Given a sample proportion x and sample size y, how can one compute the probability that the population...
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It makes sense that the expected value would simply be the sample proportion, but how would one compute the probability that the population proportion is below a certain value? For example, if a survey yielded a sample proportion of .3 with a sample size of 300, how could I compute the probability that the population proportion is below .4?
statistics
asked Jan 8 at 6:41
Richard L.Richard L.
1
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It makes sense that the expected value would simply be the sample proportion, but how would one compute the probability that the population proportion is below a certain value? For example, if a survey yielded a sample proportion of .3 with a sample size of 300, how could I compute the probability that the population proportion is below .4?
statistics
asked Jan 8 at 6:41
Richard L.Richard L.
1
$endgroup$
add a comment |
$begingroup$
It makes sense that the expected value would simply be the sample proportion, but how would one compute the probability that the population proportion is below a certain value? For example, if a survey yielded a sample proportion of .3 with a sample size of 300, how could I compute the probability that the population proportion is below .4?
statistics
asked Jan 8 at 6:41
Richard L.Richard L.
1
$endgroup$
It makes sense that the expected value would simply be the sample proportion, but how would one compute the probability that the population proportion is below a certain value? For example, if a survey yielded a sample proportion of .3 with a sample size of 300, how could I compute the probability that the population proportion is below .4?
statistics
statistics
asked Jan 8 at 6:41
Richard L.Richard L.
1
asked Jan 8 at 6:41
Richard L.Richard L.
1
asked Jan 8 at 6:41
Richard L.Richard L.
1
asked Jan 8 at 6:41
Richard L.Richard L.
1
asked Jan 8 at 6:41
Richard L.Richard L.
1
1
add a comment |
add a comment |
1 Answer
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If you are treating the population proportion as a random variable then you may need a Bayesian approach
For example, if you start with a Beta prior for the population proportion of $B(alpha,beta)$ and see $300times 0.3=90$ successes and $300-90=210$ failures, then your posterior distribution would be $B(alpha +90,beta+210)$ and have mean $frac{alpha+90}{alpha+beta+300}$
With $alpha=beta= frac12$ (a Jeffreys prior) this would give a probability of being below $0.4$ of about $0.9998378$. A uniform prior with $alpha=beta= 1$ would give about $0.9998259$ and an improper Haldane prior (with mean $0.3$) with $alpha=beta=0$ would give about $0.9998490$
answered Jan 8 at 8:28
HenryHenry
98.7k476163
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1 Answer
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$begingroup$
If you are treating the population proportion as a random variable then you may need a Bayesian approach
For example, if you start with a Beta prior for the population proportion of $B(alpha,beta)$ and see $300times 0.3=90$ successes and $300-90=210$ failures, then your posterior distribution would be $B(alpha +90,beta+210)$ and have mean $frac{alpha+90}{alpha+beta+300}$
With $alpha=beta= frac12$ (a Jeffreys prior) this would give a probability of being below $0.4$ of about $0.9998378$. A uniform prior with $alpha=beta= 1$ would give about $0.9998259$ and an improper Haldane prior (with mean $0.3$) with $alpha=beta=0$ would give about $0.9998490$
answered Jan 8 at 8:28
HenryHenry
98.7k476163
$endgroup$
add a comment |
$begingroup$
If you are treating the population proportion as a random variable then you may need a Bayesian approach
For example, if you start with a Beta prior for the population proportion of $B(alpha,beta)$ and see $300times 0.3=90$ successes and $300-90=210$ failures, then your posterior distribution would be $B(alpha +90,beta+210)$ and have mean $frac{alpha+90}{alpha+beta+300}$
With $alpha=beta= frac12$ (a Jeffreys prior) this would give a probability of being below $0.4$ of about $0.9998378$. A uniform prior with $alpha=beta= 1$ would give about $0.9998259$ and an improper Haldane prior (with mean $0.3$) with $alpha=beta=0$ would give about $0.9998490$
answered Jan 8 at 8:28
HenryHenry
98.7k476163
$endgroup$
add a comment |
$begingroup$
If you are treating the population proportion as a random variable then you may need a Bayesian approach
For example, if you start with a Beta prior for the population proportion of $B(alpha,beta)$ and see $300times 0.3=90$ successes and $300-90=210$ failures, then your posterior distribution would be $B(alpha +90,beta+210)$ and have mean $frac{alpha+90}{alpha+beta+300}$
With $alpha=beta= frac12$ (a Jeffreys prior) this would give a probability of being below $0.4$ of about $0.9998378$. A uniform prior with $alpha=beta= 1$ would give about $0.9998259$ and an improper Haldane prior (with mean $0.3$) with $alpha=beta=0$ would give about $0.9998490$
answered Jan 8 at 8:28
HenryHenry
98.7k476163
$endgroup$
If you are treating the population proportion as a random variable then you may need a Bayesian approach
For example, if you start with a Beta prior for the population proportion of $B(alpha,beta)$ and see $300times 0.3=90$ successes and $300-90=210$ failures, then your posterior distribution would be $B(alpha +90,beta+210)$ and have mean $frac{alpha+90}{alpha+beta+300}$
With $alpha=beta= frac12$ (a Jeffreys prior) this would give a probability of being below $0.4$ of about $0.9998378$. A uniform prior with $alpha=beta= 1$ would give about $0.9998259$ and an improper Haldane prior (with mean $0.3$) with $alpha=beta=0$ would give about $0.9998490$
answered Jan 8 at 8:28
HenryHenry
98.7k476163
answered Jan 8 at 8:28
HenryHenry
98.7k476163
answered Jan 8 at 8:28
HenryHenry
98.7k476163
answered Jan 8 at 8:28
HenryHenry
98.7k476163
98.7k476163
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