Determining $n$ for which the confidence interval will have level of confidence >= 0.9
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I have come to the point where there is an $Xtext{~}N(a,frac{9}{n})$ and $psi$ is its CDF, I have to determine the smallest n for which $psi(a+1)-psi(a-1)>=0.9$ I have no idea how to approach this, any help appriciated!
statistics normal-distribution confidence-interval
asked Nov 22 at 19:58
ryszard eggink
143110
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up vote
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I have come to the point where there is an $Xtext{~}N(a,frac{9}{n})$ and $psi$ is its CDF, I have to determine the smallest n for which $psi(a+1)-psi(a-1)>=0.9$ I have no idea how to approach this, any help appriciated!
statistics normal-distribution confidence-interval
asked Nov 22 at 19:58
ryszard eggink
143110
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have come to the point where there is an $Xtext{~}N(a,frac{9}{n})$ and $psi$ is its CDF, I have to determine the smallest n for which $psi(a+1)-psi(a-1)>=0.9$ I have no idea how to approach this, any help appriciated!
statistics normal-distribution confidence-interval
asked Nov 22 at 19:58
ryszard eggink
143110
I have come to the point where there is an $Xtext{~}N(a,frac{9}{n})$ and $psi$ is its CDF, I have to determine the smallest n for which $psi(a+1)-psi(a-1)>=0.9$ I have no idea how to approach this, any help appriciated!
statistics normal-distribution confidence-interval
statistics normal-distribution confidence-interval
asked Nov 22 at 19:58
ryszard eggink
143110
asked Nov 22 at 19:58
ryszard eggink
143110
asked Nov 22 at 19:58
ryszard eggink
143110
asked Nov 22 at 19:58
ryszard eggink
143110
asked Nov 22 at 19:58
ryszard eggink
143110
143110
add a comment |
add a comment |
2 Answers
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accepted
$psi(a+1)-psi(a-1)=Phi(1/sqrt{(9/n)})-Phi(-1/sqrt{(9/n)})=2Phi(1/sqrt{(9/n)})-1geq 0.9Rightarrow Phi(1/sqrt{(9/n)})geq (1+0.9)/2=0.95 Rightarrow sqrt{n}/3geq Phi^{-1}(0.95)approx 1.64 Rightarrow ngeq (3times 1.64)^2=24.20Rightarrow ngeq 25.$
Here $Phi$ is cdf of $N(0,1).$
answered Nov 22 at 20:08
John_Wick
1,104111
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Hint. $psi(a+1)-psi(a-1)=mathbf{P}(a - 1 < X < a + 1) = mathbf{P}(-1 < X - a < 1) = ...$ then use a table for standard normal.
answered Nov 22 at 20:06
Will M.
2,139213
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$psi(a+1)-psi(a-1)=Phi(1/sqrt{(9/n)})-Phi(-1/sqrt{(9/n)})=2Phi(1/sqrt{(9/n)})-1geq 0.9Rightarrow Phi(1/sqrt{(9/n)})geq (1+0.9)/2=0.95 Rightarrow sqrt{n}/3geq Phi^{-1}(0.95)approx 1.64 Rightarrow ngeq (3times 1.64)^2=24.20Rightarrow ngeq 25.$
Here $Phi$ is cdf of $N(0,1).$
answered Nov 22 at 20:08
John_Wick
1,104111
add a comment |
up vote
1
down vote
accepted
$psi(a+1)-psi(a-1)=Phi(1/sqrt{(9/n)})-Phi(-1/sqrt{(9/n)})=2Phi(1/sqrt{(9/n)})-1geq 0.9Rightarrow Phi(1/sqrt{(9/n)})geq (1+0.9)/2=0.95 Rightarrow sqrt{n}/3geq Phi^{-1}(0.95)approx 1.64 Rightarrow ngeq (3times 1.64)^2=24.20Rightarrow ngeq 25.$
Here $Phi$ is cdf of $N(0,1).$
answered Nov 22 at 20:08
John_Wick
1,104111
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$psi(a+1)-psi(a-1)=Phi(1/sqrt{(9/n)})-Phi(-1/sqrt{(9/n)})=2Phi(1/sqrt{(9/n)})-1geq 0.9Rightarrow Phi(1/sqrt{(9/n)})geq (1+0.9)/2=0.95 Rightarrow sqrt{n}/3geq Phi^{-1}(0.95)approx 1.64 Rightarrow ngeq (3times 1.64)^2=24.20Rightarrow ngeq 25.$
Here $Phi$ is cdf of $N(0,1).$
answered Nov 22 at 20:08
John_Wick
1,104111
$psi(a+1)-psi(a-1)=Phi(1/sqrt{(9/n)})-Phi(-1/sqrt{(9/n)})=2Phi(1/sqrt{(9/n)})-1geq 0.9Rightarrow Phi(1/sqrt{(9/n)})geq (1+0.9)/2=0.95 Rightarrow sqrt{n}/3geq Phi^{-1}(0.95)approx 1.64 Rightarrow ngeq (3times 1.64)^2=24.20Rightarrow ngeq 25.$
Here $Phi$ is cdf of $N(0,1).$
answered Nov 22 at 20:08
John_Wick
1,104111
answered Nov 22 at 20:08
John_Wick
1,104111
answered Nov 22 at 20:08
John_Wick
1,104111
answered Nov 22 at 20:08
John_Wick
1,104111
1,104111
add a comment |
add a comment |
up vote
1
down vote
Hint. $psi(a+1)-psi(a-1)=mathbf{P}(a - 1 < X < a + 1) = mathbf{P}(-1 < X - a < 1) = ...$ then use a table for standard normal.
answered Nov 22 at 20:06
Will M.
2,139213
add a comment |
up vote
1
down vote
Hint. $psi(a+1)-psi(a-1)=mathbf{P}(a - 1 < X < a + 1) = mathbf{P}(-1 < X - a < 1) = ...$ then use a table for standard normal.
answered Nov 22 at 20:06
Will M.
2,139213
add a comment |
up vote
1
down vote
up vote
1
down vote
Hint. $psi(a+1)-psi(a-1)=mathbf{P}(a - 1 < X < a + 1) = mathbf{P}(-1 < X - a < 1) = ...$ then use a table for standard normal.
answered Nov 22 at 20:06
Will M.
2,139213
Hint. $psi(a+1)-psi(a-1)=mathbf{P}(a - 1 < X < a + 1) = mathbf{P}(-1 < X - a < 1) = ...$ then use a table for standard normal.
answered Nov 22 at 20:06
Will M.
2,139213
answered Nov 22 at 20:06
Will M.
2,139213
answered Nov 22 at 20:06
Will M.
2,139213
answered Nov 22 at 20:06
Will M.
2,139213
2,139213
add a comment |
add a comment |
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