be less than
or equal to x.
Þ(x) X Þ(x) X Þ(x)
0.00 0.5000 0.50 0.6915 1.00
0.05 0.5199 0.55 0.7088 1.05
0.10 0.5398 0.60 0.7257 1.10
0.15 0.5596 0.65 0.7422 1.15
0.20 0.5793 0.70 0.7580 1.20
0.25 0.5987 0.75
0.30 0.6179 0.80
0.35 0.6368 0.85
0.40 0.6554 0.90
0.45 0.6736 0.95
0.7734 1.25
0.7881 1.30
0.8023 1.35
0.8159 1.40
0.8289 1.45
0.50 0.6915 1.00 0.8413
Þ(x)
X
Þ(x) X
Þ(x)
0.8413 1.50 0.9332 2.00 0.9772 2.50
0.8531 1.55 0.9394 2.05 0.9798 2.55
0.8643 1.60 0.9452 2.10 0.9821 2.60
0.8749 1.65 0.9505 2.15 0.9842 2.65
0.8849 1.70 0.9554 2.20 0.9861 2.70
X
0.8944 1.75 0.9599 2.25
0.9032 1.80 0.9641 2.30
0.9115 1.85 0.9678 2.35
0.9192 1.90 0.9713 2.40
0.9265 1.95
0.9744 2.45
X
Þ(x)
0.9938
0.9946
0.9953
0.9960
0.9965
0.9878 2.75 0.9970
0.9893 2.80 0.9974
0.9906 2.85 0.9978
0.9918 2.90 0.9981
0.9929 2.95
0.9984
1.50 0.9332 2.00 0.9772 2.50 0.9938 3.00
0.9987
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