Constructing an upper confidence limit for $σ^2$
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Suppose that $X_1, ..., X_n$ form a random sample from the normal distribution with unknown mean µ and
unknown standard deviation σ.
Construct an upper confidence limit U(X1, ..., Xn) for $σ^2$ such that
$P[σ^2 < U(X_1, ..., X_n)]= 0.99$.
What I've done is so far
Since $dfrac {(n-1)s^2}{σ^2}$~ $X^2(n-1)$,
$P(X^2_.99 <dfrac {(n-1)s^2}{σ^2})$=$P(σ^2<dfrac {(n-1)s^2}{X^2_.99})$
Therefore $U(X_1, ..., X_n)=dfrac {(n-1)s^2}{X^2_.99}$
Now is it correctly done?
sampling estimation-theory confidence-interval interval-arithmetic
asked Dec 2 '18 at 6:30
NewtNewt
207
$endgroup$
add a comment |
$begingroup$
Suppose that $X_1, ..., X_n$ form a random sample from the normal distribution with unknown mean µ and
unknown standard deviation σ.
Construct an upper confidence limit U(X1, ..., Xn) for $σ^2$ such that
$P[σ^2 < U(X_1, ..., X_n)]= 0.99$.
What I've done is so far
Since $dfrac {(n-1)s^2}{σ^2}$~ $X^2(n-1)$,
$P(X^2_.99 <dfrac {(n-1)s^2}{σ^2})$=$P(σ^2<dfrac {(n-1)s^2}{X^2_.99})$
Therefore $U(X_1, ..., X_n)=dfrac {(n-1)s^2}{X^2_.99}$
Now is it correctly done?
sampling estimation-theory confidence-interval interval-arithmetic
asked Dec 2 '18 at 6:30
NewtNewt
207
$endgroup$
$begingroup$
Don't forget the mean $mu$ which is unknown too
$endgroup$
– Charles Madeline
Dec 2 '18 at 6:36
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@CharlesMadeline What does μ have to do with this? $dfrac {(n-1)s^2}{X^2_.99}$ does not contain μ
$endgroup$
– Newt
Dec 2 '18 at 6:39
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Yes, your answer is perfect.
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– John_Wick
Dec 2 '18 at 7:20
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Your normal variables need to have mean $0$ for the sum to follow a chi-squared law en.m.wikipedia.org/wiki/Chi-squared_distribution
$endgroup$
– Charles Madeline
Dec 2 '18 at 16:43
add a comment |
$begingroup$
Suppose that $X_1, ..., X_n$ form a random sample from the normal distribution with unknown mean µ and
unknown standard deviation σ.
Construct an upper confidence limit U(X1, ..., Xn) for $σ^2$ such that
$P[σ^2 < U(X_1, ..., X_n)]= 0.99$.
What I've done is so far
Since $dfrac {(n-1)s^2}{σ^2}$~ $X^2(n-1)$,
$P(X^2_.99 <dfrac {(n-1)s^2}{σ^2})$=$P(σ^2<dfrac {(n-1)s^2}{X^2_.99})$
Therefore $U(X_1, ..., X_n)=dfrac {(n-1)s^2}{X^2_.99}$
Now is it correctly done?
sampling estimation-theory confidence-interval interval-arithmetic
asked Dec 2 '18 at 6:30
NewtNewt
207
$endgroup$
Suppose that $X_1, ..., X_n$ form a random sample from the normal distribution with unknown mean µ and
unknown standard deviation σ.
Construct an upper confidence limit U(X1, ..., Xn) for $σ^2$ such that
$P[σ^2 < U(X_1, ..., X_n)]= 0.99$.
What I've done is so far
Since $dfrac {(n-1)s^2}{σ^2}$~ $X^2(n-1)$,
$P(X^2_.99 <dfrac {(n-1)s^2}{σ^2})$=$P(σ^2<dfrac {(n-1)s^2}{X^2_.99})$
Therefore $U(X_1, ..., X_n)=dfrac {(n-1)s^2}{X^2_.99}$
Now is it correctly done?
sampling estimation-theory confidence-interval interval-arithmetic
sampling estimation-theory confidence-interval interval-arithmetic
asked Dec 2 '18 at 6:30
NewtNewt
207
asked Dec 2 '18 at 6:30
NewtNewt
207
edited Dec 2 '18 at 6:38
Newt
asked Dec 2 '18 at 6:30
NewtNewt
207
asked Dec 2 '18 at 6:30
NewtNewt
207
asked Dec 2 '18 at 6:30
NewtNewt
207
207
$begingroup$
Don't forget the mean $mu$ which is unknown too
$endgroup$
– Charles Madeline
Dec 2 '18 at 6:36
$begingroup$
@CharlesMadeline What does μ have to do with this? $dfrac {(n-1)s^2}{X^2_.99}$ does not contain μ
$endgroup$
– Newt
Dec 2 '18 at 6:39
$begingroup$
Yes, your answer is perfect.
$endgroup$
– John_Wick
Dec 2 '18 at 7:20
$begingroup$
Your normal variables need to have mean $0$ for the sum to follow a chi-squared law en.m.wikipedia.org/wiki/Chi-squared_distribution
$endgroup$
– Charles Madeline
Dec 2 '18 at 16:43
add a comment |
$begingroup$
Don't forget the mean $mu$ which is unknown too
$endgroup$
– Charles Madeline
Dec 2 '18 at 6:36
$begingroup$
@CharlesMadeline What does μ have to do with this? $dfrac {(n-1)s^2}{X^2_.99}$ does not contain μ
$endgroup$
– Newt
Dec 2 '18 at 6:39
$begingroup$
Yes, your answer is perfect.
$endgroup$
– John_Wick
Dec 2 '18 at 7:20
$begingroup$
Your normal variables need to have mean $0$ for the sum to follow a chi-squared law en.m.wikipedia.org/wiki/Chi-squared_distribution
$endgroup$
– Charles Madeline
Dec 2 '18 at 16:43
$begingroup$
Don't forget the mean $mu$ which is unknown too
$endgroup$
– Charles Madeline
Dec 2 '18 at 6:36
$begingroup$
Don't forget the mean $mu$ which is unknown too
$endgroup$
– Charles Madeline
Dec 2 '18 at 6:36
$begingroup$
@CharlesMadeline What does μ have to do with this? $dfrac {(n-1)s^2}{X^2_.99}$ does not contain μ
$endgroup$
– Newt
Dec 2 '18 at 6:39
$begingroup$
@CharlesMadeline What does μ have to do with this? $dfrac {(n-1)s^2}{X^2_.99}$ does not contain μ
$endgroup$
– Newt
Dec 2 '18 at 6:39
$begingroup$
Yes, your answer is perfect.
$endgroup$
– John_Wick
Dec 2 '18 at 7:20
$begingroup$
Yes, your answer is perfect.
$endgroup$
– John_Wick
Dec 2 '18 at 7:20
$begingroup$
Your normal variables need to have mean $0$ for the sum to follow a chi-squared law en.m.wikipedia.org/wiki/Chi-squared_distribution
$endgroup$
– Charles Madeline
Dec 2 '18 at 16:43
$begingroup$
Your normal variables need to have mean $0$ for the sum to follow a chi-squared law en.m.wikipedia.org/wiki/Chi-squared_distribution
$endgroup$
– Charles Madeline
Dec 2 '18 at 16:43
add a comment |
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$begingroup$
Don't forget the mean $mu$ which is unknown too
$endgroup$
– Charles Madeline
Dec 2 '18 at 6:36
$begingroup$
@CharlesMadeline What does μ have to do with this? $dfrac {(n-1)s^2}{X^2_.99}$ does not contain μ
$endgroup$
– Newt
Dec 2 '18 at 6:39
$begingroup$
Yes, your answer is perfect.
$endgroup$
– John_Wick
Dec 2 '18 at 7:20
$begingroup$
Your normal variables need to have mean $0$ for the sum to follow a chi-squared law en.m.wikipedia.org/wiki/Chi-squared_distribution
$endgroup$
– Charles Madeline
Dec 2 '18 at 16:43