Find the sample size so that the probability that the mean sample does not differ from the population mean
up vote
0
down vote
favorite
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 at 17:57
saulspatz
13.5k21327
asked Nov 24 at 17:18
Bvss12
1,724617
add a comment |
up vote
0
down vote
favorite
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 at 17:57
saulspatz
13.5k21327
asked Nov 24 at 17:18
Bvss12
1,724617
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
– Sean Roberson
Nov 24 at 17:30
@SeanRoberson and how can i calculate $sigma^2$
– Bvss12
Nov 24 at 17:36
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
– saulspatz
Nov 24 at 18:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 17:57
saulspatz
13.5k21327
asked Nov 24 at 17:18
Bvss12
1,724617
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 at 17:57
saulspatz
13.5k21327
asked Nov 24 at 17:18
Bvss12
1,724617
edited Nov 24 at 17:57
saulspatz
13.5k21327
asked Nov 24 at 17:18
Bvss12
1,724617
edited Nov 24 at 17:57
saulspatz
13.5k21327
edited Nov 24 at 17:57
saulspatz
13.5k21327
edited Nov 24 at 17:57
saulspatz
13.5k21327
13.5k21327
asked Nov 24 at 17:18
Bvss12
1,724617
asked Nov 24 at 17:18
Bvss12
1,724617
asked Nov 24 at 17:18
Bvss12
1,724617
1,724617
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
– Sean Roberson
Nov 24 at 17:30
@SeanRoberson and how can i calculate $sigma^2$
– Bvss12
Nov 24 at 17:36
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
– saulspatz
Nov 24 at 18:00
add a comment |
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
– Sean Roberson
Nov 24 at 17:30
@SeanRoberson and how can i calculate $sigma^2$
– Bvss12
Nov 24 at 17:36
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
– saulspatz
Nov 24 at 18:00
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
– Sean Roberson
Nov 24 at 17:30
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
– Sean Roberson
Nov 24 at 17:30
@SeanRoberson and how can i calculate $sigma^2$
– Bvss12
Nov 24 at 17:36
@SeanRoberson and how can i calculate $sigma^2$
– Bvss12
Nov 24 at 17:36
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
– saulspatz
Nov 24 at 18:00
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
– saulspatz
Nov 24 at 18:00
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011825%2ffind-the-sample-size-so-that-the-probability-that-the-mean-sample-does-not-diffe%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
You have numbers backwards. The probability is $0.95$ and the tolerance is $0.25$.
– Sean Roberson
Nov 24 at 17:30
@SeanRoberson and how can i calculate $sigma^2$
– Bvss12
Nov 24 at 17:36
You don't seem to have understood Sean's comment, or at least you haven't edited the question to reflect it. You want $$P(|x-mu|le .25sigma)=.95$$
– saulspatz
Nov 24 at 18:00