Confidence Interval question for amount of experiments one should do.












0














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







share|cite|improve this question




















  • 2




    You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
    – Henry
    Jan 3 at 22:37








  • 1




    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...
    – LoveTooNap29
    Jan 3 at 23:15
















0














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







share|cite|improve this question




















  • 2




    You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
    – Henry
    Jan 3 at 22:37








  • 1




    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...
    – LoveTooNap29
    Jan 3 at 23:15














0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 5 at 23:18







Wesley Strik

















asked Jan 3 at 22:24









Wesley StrikWesley Strik

1,617423




1,617423








  • 2




    You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
    – Henry
    Jan 3 at 22:37








  • 1




    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...
    – LoveTooNap29
    Jan 3 at 23:15














  • 2




    You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
    – Henry
    Jan 3 at 22:37








  • 1




    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...
    – LoveTooNap29
    Jan 3 at 23:15








2




2




You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
– Henry
Jan 3 at 22:37






You may want to use the fact that $x(1-x)$ is maximised when $x=frac12$
– Henry
Jan 3 at 22:37






1




1




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...
– LoveTooNap29
Jan 3 at 23:15




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...
– LoveTooNap29
Jan 3 at 23:15










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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:
$$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$






share|cite|improve this answer























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    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:
    $$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$






    share|cite|improve this answer




























      0














      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:
      $$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$






      share|cite|improve this answer


























        0












        0








        0






        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:
        $$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$






        share|cite|improve this answer














        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:
        $$ N_2 geq 1.645^2 cdot frac{1}{4} cdot 10^6 approx 676506 $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 5 at 23:01

























        answered Jan 3 at 23:13









        Wesley StrikWesley Strik

        1,617423




        1,617423






























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