Mixing
Confidence Interval question for amount of experiments one should do.
$begingroup$
Before posing this question, the lecture notes I am reading discussed games, probability, the binomial distribution and central limit theorem. It usually assumes some form of game when it asks something. This question has been confusing me a bit:
How many experiments do we need to perform to estimate with $90$% confidence a winning
probability within an accuracy of $1$%? And if we want an accuracy of $0.1$%?
We want to estimate $p$, let us use $hat{p}$. Normally the confidence interval for $90$% certainty for a binomial distribution (win/lose) is given by:
$$(hat{p}-frac{ 1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}},hat{p}+frac{ 1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}} ) $$
Here we have that the last bit is the uncertainty, we want this to be equal to $1$% so $0.01$, we then get that:
$$ 0.01=frac{1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}} $$
We get that we should at least pick:
$$ N geq 1.645^2hat{p}(1-hat{p}) cdot 10^4$$
Is it true that I cannot directly compute $hat{p}$ from how the answer is phrased and this would be the best answer?
The answer for the second question will be:
$$ N geq 1.645^2hat{p}(1-hat{p}) cdot 10^6$$
By symmetry of the zeros of $f(x)=x(1-x)$ at $x=0$ and $x=1$, we know that this estimate for $N$ is maximised whenever $hat{p}= frac{1}{2}$. This would indeed correspond to the worst case scenario for an estimator.
$$ N_1 geq 1.645^2 cdot frac{1}{4} cdot 10^4 approx 6765 $$
And similarly:
$$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$
statistics proof-verification binomial-distribution confidence-interval
asked Jan 3 at 22:24
Wesley StrikWesley Strik
1,653423
$endgroup$
add a comment |
$begingroup$
Before posing this question, the lecture notes I am reading discussed games, probability, the binomial distribution and central limit theorem. It usually assumes some form of game when it asks something. This question has been confusing me a bit:
How many experiments do we need to perform to estimate with $90$% confidence a winning
probability within an accuracy of $1$%? And if we want an accuracy of $0.1$%?
We want to estimate $p$, let us use $hat{p}$. Normally the confidence interval for $90$% certainty for a binomial distribution (win/lose) is given by:
$$(hat{p}-frac{ 1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}},hat{p}+frac{ 1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}} ) $$
Here we have that the last bit is the uncertainty, we want this to be equal to $1$% so $0.01$, we then get that:
$$ 0.01=frac{1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}} $$
We get that we should at least pick:
$$ N geq 1.645^2hat{p}(1-hat{p}) cdot 10^4$$
Is it true that I cannot directly compute $hat{p}$ from how the answer is phrased and this would be the best answer?
The answer for the second question will be:
$$ N geq 1.645^2hat{p}(1-hat{p}) cdot 10^6$$
By symmetry of the zeros of $f(x)=x(1-x)$ at $x=0$ and $x=1$, we know that this estimate for $N$ is maximised whenever $hat{p}= frac{1}{2}$. This would indeed correspond to the worst case scenario for an estimator.
$$ N_1 geq 1.645^2 cdot frac{1}{4} cdot 10^4 approx 6765 $$
And similarly:
$$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$
statistics proof-verification binomial-distribution confidence-interval
asked Jan 3 at 22:24
Wesley StrikWesley Strik
1,653423
$endgroup$
2
$begingroup$
You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
$endgroup$
– Henry
Jan 3 at 22:37
1
$begingroup$
Consider a Bernoulli game and suppose $N$ games are played and the outcomes recorded as the variates $X_1=x_1,dotsc, X_N=x_N$. Then an estimate of $p$ is given by $$hat{p}=frac{X_1+dotsc +X_N}{N},$$ so that actually $hat{p}$ is a function of the sample size $N$. If we want the standard error of a $90%$ confidence interval to be equal to $0.01$, then, indeed, we must have $N$ at least $geq 2.576^2 hat{p}(1-hat{p})10^4$—but the RHS still depends on $N$ here! So you must use Henry's suggestion...
$endgroup$
– LoveTooNap29
Jan 3 at 23:15
add a comment |
$begingroup$
Before posing this question, the lecture notes I am reading discussed games, probability, the binomial distribution and central limit theorem. It usually assumes some form of game when it asks something. This question has been confusing me a bit:
How many experiments do we need to perform to estimate with $90$% confidence a winning
probability within an accuracy of $1$%? And if we want an accuracy of $0.1$%?
We want to estimate $p$, let us use $hat{p}$. Normally the confidence interval for $90$% certainty for a binomial distribution (win/lose) is given by:
$$(hat{p}-frac{ 1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}},hat{p}+frac{ 1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}} ) $$
Here we have that the last bit is the uncertainty, we want this to be equal to $1$% so $0.01$, we then get that:
$$ 0.01=frac{1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}} $$
We get that we should at least pick:
$$ N geq 1.645^2hat{p}(1-hat{p}) cdot 10^4$$
Is it true that I cannot directly compute $hat{p}$ from how the answer is phrased and this would be the best answer?
The answer for the second question will be:
$$ N geq 1.645^2hat{p}(1-hat{p}) cdot 10^6$$
By symmetry of the zeros of $f(x)=x(1-x)$ at $x=0$ and $x=1$, we know that this estimate for $N$ is maximised whenever $hat{p}= frac{1}{2}$. This would indeed correspond to the worst case scenario for an estimator.
$$ N_1 geq 1.645^2 cdot frac{1}{4} cdot 10^4 approx 6765 $$
And similarly:
$$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$
statistics proof-verification binomial-distribution confidence-interval
asked Jan 3 at 22:24
Wesley StrikWesley Strik
1,653423
$endgroup$
Before posing this question, the lecture notes I am reading discussed games, probability, the binomial distribution and central limit theorem. It usually assumes some form of game when it asks something. This question has been confusing me a bit:
How many experiments do we need to perform to estimate with $90$% confidence a winning
probability within an accuracy of $1$%? And if we want an accuracy of $0.1$%?
We want to estimate $p$, let us use $hat{p}$. Normally the confidence interval for $90$% certainty for a binomial distribution (win/lose) is given by:
$$(hat{p}-frac{ 1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}},hat{p}+frac{ 1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}} ) $$
Here we have that the last bit is the uncertainty, we want this to be equal to $1$% so $0.01$, we then get that:
$$ 0.01=frac{1.645 sqrt{hat{p}(1-hat{p})}}{sqrt{N}} $$
We get that we should at least pick:
$$ N geq 1.645^2hat{p}(1-hat{p}) cdot 10^4$$
Is it true that I cannot directly compute $hat{p}$ from how the answer is phrased and this would be the best answer?
The answer for the second question will be:
$$ N geq 1.645^2hat{p}(1-hat{p}) cdot 10^6$$
By symmetry of the zeros of $f(x)=x(1-x)$ at $x=0$ and $x=1$, we know that this estimate for $N$ is maximised whenever $hat{p}= frac{1}{2}$. This would indeed correspond to the worst case scenario for an estimator.
$$ N_1 geq 1.645^2 cdot frac{1}{4} cdot 10^4 approx 6765 $$
And similarly:
$$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$
statistics proof-verification binomial-distribution confidence-interval
statistics proof-verification binomial-distribution confidence-interval
asked Jan 3 at 22:24
Wesley StrikWesley Strik
1,653423
asked Jan 3 at 22:24
Wesley StrikWesley Strik
1,653423
edited Jan 5 at 23:18
Wesley Strik
asked Jan 3 at 22:24
Wesley StrikWesley Strik
1,653423
asked Jan 3 at 22:24
Wesley StrikWesley Strik
1,653423
asked Jan 3 at 22:24
Wesley StrikWesley Strik
1,653423
1,653423
2
$begingroup$
You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
$endgroup$
– Henry
Jan 3 at 22:37
1
$begingroup$
Consider a Bernoulli game and suppose $N$ games are played and the outcomes recorded as the variates $X_1=x_1,dotsc, X_N=x_N$. Then an estimate of $p$ is given by $$hat{p}=frac{X_1+dotsc +X_N}{N},$$ so that actually $hat{p}$ is a function of the sample size $N$. If we want the standard error of a $90%$ confidence interval to be equal to $0.01$, then, indeed, we must have $N$ at least $geq 2.576^2 hat{p}(1-hat{p})10^4$—but the RHS still depends on $N$ here! So you must use Henry's suggestion...
$endgroup$
– LoveTooNap29
Jan 3 at 23:15
add a comment |
2
$begingroup$
You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
$endgroup$
– Henry
Jan 3 at 22:37
1
$begingroup$
Consider a Bernoulli game and suppose $N$ games are played and the outcomes recorded as the variates $X_1=x_1,dotsc, X_N=x_N$. Then an estimate of $p$ is given by $$hat{p}=frac{X_1+dotsc +X_N}{N},$$ so that actually $hat{p}$ is a function of the sample size $N$. If we want the standard error of a $90%$ confidence interval to be equal to $0.01$, then, indeed, we must have $N$ at least $geq 2.576^2 hat{p}(1-hat{p})10^4$—but the RHS still depends on $N$ here! So you must use Henry's suggestion...
$endgroup$
– LoveTooNap29
Jan 3 at 23:15
2
2
$begingroup$
You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
$endgroup$
– Henry
Jan 3 at 22:37
$begingroup$
You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
$endgroup$
– Henry
Jan 3 at 22:37
1
1
$begingroup$
Consider a Bernoulli game and suppose $N$ games are played and the outcomes recorded as the variates $X_1=x_1,dotsc, X_N=x_N$. Then an estimate of $p$ is given by $$hat{p}=frac{X_1+dotsc +X_N}{N},$$ so that actually $hat{p}$ is a function of the sample size $N$. If we want the standard error of a $90%$ confidence interval to be equal to $0.01$, then, indeed, we must have $N$ at least $geq 2.576^2 hat{p}(1-hat{p})10^4$—but the RHS still depends on $N$ here! So you must use Henry's suggestion...
$endgroup$
– LoveTooNap29
Jan 3 at 23:15
$begingroup$
Consider a Bernoulli game and suppose $N$ games are played and the outcomes recorded as the variates $X_1=x_1,dotsc, X_N=x_N$. Then an estimate of $p$ is given by $$hat{p}=frac{X_1+dotsc +X_N}{N},$$ so that actually $hat{p}$ is a function of the sample size $N$. If we want the standard error of a $90%$ confidence interval to be equal to $0.01$, then, indeed, we must have $N$ at least $geq 2.576^2 hat{p}(1-hat{p})10^4$—but the RHS still depends on $N$ here! So you must use Henry's suggestion...
$endgroup$
– LoveTooNap29
Jan 3 at 23:15
add a comment |
1 Answer
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$begingroup$
By symmetry of the zeros of $f(x)=x(1-x)$, we know that this estimate for $N$ is maximised whenever $hat{p}= frac{1}{2}$
$$ N_1 geq 1.645^2 cdot frac{1}{4} cdot 10^4 approx 6765 $$
And similarly:
$$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$
answered Jan 3 at 23:13
Wesley StrikWesley Strik
1,653423
$endgroup$
add a comment |
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$begingroup$
By symmetry of the zeros of $f(x)=x(1-x)$, we know that this estimate for $N$ is maximised whenever $hat{p}= frac{1}{2}$
$$ N_1 geq 1.645^2 cdot frac{1}{4} cdot 10^4 approx 6765 $$
And similarly:
$$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$
answered Jan 3 at 23:13
Wesley StrikWesley Strik
1,653423
$endgroup$
add a comment |
$begingroup$
By symmetry of the zeros of $f(x)=x(1-x)$, we know that this estimate for $N$ is maximised whenever $hat{p}= frac{1}{2}$
$$ N_1 geq 1.645^2 cdot frac{1}{4} cdot 10^4 approx 6765 $$
And similarly:
$$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$
answered Jan 3 at 23:13
Wesley StrikWesley Strik
1,653423
$endgroup$
add a comment |
$begingroup$
By symmetry of the zeros of $f(x)=x(1-x)$, we know that this estimate for $N$ is maximised whenever $hat{p}= frac{1}{2}$
$$ N_1 geq 1.645^2 cdot frac{1}{4} cdot 10^4 approx 6765 $$
And similarly:
$$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$
answered Jan 3 at 23:13
Wesley StrikWesley Strik
1,653423
$endgroup$
By symmetry of the zeros of $f(x)=x(1-x)$, we know that this estimate for $N$ is maximised whenever $hat{p}= frac{1}{2}$
$$ N_1 geq 1.645^2 cdot frac{1}{4} cdot 10^4 approx 6765 $$
And similarly:
$$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$
answered Jan 3 at 23:13
Wesley StrikWesley Strik
1,653423
edited Jan 5 at 23:01
answered Jan 3 at 23:13
Wesley StrikWesley Strik
1,653423
answered Jan 3 at 23:13
Wesley StrikWesley Strik
1,653423
answered Jan 3 at 23:13
Wesley StrikWesley Strik
1,653423
1,653423
add a comment |
add a comment |
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$begingroup$
You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
$endgroup$
– Henry
Jan 3 at 22:37
1
$begingroup$
Consider a Bernoulli game and suppose $N$ games are played and the outcomes recorded as the variates $X_1=x_1,dotsc, X_N=x_N$. Then an estimate of $p$ is given by $$hat{p}=frac{X_1+dotsc +X_N}{N},$$ so that actually $hat{p}$ is a function of the sample size $N$. If we want the standard error of a $90%$ confidence interval to be equal to $0.01$, then, indeed, we must have $N$ at least $geq 2.576^2 hat{p}(1-hat{p})10^4$—but the RHS still depends on $N$ here! So you must use Henry's suggestion...
$endgroup$
– LoveTooNap29
Jan 3 at 23:15