A r.s. of size n is taken from a normal r.v. X~N(μ, 1.5). To be 95% confident that the error between X̄ and...
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Suppose a random sample of size n is taken from a normal random variable X~N(μ, 1.5). To be 95% confident that the error between X̄ and the unknown population mean μ is at most .85, how large of a sample needs to be taken?
Does this mean that the confidence itnerval is of size .85*2?
I know what the formulas are for a confidence interval for the mean, but I don't know where to start with this one.
statistics confidence-interval
asked Nov 25 '18 at 21:45
kmediatekmediate
115
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add a comment |
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Suppose a random sample of size n is taken from a normal random variable X~N(μ, 1.5). To be 95% confident that the error between X̄ and the unknown population mean μ is at most .85, how large of a sample needs to be taken?
Does this mean that the confidence itnerval is of size .85*2?
I know what the formulas are for a confidence interval for the mean, but I don't know where to start with this one.
statistics confidence-interval
asked Nov 25 '18 at 21:45
kmediatekmediate
115
$endgroup$
add a comment |
$begingroup$
Suppose a random sample of size n is taken from a normal random variable X~N(μ, 1.5). To be 95% confident that the error between X̄ and the unknown population mean μ is at most .85, how large of a sample needs to be taken?
Does this mean that the confidence itnerval is of size .85*2?
I know what the formulas are for a confidence interval for the mean, but I don't know where to start with this one.
statistics confidence-interval
asked Nov 25 '18 at 21:45
kmediatekmediate
115
$endgroup$
Suppose a random sample of size n is taken from a normal random variable X~N(μ, 1.5). To be 95% confident that the error between X̄ and the unknown population mean μ is at most .85, how large of a sample needs to be taken?
Does this mean that the confidence itnerval is of size .85*2?
I know what the formulas are for a confidence interval for the mean, but I don't know where to start with this one.
statistics confidence-interval
statistics confidence-interval
asked Nov 25 '18 at 21:45
kmediatekmediate
115
asked Nov 25 '18 at 21:45
kmediatekmediate
115
asked Nov 25 '18 at 21:45
kmediatekmediate
115
asked Nov 25 '18 at 21:45
kmediatekmediate
115
asked Nov 25 '18 at 21:45
kmediatekmediate
115
115
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2 Answers
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$bar{X}$ is distributed like a $N(mu, 1.5 / n)$ (the way to think clearly about this is to remember that adding independent normals adds their means and variances, and scaling a random variable by $lambda$ multiplies its variance by $lambda^2$).
You can translate everything by $-mu$.
Therefore, you want to choose $n$ so that the $P( A_n in [ - .85, .85]) geq .95$, where $A_n sim N(0, 1.5/n)$. ($A_n$ is the distribution of $bar{X}$ when taking $n$ samples.)
From the $68-95-99.7$ rule, you basically want $.85$ to be 2 standard deviations - now, the standard deviation of $A_N$ is $sqrt{1.5/n}$...
(You can compute it more exactly also.)
Does that help?
answered Nov 25 '18 at 21:57
LorenzoLorenzo
11.7k31638
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Quite simply, the desired margin of error, $$0.85 = text{ME} = z_{alpha/2}^* frac{sigma}{sqrt{n}},$$ or equivalently, $$n = left(frac{z_{alpha/2}^* sigma}{text{ME}}right)^{!2},$$ where $z_{alpha/2}^*$ is the upper $alpha/2$ quantile of the standard normal distribution, which for a $95%$ confidence interval corresponds to $alpha = 0.05$ and $z_{.025}^* approx 1.96$; $sigma = 1.5$ is the population standard deviation, and $n$ is the sample size. The rest is simply substitution and calculation. Note that if the result is not an integer, one must round up to the nearest integer.
answered Nov 25 '18 at 22:09
heropupheropup
62.9k66099
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2 Answers
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2 Answers
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active
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$begingroup$
$bar{X}$ is distributed like a $N(mu, 1.5 / n)$ (the way to think clearly about this is to remember that adding independent normals adds their means and variances, and scaling a random variable by $lambda$ multiplies its variance by $lambda^2$).
You can translate everything by $-mu$.
Therefore, you want to choose $n$ so that the $P( A_n in [ - .85, .85]) geq .95$, where $A_n sim N(0, 1.5/n)$. ($A_n$ is the distribution of $bar{X}$ when taking $n$ samples.)
From the $68-95-99.7$ rule, you basically want $.85$ to be 2 standard deviations - now, the standard deviation of $A_N$ is $sqrt{1.5/n}$...
(You can compute it more exactly also.)
Does that help?
answered Nov 25 '18 at 21:57
LorenzoLorenzo
11.7k31638
$endgroup$
add a comment |
$begingroup$
$bar{X}$ is distributed like a $N(mu, 1.5 / n)$ (the way to think clearly about this is to remember that adding independent normals adds their means and variances, and scaling a random variable by $lambda$ multiplies its variance by $lambda^2$).
You can translate everything by $-mu$.
Therefore, you want to choose $n$ so that the $P( A_n in [ - .85, .85]) geq .95$, where $A_n sim N(0, 1.5/n)$. ($A_n$ is the distribution of $bar{X}$ when taking $n$ samples.)
From the $68-95-99.7$ rule, you basically want $.85$ to be 2 standard deviations - now, the standard deviation of $A_N$ is $sqrt{1.5/n}$...
(You can compute it more exactly also.)
Does that help?
answered Nov 25 '18 at 21:57
LorenzoLorenzo
11.7k31638
$endgroup$
add a comment |
$begingroup$
$bar{X}$ is distributed like a $N(mu, 1.5 / n)$ (the way to think clearly about this is to remember that adding independent normals adds their means and variances, and scaling a random variable by $lambda$ multiplies its variance by $lambda^2$).
You can translate everything by $-mu$.
Therefore, you want to choose $n$ so that the $P( A_n in [ - .85, .85]) geq .95$, where $A_n sim N(0, 1.5/n)$. ($A_n$ is the distribution of $bar{X}$ when taking $n$ samples.)
From the $68-95-99.7$ rule, you basically want $.85$ to be 2 standard deviations - now, the standard deviation of $A_N$ is $sqrt{1.5/n}$...
(You can compute it more exactly also.)
Does that help?
answered Nov 25 '18 at 21:57
LorenzoLorenzo
11.7k31638
$endgroup$
$bar{X}$ is distributed like a $N(mu, 1.5 / n)$ (the way to think clearly about this is to remember that adding independent normals adds their means and variances, and scaling a random variable by $lambda$ multiplies its variance by $lambda^2$).
You can translate everything by $-mu$.
Therefore, you want to choose $n$ so that the $P( A_n in [ - .85, .85]) geq .95$, where $A_n sim N(0, 1.5/n)$. ($A_n$ is the distribution of $bar{X}$ when taking $n$ samples.)
From the $68-95-99.7$ rule, you basically want $.85$ to be 2 standard deviations - now, the standard deviation of $A_N$ is $sqrt{1.5/n}$...
(You can compute it more exactly also.)
Does that help?
answered Nov 25 '18 at 21:57
LorenzoLorenzo
11.7k31638
answered Nov 25 '18 at 21:57
LorenzoLorenzo
11.7k31638
answered Nov 25 '18 at 21:57
LorenzoLorenzo
11.7k31638
answered Nov 25 '18 at 21:57
LorenzoLorenzo
11.7k31638
11.7k31638
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$begingroup$
Quite simply, the desired margin of error, $$0.85 = text{ME} = z_{alpha/2}^* frac{sigma}{sqrt{n}},$$ or equivalently, $$n = left(frac{z_{alpha/2}^* sigma}{text{ME}}right)^{!2},$$ where $z_{alpha/2}^*$ is the upper $alpha/2$ quantile of the standard normal distribution, which for a $95%$ confidence interval corresponds to $alpha = 0.05$ and $z_{.025}^* approx 1.96$; $sigma = 1.5$ is the population standard deviation, and $n$ is the sample size. The rest is simply substitution and calculation. Note that if the result is not an integer, one must round up to the nearest integer.
answered Nov 25 '18 at 22:09
heropupheropup
62.9k66099
$endgroup$
add a comment |
$begingroup$
Quite simply, the desired margin of error, $$0.85 = text{ME} = z_{alpha/2}^* frac{sigma}{sqrt{n}},$$ or equivalently, $$n = left(frac{z_{alpha/2}^* sigma}{text{ME}}right)^{!2},$$ where $z_{alpha/2}^*$ is the upper $alpha/2$ quantile of the standard normal distribution, which for a $95%$ confidence interval corresponds to $alpha = 0.05$ and $z_{.025}^* approx 1.96$; $sigma = 1.5$ is the population standard deviation, and $n$ is the sample size. The rest is simply substitution and calculation. Note that if the result is not an integer, one must round up to the nearest integer.
answered Nov 25 '18 at 22:09
heropupheropup
62.9k66099
$endgroup$
add a comment |
$begingroup$
Quite simply, the desired margin of error, $$0.85 = text{ME} = z_{alpha/2}^* frac{sigma}{sqrt{n}},$$ or equivalently, $$n = left(frac{z_{alpha/2}^* sigma}{text{ME}}right)^{!2},$$ where $z_{alpha/2}^*$ is the upper $alpha/2$ quantile of the standard normal distribution, which for a $95%$ confidence interval corresponds to $alpha = 0.05$ and $z_{.025}^* approx 1.96$; $sigma = 1.5$ is the population standard deviation, and $n$ is the sample size. The rest is simply substitution and calculation. Note that if the result is not an integer, one must round up to the nearest integer.
answered Nov 25 '18 at 22:09
heropupheropup
62.9k66099
$endgroup$
Quite simply, the desired margin of error, $$0.85 = text{ME} = z_{alpha/2}^* frac{sigma}{sqrt{n}},$$ or equivalently, $$n = left(frac{z_{alpha/2}^* sigma}{text{ME}}right)^{!2},$$ where $z_{alpha/2}^*$ is the upper $alpha/2$ quantile of the standard normal distribution, which for a $95%$ confidence interval corresponds to $alpha = 0.05$ and $z_{.025}^* approx 1.96$; $sigma = 1.5$ is the population standard deviation, and $n$ is the sample size. The rest is simply substitution and calculation. Note that if the result is not an integer, one must round up to the nearest integer.
answered Nov 25 '18 at 22:09
heropupheropup
62.9k66099
answered Nov 25 '18 at 22:09
heropupheropup
62.9k66099
answered Nov 25 '18 at 22:09
heropupheropup
62.9k66099
answered Nov 25 '18 at 22:09
heropupheropup
62.9k66099
62.9k66099
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