Type 2 Error Question - How to calculate for a two tailed?
$begingroup$
The modulus of rupture (MOR) for a particular grade of pencil lead is known to have
a standard deviation of 250 psi. Process standards call for a target value of 6500 psi
for the true mean MOR. For each batch, an inspector tests a random sample of 16
leads. Management wishes to detect any change in the true mean MOR. (Assume normal
distribution.)
QUESTION : Find the probability of type II error of the test when the true mean MOR is 6400.
How do we solve for two tails? i am able to get 0.4821 but with 1 tail method.
6397-6400/62.5 = 0.045
than using normcdf on MATLAB, i got this value
probability statistics data-analysis hypothesis-testing
asked Apr 27 '16 at 1:12
user334902user334902
115
$endgroup$
add a comment |
$begingroup$
The modulus of rupture (MOR) for a particular grade of pencil lead is known to have
a standard deviation of 250 psi. Process standards call for a target value of 6500 psi
for the true mean MOR. For each batch, an inspector tests a random sample of 16
leads. Management wishes to detect any change in the true mean MOR. (Assume normal
distribution.)
QUESTION : Find the probability of type II error of the test when the true mean MOR is 6400.
How do we solve for two tails? i am able to get 0.4821 but with 1 tail method.
6397-6400/62.5 = 0.045
than using normcdf on MATLAB, i got this value
probability statistics data-analysis hypothesis-testing
asked Apr 27 '16 at 1:12
user334902user334902
115
$endgroup$
add a comment |
$begingroup$
The modulus of rupture (MOR) for a particular grade of pencil lead is known to have
a standard deviation of 250 psi. Process standards call for a target value of 6500 psi
for the true mean MOR. For each batch, an inspector tests a random sample of 16
leads. Management wishes to detect any change in the true mean MOR. (Assume normal
distribution.)
QUESTION : Find the probability of type II error of the test when the true mean MOR is 6400.
How do we solve for two tails? i am able to get 0.4821 but with 1 tail method.
6397-6400/62.5 = 0.045
than using normcdf on MATLAB, i got this value
probability statistics data-analysis hypothesis-testing
asked Apr 27 '16 at 1:12
user334902user334902
115
$endgroup$
The modulus of rupture (MOR) for a particular grade of pencil lead is known to have
a standard deviation of 250 psi. Process standards call for a target value of 6500 psi
for the true mean MOR. For each batch, an inspector tests a random sample of 16
leads. Management wishes to detect any change in the true mean MOR. (Assume normal
distribution.)
QUESTION : Find the probability of type II error of the test when the true mean MOR is 6400.
How do we solve for two tails? i am able to get 0.4821 but with 1 tail method.
6397-6400/62.5 = 0.045
than using normcdf on MATLAB, i got this value
probability statistics data-analysis hypothesis-testing
probability statistics data-analysis hypothesis-testing
asked Apr 27 '16 at 1:12
user334902user334902
115
asked Apr 27 '16 at 1:12
user334902user334902
115
asked Apr 27 '16 at 1:12
user334902user334902
115
asked Apr 27 '16 at 1:12
user334902user334902
115
asked Apr 27 '16 at 1:12
user334902user334902
115
115
add a comment |
add a comment |
1 Answer
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$begingroup$
You have not said what significance level you are using, but it seems that what you did for one tail used $5%$ and
- standard error of the mean $dfrac{250}{sqrt{16}} = 62.5$
- If the true population mean is $6500$ then there is a $5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.05) approx 6397$
- If the true population mean is $6400$ then there is about a $Phileft(dfrac{6397-6400}{62.5}right) approx 48.21%$ probability that the the sample mean will be below $6397$
You need to do something similar for two tails:
- If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.025) approx 6377.5$
- If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be above $6500 + 62.5 Phi^{-1}(0.975) approx 6622.5$
- If the true population mean is $6400$ then there is about a $Phileft(dfrac{6377.5-6400}{62.5}right) approx 35.94%$ probability that the the sample mean will be below $6377.5$
- If the true population mean is $6400$ then there is about a $1-Phileft(dfrac{6622.5-6400}{62.5}right) approx 0.02%$ probability that the the sample mean will be above $6622.5$
- So the probability of a type II error of the test when the true mean is $6400$ is about $35.96%$
answered Apr 28 '16 at 7:59
HenryHenry
101k481168
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
You have not said what significance level you are using, but it seems that what you did for one tail used $5%$ and
- standard error of the mean $dfrac{250}{sqrt{16}} = 62.5$
- If the true population mean is $6500$ then there is a $5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.05) approx 6397$
- If the true population mean is $6400$ then there is about a $Phileft(dfrac{6397-6400}{62.5}right) approx 48.21%$ probability that the the sample mean will be below $6397$
You need to do something similar for two tails:
- If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.025) approx 6377.5$
- If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be above $6500 + 62.5 Phi^{-1}(0.975) approx 6622.5$
- If the true population mean is $6400$ then there is about a $Phileft(dfrac{6377.5-6400}{62.5}right) approx 35.94%$ probability that the the sample mean will be below $6377.5$
- If the true population mean is $6400$ then there is about a $1-Phileft(dfrac{6622.5-6400}{62.5}right) approx 0.02%$ probability that the the sample mean will be above $6622.5$
- So the probability of a type II error of the test when the true mean is $6400$ is about $35.96%$
answered Apr 28 '16 at 7:59
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
You have not said what significance level you are using, but it seems that what you did for one tail used $5%$ and
- standard error of the mean $dfrac{250}{sqrt{16}} = 62.5$
- If the true population mean is $6500$ then there is a $5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.05) approx 6397$
- If the true population mean is $6400$ then there is about a $Phileft(dfrac{6397-6400}{62.5}right) approx 48.21%$ probability that the the sample mean will be below $6397$
You need to do something similar for two tails:
- If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.025) approx 6377.5$
- If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be above $6500 + 62.5 Phi^{-1}(0.975) approx 6622.5$
- If the true population mean is $6400$ then there is about a $Phileft(dfrac{6377.5-6400}{62.5}right) approx 35.94%$ probability that the the sample mean will be below $6377.5$
- If the true population mean is $6400$ then there is about a $1-Phileft(dfrac{6622.5-6400}{62.5}right) approx 0.02%$ probability that the the sample mean will be above $6622.5$
- So the probability of a type II error of the test when the true mean is $6400$ is about $35.96%$
answered Apr 28 '16 at 7:59
HenryHenry
101k481168
$endgroup$
add a comment |
$begingroup$
You have not said what significance level you are using, but it seems that what you did for one tail used $5%$ and
- standard error of the mean $dfrac{250}{sqrt{16}} = 62.5$
- If the true population mean is $6500$ then there is a $5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.05) approx 6397$
- If the true population mean is $6400$ then there is about a $Phileft(dfrac{6397-6400}{62.5}right) approx 48.21%$ probability that the the sample mean will be below $6397$
You need to do something similar for two tails:
- If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.025) approx 6377.5$
- If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be above $6500 + 62.5 Phi^{-1}(0.975) approx 6622.5$
- If the true population mean is $6400$ then there is about a $Phileft(dfrac{6377.5-6400}{62.5}right) approx 35.94%$ probability that the the sample mean will be below $6377.5$
- If the true population mean is $6400$ then there is about a $1-Phileft(dfrac{6622.5-6400}{62.5}right) approx 0.02%$ probability that the the sample mean will be above $6622.5$
- So the probability of a type II error of the test when the true mean is $6400$ is about $35.96%$
answered Apr 28 '16 at 7:59
HenryHenry
101k481168
$endgroup$
You have not said what significance level you are using, but it seems that what you did for one tail used $5%$ and
- standard error of the mean $dfrac{250}{sqrt{16}} = 62.5$
- If the true population mean is $6500$ then there is a $5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.05) approx 6397$
- If the true population mean is $6400$ then there is about a $Phileft(dfrac{6397-6400}{62.5}right) approx 48.21%$ probability that the the sample mean will be below $6397$
You need to do something similar for two tails:
- If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be below $6500 + 62.5 Phi^{-1}(0.025) approx 6377.5$
- If the true population mean is $6500$ then there is a $2.5%$ probability that the the sample mean will be above $6500 + 62.5 Phi^{-1}(0.975) approx 6622.5$
- If the true population mean is $6400$ then there is about a $Phileft(dfrac{6377.5-6400}{62.5}right) approx 35.94%$ probability that the the sample mean will be below $6377.5$
- If the true population mean is $6400$ then there is about a $1-Phileft(dfrac{6622.5-6400}{62.5}right) approx 0.02%$ probability that the the sample mean will be above $6622.5$
- So the probability of a type II error of the test when the true mean is $6400$ is about $35.96%$
answered Apr 28 '16 at 7:59
HenryHenry
101k481168
answered Apr 28 '16 at 7:59
HenryHenry
101k481168
answered Apr 28 '16 at 7:59
HenryHenry
101k481168
answered Apr 28 '16 at 7:59
HenryHenry
101k481168
101k481168
add a comment |
add a comment |
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