Find the sample size so that the probability that the mean sample does not differ from the population mean
$begingroup$
A researcher wishes to estimate the mean of a population using a sufficiently large sample so that the probability that the sample mean does not differ from the population mean by more than 25% of the standard deviation is 0.95. What size should the sample be taken?
My attempt
I think we need solve this using law of big numbers.
We know
$$P(|bar{x}-mu|>0.95)leq frac{25}{100}$$
this is equivalent to
$$P(|bar{x}-mu|leq 0.95) > 1-frac{25}{100}$$
Let $epsilon=0.95$ and $alpha=1-frac{25}{100}$ then $1-alpha=frac{25}{100}$
The by law of big numbers we have:
$$nleq frac{sigma^2}{epsilon^2(1-alpha)}$$
Here i'm stuck because i don't know how to find $sigma^2$ (variance)
probability
edited Nov 24 '18 at 17:57
saulspatz
14.2k21329
asked Nov 24 '18 at 17:18
Bvss12Bvss12
1,785618
$endgroup$
add a comment |
$begingroup$
A researcher wishes to estimate the mean of a population using a sufficiently large sample so that the probability that the sample mean does not differ from the population mean by more than 25% of the standard deviation is 0.95. What size should the sample be taken?
My attempt
I think we need solve this using law of big numbers.
We know
$$P(|bar{x}-mu|>0.95)leq frac{25}{100}$$
this is equivalent to
$$P(|bar{x}-mu|leq 0.95) > 1-frac{25}{100}$$
Let $epsilon=0.95$ and $alpha=1-frac{25}{100}$ then $1-alpha=frac{25}{100}$
The by law of big numbers we have:
$$nleq frac{sigma^2}{epsilon^2(1-alpha)}$$
Here i'm stuck because i don't know how to find $sigma^2$ (variance)
probability
edited Nov 24 '18 at 17:57
saulspatz
14.2k21329
asked Nov 24 '18 at 17:18
Bvss12Bvss12
1,785618
$endgroup$
$begingroup$
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
$endgroup$
– Sean Roberson
Nov 24 '18 at 17:30
$begingroup$
@SeanRoberson and how can i calculate $sigma^2$
$endgroup$
– Bvss12
Nov 24 '18 at 17:36
$begingroup$
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
$endgroup$
– saulspatz
Nov 24 '18 at 18:00
add a comment |
$begingroup$
A researcher wishes to estimate the mean of a population using a sufficiently large sample so that the probability that the sample mean does not differ from the population mean by more than 25% of the standard deviation is 0.95. What size should the sample be taken?
My attempt
I think we need solve this using law of big numbers.
We know
$$P(|bar{x}-mu|>0.95)leq frac{25}{100}$$
this is equivalent to
$$P(|bar{x}-mu|leq 0.95) > 1-frac{25}{100}$$
Let $epsilon=0.95$ and $alpha=1-frac{25}{100}$ then $1-alpha=frac{25}{100}$
The by law of big numbers we have:
$$nleq frac{sigma^2}{epsilon^2(1-alpha)}$$
Here i'm stuck because i don't know how to find $sigma^2$ (variance)
probability
edited Nov 24 '18 at 17:57
saulspatz
14.2k21329
asked Nov 24 '18 at 17:18
Bvss12Bvss12
1,785618
$endgroup$
A researcher wishes to estimate the mean of a population using a sufficiently large sample so that the probability that the sample mean does not differ from the population mean by more than 25% of the standard deviation is 0.95. What size should the sample be taken?
My attempt
I think we need solve this using law of big numbers.
We know
$$P(|bar{x}-mu|>0.95)leq frac{25}{100}$$
this is equivalent to
$$P(|bar{x}-mu|leq 0.95) > 1-frac{25}{100}$$
Let $epsilon=0.95$ and $alpha=1-frac{25}{100}$ then $1-alpha=frac{25}{100}$
The by law of big numbers we have:
$$nleq frac{sigma^2}{epsilon^2(1-alpha)}$$
Here i'm stuck because i don't know how to find $sigma^2$ (variance)
probability
probability
edited Nov 24 '18 at 17:57
saulspatz
14.2k21329
asked Nov 24 '18 at 17:18
Bvss12Bvss12
1,785618
edited Nov 24 '18 at 17:57
saulspatz
14.2k21329
asked Nov 24 '18 at 17:18
Bvss12Bvss12
1,785618
edited Nov 24 '18 at 17:57
saulspatz
14.2k21329
edited Nov 24 '18 at 17:57
saulspatz
14.2k21329
edited Nov 24 '18 at 17:57
saulspatz
14.2k21329
14.2k21329
asked Nov 24 '18 at 17:18
Bvss12Bvss12
1,785618
asked Nov 24 '18 at 17:18
Bvss12Bvss12
1,785618
asked Nov 24 '18 at 17:18
Bvss12Bvss12
1,785618
1,785618
$begingroup$
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
$endgroup$
– Sean Roberson
Nov 24 '18 at 17:30
$begingroup$
@SeanRoberson and how can i calculate $sigma^2$
$endgroup$
– Bvss12
Nov 24 '18 at 17:36
$begingroup$
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
$endgroup$
– saulspatz
Nov 24 '18 at 18:00
add a comment |
$begingroup$
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
$endgroup$
– Sean Roberson
Nov 24 '18 at 17:30
$begingroup$
@SeanRoberson and how can i calculate $sigma^2$
$endgroup$
– Bvss12
Nov 24 '18 at 17:36
$begingroup$
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
$endgroup$
– saulspatz
Nov 24 '18 at 18:00
$begingroup$
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
$endgroup$
– Sean Roberson
Nov 24 '18 at 17:30
$begingroup$
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
$endgroup$
– Sean Roberson
Nov 24 '18 at 17:30
$begingroup$
@SeanRoberson and how can i calculate $sigma^2$
$endgroup$
– Bvss12
Nov 24 '18 at 17:36
$begingroup$
@SeanRoberson and how can i calculate $sigma^2$
$endgroup$
– Bvss12
Nov 24 '18 at 17:36
$begingroup$
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
$endgroup$
– saulspatz
Nov 24 '18 at 18:00
$begingroup$
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
$endgroup$
– saulspatz
Nov 24 '18 at 18:00
add a comment |
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$begingroup$
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
$endgroup$
– Sean Roberson
Nov 24 '18 at 17:30
$begingroup$
@SeanRoberson and how can i calculate $sigma^2$
$endgroup$
– Bvss12
Nov 24 '18 at 17:36
$begingroup$
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
$endgroup$
– saulspatz
Nov 24 '18 at 18:00