Minimum Sample size between two samples for a specified confidence interval/level with sigma known
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Hi math stack exchange,
I came across the following question and found it quite interesting and am struggling to solve it. I haven't seen anything like it because it is both two sample and has sigma known. If anyone could give me some insight as to how to begin looking at the problem would be appreciated
Let $X_1, ldots , X_n$ and $Y_1, ldots , Y_n$ be independent random samples from the $N(mu_{X}, 2)$ and
$N(mu_{Y} , 2)$ distributions, respectively. Calculate the minimal value of $n$ so that one can be
95% confident the interval
$[bar{X} − bar{Y}− 1, bar{X} − bar{Y}+ 1]$
contains the true value of $mu_X − mu_Y$.
My idea is to set up the usual confidence interval for one side for a sigma unknown as the following form:
$1= sqrt{(sigma^2/n) +(sigma^2/n)}cdot 1.96$
$1= sigmacdot sqrt{1/n +1/n}cdot 1.96$
$1=2cdot(2/n)cdot 1.96$ ($2$ here at the start subbed in for sigma
And then solve for $n$?
Does this approach work?
Thanks
statistics statistical-inference confidence-interval interval-arithmetic
edited Jan 11 at 0:04
NCh
6,8273825
asked Jan 10 at 9:41
JamesJames
61
$endgroup$
add a comment |
$begingroup$
Hi math stack exchange,
I came across the following question and found it quite interesting and am struggling to solve it. I haven't seen anything like it because it is both two sample and has sigma known. If anyone could give me some insight as to how to begin looking at the problem would be appreciated
Let $X_1, ldots , X_n$ and $Y_1, ldots , Y_n$ be independent random samples from the $N(mu_{X}, 2)$ and
$N(mu_{Y} , 2)$ distributions, respectively. Calculate the minimal value of $n$ so that one can be
95% confident the interval
$[bar{X} − bar{Y}− 1, bar{X} − bar{Y}+ 1]$
contains the true value of $mu_X − mu_Y$.
My idea is to set up the usual confidence interval for one side for a sigma unknown as the following form:
$1= sqrt{(sigma^2/n) +(sigma^2/n)}cdot 1.96$
$1= sigmacdot sqrt{1/n +1/n}cdot 1.96$
$1=2cdot(2/n)cdot 1.96$ ($2$ here at the start subbed in for sigma
And then solve for $n$?
Does this approach work?
Thanks
statistics statistical-inference confidence-interval interval-arithmetic
edited Jan 11 at 0:04
NCh
6,8273825
asked Jan 10 at 9:41
JamesJames
61
$endgroup$
1
$begingroup$
Yes, it is a good approach. Note that you lost sqrt in the last equation.
$endgroup$
– NCh
Jan 11 at 0:05
$begingroup$
You can edit your existing posts to get them 'out of hold' by adding context as you have done here, instead of a new post.
$endgroup$
– StubbornAtom
Jan 11 at 19:06
add a comment |
$begingroup$
Hi math stack exchange,
I came across the following question and found it quite interesting and am struggling to solve it. I haven't seen anything like it because it is both two sample and has sigma known. If anyone could give me some insight as to how to begin looking at the problem would be appreciated
Let $X_1, ldots , X_n$ and $Y_1, ldots , Y_n$ be independent random samples from the $N(mu_{X}, 2)$ and
$N(mu_{Y} , 2)$ distributions, respectively. Calculate the minimal value of $n$ so that one can be
95% confident the interval
$[bar{X} − bar{Y}− 1, bar{X} − bar{Y}+ 1]$
contains the true value of $mu_X − mu_Y$.
My idea is to set up the usual confidence interval for one side for a sigma unknown as the following form:
$1= sqrt{(sigma^2/n) +(sigma^2/n)}cdot 1.96$
$1= sigmacdot sqrt{1/n +1/n}cdot 1.96$
$1=2cdot(2/n)cdot 1.96$ ($2$ here at the start subbed in for sigma
And then solve for $n$?
Does this approach work?
Thanks
statistics statistical-inference confidence-interval interval-arithmetic
edited Jan 11 at 0:04
NCh
6,8273825
asked Jan 10 at 9:41
JamesJames
61
$endgroup$
Hi math stack exchange,
I came across the following question and found it quite interesting and am struggling to solve it. I haven't seen anything like it because it is both two sample and has sigma known. If anyone could give me some insight as to how to begin looking at the problem would be appreciated
Let $X_1, ldots , X_n$ and $Y_1, ldots , Y_n$ be independent random samples from the $N(mu_{X}, 2)$ and
$N(mu_{Y} , 2)$ distributions, respectively. Calculate the minimal value of $n$ so that one can be
95% confident the interval
$[bar{X} − bar{Y}− 1, bar{X} − bar{Y}+ 1]$
contains the true value of $mu_X − mu_Y$.
My idea is to set up the usual confidence interval for one side for a sigma unknown as the following form:
$1= sqrt{(sigma^2/n) +(sigma^2/n)}cdot 1.96$
$1= sigmacdot sqrt{1/n +1/n}cdot 1.96$
$1=2cdot(2/n)cdot 1.96$ ($2$ here at the start subbed in for sigma
And then solve for $n$?
Does this approach work?
Thanks
statistics statistical-inference confidence-interval interval-arithmetic
statistics statistical-inference confidence-interval interval-arithmetic
edited Jan 11 at 0:04
NCh
6,8273825
asked Jan 10 at 9:41
JamesJames
61
edited Jan 11 at 0:04
NCh
6,8273825
asked Jan 10 at 9:41
JamesJames
61
edited Jan 11 at 0:04
NCh
6,8273825
edited Jan 11 at 0:04
NCh
6,8273825
edited Jan 11 at 0:04
NCh
6,8273825
6,8273825
asked Jan 10 at 9:41
JamesJames
61
asked Jan 10 at 9:41
JamesJames
61
asked Jan 10 at 9:41
JamesJames
61
61
1
$begingroup$
Yes, it is a good approach. Note that you lost sqrt in the last equation.
$endgroup$
– NCh
Jan 11 at 0:05
$begingroup$
You can edit your existing posts to get them 'out of hold' by adding context as you have done here, instead of a new post.
$endgroup$
– StubbornAtom
Jan 11 at 19:06
add a comment |
1
$begingroup$
Yes, it is a good approach. Note that you lost sqrt in the last equation.
$endgroup$
– NCh
Jan 11 at 0:05
$begingroup$
You can edit your existing posts to get them 'out of hold' by adding context as you have done here, instead of a new post.
$endgroup$
– StubbornAtom
Jan 11 at 19:06
1
1
$begingroup$
Yes, it is a good approach. Note that you lost sqrt in the last equation.
$endgroup$
– NCh
Jan 11 at 0:05
$begingroup$
Yes, it is a good approach. Note that you lost sqrt in the last equation.
$endgroup$
– NCh
Jan 11 at 0:05
$begingroup$
You can edit your existing posts to get them 'out of hold' by adding context as you have done here, instead of a new post.
$endgroup$
– StubbornAtom
Jan 11 at 19:06
$begingroup$
You can edit your existing posts to get them 'out of hold' by adding context as you have done here, instead of a new post.
$endgroup$
– StubbornAtom
Jan 11 at 19:06
add a comment |
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$begingroup$
Yes, it is a good approach. Note that you lost sqrt in the last equation.
$endgroup$
– NCh
Jan 11 at 0:05
$begingroup$
You can edit your existing posts to get them 'out of hold' by adding context as you have done here, instead of a new post.
$endgroup$
– StubbornAtom
Jan 11 at 19:06