esis HO: mu 200 again1st the altermative Ha: mu opose n 100 samples X1, ... , Xn were selected from a normal distribution with mean mu and variance 900. Determine the rejection region for this test at the level of significance alpha=D0.0125 Let T = (bar{X}- 1200) / (30/sqrt{n} ) where bar{X} = (X1, ... , X,)/n is the sample mean sqrt{n} is the square root of n. OT> 2.24
esis HO: mu 200 again1st the altermative Ha: mu opose n 100 samples X1, ... , Xn were selected from a normal distribution with mean mu and variance 900. Determine the rejection region for this test at the level of significance alpha=D0.0125 Let T = (bar{X}- 1200) / (30/sqrt{n} ) where bar{X} = (X1, ... , X,)/n is the sample mean sqrt{n} is the square root of n. OT> 2.24
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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![According to a researcher, the average level of mercury uptake in wading birds in Everglades has
declined over the past several years. Ten years ago, the average level was 15 parts per million (ppm).
Suppose we are interested in testing whether the average level today is less than 15 ppm. Describe
the type I error for the test of hypothesis.
O The mean mercury level is equal to 15 ppm when in fact the mean is less than 15 ppm.
O The mean mercury level is less than 15 ppm when in fact the mean is equal to 15 ppm.
O The mean mercury level is greater than 15 ppm when in fact the mean is equal to 15 ppm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe7daf6a4-672c-4afd-805d-57a51bb17cdb%2Fa082cb8d-1d1e-470d-90c3-baef3cb596f6%2Favpwx5m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:According to a researcher, the average level of mercury uptake in wading birds in Everglades has
declined over the past several years. Ten years ago, the average level was 15 parts per million (ppm).
Suppose we are interested in testing whether the average level today is less than 15 ppm. Describe
the type I error for the test of hypothesis.
O The mean mercury level is equal to 15 ppm when in fact the mean is less than 15 ppm.
O The mean mercury level is less than 15 ppm when in fact the mean is equal to 15 ppm.
O The mean mercury level is greater than 15 ppm when in fact the mean is equal to 15 ppm.
![Consider the test of hypothesis HO: mu=1200 against the alternative Ha: mu < 1200. Suppose n% D
100 samples X1, ... , Xn were selected from a normal distribution with mean mu and variance 900.
Determine the rejection region for this test at the level of significance alpha=0.0125
Let T = (bar{X} - 1200) / (30/sqrt{n} )
where
bar{X} = (X1,... , Xn)/n is the sample mean
sqrt{n} is the square root of n.
OT> 2.24
OT< -2.24
OT< -1.96
OT> 1.96](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe7daf6a4-672c-4afd-805d-57a51bb17cdb%2Fa082cb8d-1d1e-470d-90c3-baef3cb596f6%2Fiub4qm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the test of hypothesis HO: mu=1200 against the alternative Ha: mu < 1200. Suppose n% D
100 samples X1, ... , Xn were selected from a normal distribution with mean mu and variance 900.
Determine the rejection region for this test at the level of significance alpha=0.0125
Let T = (bar{X} - 1200) / (30/sqrt{n} )
where
bar{X} = (X1,... , Xn)/n is the sample mean
sqrt{n} is the square root of n.
OT> 2.24
OT< -2.24
OT< -1.96
OT> 1.96
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