Hypothesis testing of binomialy distributed data












2












$begingroup$


I haven't done any hypothesis testing for years since I left school and I just wanted to refresh my memory of it.



The hypothesis is stated as following: Assume that average high school student has a dropout rate of 70%. Alternative hypothesis would be that the dropout rate is less than 70%. Since student can only either stay in school or leave we can model that using Binomial distribution, when we sample.



Thus, we have:



Suppose $theta$ is the probability that a student stays in school.
Then



$$H_0 : theta = 0.7 quad text{vs.} quad H_a : theta < 0.7.$$



The test statistic we will use is based on the binomial distribution. $X$ is the number of students in $n$ cases that stayed in school, then $$X mid H_0 sim operatorname{Binomial}(n, theta = 0.7).$$



Then I sample, say, $100$ students and count how many of them actually stayed in school. We observe that $57$ of them stay in school. Then, $p$ value would be
$$p = Pr[X le 57 mid H_0] = 0.00396779$$
Can we use normal approximation in this case? Also, what type of test would I need to use then? Left-sided?



Thanks!






probability hypothesis-testing







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  • $begingroup$
    $p(H_0)=0,$ right variant $H_0: theta geq 0.7$.
    $endgroup$
    – Yuri Negometyanov
    Dec 13 '18 at 1:54
















2












$begingroup$


I haven't done any hypothesis testing for years since I left school and I just wanted to refresh my memory of it.



The hypothesis is stated as following: Assume that average high school student has a dropout rate of 70%. Alternative hypothesis would be that the dropout rate is less than 70%. Since student can only either stay in school or leave we can model that using Binomial distribution, when we sample.



Thus, we have:



Suppose $theta$ is the probability that a student stays in school.
Then



$$H_0 : theta = 0.7 quad text{vs.} quad H_a : theta < 0.7.$$



The test statistic we will use is based on the binomial distribution. $X$ is the number of students in $n$ cases that stayed in school, then $$X mid H_0 sim operatorname{Binomial}(n, theta = 0.7).$$



Then I sample, say, $100$ students and count how many of them actually stayed in school. We observe that $57$ of them stay in school. Then, $p$ value would be
$$p = Pr[X le 57 mid H_0] = 0.00396779$$
Can we use normal approximation in this case? Also, what type of test would I need to use then? Left-sided?



Thanks!






probability hypothesis-testing







share|cite|improve this question









$endgroup$












  • $begingroup$
    $p(H_0)=0,$ right variant $H_0: theta geq 0.7$.
    $endgroup$
    – Yuri Negometyanov
    Dec 13 '18 at 1:54














2












2








2





$begingroup$


I haven't done any hypothesis testing for years since I left school and I just wanted to refresh my memory of it.



The hypothesis is stated as following: Assume that average high school student has a dropout rate of 70%. Alternative hypothesis would be that the dropout rate is less than 70%. Since student can only either stay in school or leave we can model that using Binomial distribution, when we sample.



Thus, we have:



Suppose $theta$ is the probability that a student stays in school.
Then



$$H_0 : theta = 0.7 quad text{vs.} quad H_a : theta < 0.7.$$



The test statistic we will use is based on the binomial distribution. $X$ is the number of students in $n$ cases that stayed in school, then $$X mid H_0 sim operatorname{Binomial}(n, theta = 0.7).$$



Then I sample, say, $100$ students and count how many of them actually stayed in school. We observe that $57$ of them stay in school. Then, $p$ value would be
$$p = Pr[X le 57 mid H_0] = 0.00396779$$
Can we use normal approximation in this case? Also, what type of test would I need to use then? Left-sided?



Thanks!






probability hypothesis-testing







share|cite|improve this question









$endgroup$




I haven't done any hypothesis testing for years since I left school and I just wanted to refresh my memory of it.



The hypothesis is stated as following: Assume that average high school student has a dropout rate of 70%. Alternative hypothesis would be that the dropout rate is less than 70%. Since student can only either stay in school or leave we can model that using Binomial distribution, when we sample.



Thus, we have:



Suppose $theta$ is the probability that a student stays in school.
Then



$$H_0 : theta = 0.7 quad text{vs.} quad H_a : theta < 0.7.$$



The test statistic we will use is based on the binomial distribution. $X$ is the number of students in $n$ cases that stayed in school, then $$X mid H_0 sim operatorname{Binomial}(n, theta = 0.7).$$



Then I sample, say, $100$ students and count how many of them actually stayed in school. We observe that $57$ of them stay in school. Then, $p$ value would be
$$p = Pr[X le 57 mid H_0] = 0.00396779$$
Can we use normal approximation in this case? Also, what type of test would I need to use then? Left-sided?



Thanks!






probability hypothesis-testing




probability hypothesis-testing






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asked Dec 9 '18 at 15:53









ZakkeryZakkery

634




634












  • $begingroup$
    $p(H_0)=0,$ right variant $H_0: theta geq 0.7$.
    $endgroup$
    – Yuri Negometyanov
    Dec 13 '18 at 1:54


















  • $begingroup$
    $p(H_0)=0,$ right variant $H_0: theta geq 0.7$.
    $endgroup$
    – Yuri Negometyanov
    Dec 13 '18 at 1:54
















$begingroup$
$p(H_0)=0,$ right variant $H_0: theta geq 0.7$.
$endgroup$
– Yuri Negometyanov
Dec 13 '18 at 1:54




$begingroup$
$p(H_0)=0,$ right variant $H_0: theta geq 0.7$.
$endgroup$
– Yuri Negometyanov
Dec 13 '18 at 1:54










1 Answer
1






active

oldest

votes


















1





+50







$begingroup$

If we were to apply a normal approximation, we would model X as



$$X|H_{0}simmathcal{N}left(ntheta,nthetaleft(1-thetaright)right)$$



$$X|H_{0}simmathcal{N}(70,21)$$



$$sigma=sqrt{21}$$



Thus, our p-value, under normal approximation, would be $mathbb{P}(Z<frac{left(57-70right)}{sqrt{21}})$



This is a one sided Z-test, and we get a p value of 0.002327, which as you can see is quite a bit lower than yours(but is a nice quick order of magnitude approximation if you're away from a computer). It is well known that normal approximations to binomial give p values that smaller than they should be when we are deep into the tails. With only 100 people, you really don't need to apply a normal approximation






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What if I have, let's say 10000 people? What is an appropriate condition to apply normal approximation?
    $endgroup$
    – Zakkery
    Dec 12 '18 at 19:20










  • $begingroup$
    A stirling approximation would be better, $$left(begin{array}{c} n\ i end{array}right)approxfrac{n^{i}}{i!}$$ $$mathbb{P}left(X<mright)=sum_{i=0}^{m}left(begin{array}{c} n\ i end{array}right)p^{i}left(1-pright)^{n-i} approxsum_{i=0}^{m}frac{n^{k}}{i!}p^{i}left(1-pright)^{n-i}$$ Should be a little bit easier than binomial to compute. A normal won't be bad as long as you aren't too too far in the tail--just keep in mind that if you have something like $p=10^{-6}$, you're probably underestimating the p value by using a normal approximation.
    $endgroup$
    – Alex
    Dec 12 '18 at 19:49












  • $begingroup$
    Also, to answer the question of when the normal approximation is good, it will be good with a sample size of 10,000 up to very deep into the tails, and it is more accurate when $theta$ is closer to 0.5 as the distribution will be more symmetric.
    $endgroup$
    – Alex
    Dec 12 '18 at 23:20











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1





+50







$begingroup$

If we were to apply a normal approximation, we would model X as



$$X|H_{0}simmathcal{N}left(ntheta,nthetaleft(1-thetaright)right)$$



$$X|H_{0}simmathcal{N}(70,21)$$



$$sigma=sqrt{21}$$



Thus, our p-value, under normal approximation, would be $mathbb{P}(Z<frac{left(57-70right)}{sqrt{21}})$



This is a one sided Z-test, and we get a p value of 0.002327, which as you can see is quite a bit lower than yours(but is a nice quick order of magnitude approximation if you're away from a computer). It is well known that normal approximations to binomial give p values that smaller than they should be when we are deep into the tails. With only 100 people, you really don't need to apply a normal approximation






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What if I have, let's say 10000 people? What is an appropriate condition to apply normal approximation?
    $endgroup$
    – Zakkery
    Dec 12 '18 at 19:20










  • $begingroup$
    A stirling approximation would be better, $$left(begin{array}{c} n\ i end{array}right)approxfrac{n^{i}}{i!}$$ $$mathbb{P}left(X<mright)=sum_{i=0}^{m}left(begin{array}{c} n\ i end{array}right)p^{i}left(1-pright)^{n-i} approxsum_{i=0}^{m}frac{n^{k}}{i!}p^{i}left(1-pright)^{n-i}$$ Should be a little bit easier than binomial to compute. A normal won't be bad as long as you aren't too too far in the tail--just keep in mind that if you have something like $p=10^{-6}$, you're probably underestimating the p value by using a normal approximation.
    $endgroup$
    – Alex
    Dec 12 '18 at 19:49












  • $begingroup$
    Also, to answer the question of when the normal approximation is good, it will be good with a sample size of 10,000 up to very deep into the tails, and it is more accurate when $theta$ is closer to 0.5 as the distribution will be more symmetric.
    $endgroup$
    – Alex
    Dec 12 '18 at 23:20
















1





+50







$begingroup$

If we were to apply a normal approximation, we would model X as



$$X|H_{0}simmathcal{N}left(ntheta,nthetaleft(1-thetaright)right)$$



$$X|H_{0}simmathcal{N}(70,21)$$



$$sigma=sqrt{21}$$



Thus, our p-value, under normal approximation, would be $mathbb{P}(Z<frac{left(57-70right)}{sqrt{21}})$



This is a one sided Z-test, and we get a p value of 0.002327, which as you can see is quite a bit lower than yours(but is a nice quick order of magnitude approximation if you're away from a computer). It is well known that normal approximations to binomial give p values that smaller than they should be when we are deep into the tails. With only 100 people, you really don't need to apply a normal approximation






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What if I have, let's say 10000 people? What is an appropriate condition to apply normal approximation?
    $endgroup$
    – Zakkery
    Dec 12 '18 at 19:20










  • $begingroup$
    A stirling approximation would be better, $$left(begin{array}{c} n\ i end{array}right)approxfrac{n^{i}}{i!}$$ $$mathbb{P}left(X<mright)=sum_{i=0}^{m}left(begin{array}{c} n\ i end{array}right)p^{i}left(1-pright)^{n-i} approxsum_{i=0}^{m}frac{n^{k}}{i!}p^{i}left(1-pright)^{n-i}$$ Should be a little bit easier than binomial to compute. A normal won't be bad as long as you aren't too too far in the tail--just keep in mind that if you have something like $p=10^{-6}$, you're probably underestimating the p value by using a normal approximation.
    $endgroup$
    – Alex
    Dec 12 '18 at 19:49












  • $begingroup$
    Also, to answer the question of when the normal approximation is good, it will be good with a sample size of 10,000 up to very deep into the tails, and it is more accurate when $theta$ is closer to 0.5 as the distribution will be more symmetric.
    $endgroup$
    – Alex
    Dec 12 '18 at 23:20














1





+50







1





+50



1




+50



$begingroup$

If we were to apply a normal approximation, we would model X as



$$X|H_{0}simmathcal{N}left(ntheta,nthetaleft(1-thetaright)right)$$



$$X|H_{0}simmathcal{N}(70,21)$$



$$sigma=sqrt{21}$$



Thus, our p-value, under normal approximation, would be $mathbb{P}(Z<frac{left(57-70right)}{sqrt{21}})$



This is a one sided Z-test, and we get a p value of 0.002327, which as you can see is quite a bit lower than yours(but is a nice quick order of magnitude approximation if you're away from a computer). It is well known that normal approximations to binomial give p values that smaller than they should be when we are deep into the tails. With only 100 people, you really don't need to apply a normal approximation






share|cite|improve this answer









$endgroup$



If we were to apply a normal approximation, we would model X as



$$X|H_{0}simmathcal{N}left(ntheta,nthetaleft(1-thetaright)right)$$



$$X|H_{0}simmathcal{N}(70,21)$$



$$sigma=sqrt{21}$$



Thus, our p-value, under normal approximation, would be $mathbb{P}(Z<frac{left(57-70right)}{sqrt{21}})$



This is a one sided Z-test, and we get a p value of 0.002327, which as you can see is quite a bit lower than yours(but is a nice quick order of magnitude approximation if you're away from a computer). It is well known that normal approximations to binomial give p values that smaller than they should be when we are deep into the tails. With only 100 people, you really don't need to apply a normal approximation







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 17:21









AlexAlex

17713




17713












  • $begingroup$
    What if I have, let's say 10000 people? What is an appropriate condition to apply normal approximation?
    $endgroup$
    – Zakkery
    Dec 12 '18 at 19:20










  • $begingroup$
    A stirling approximation would be better, $$left(begin{array}{c} n\ i end{array}right)approxfrac{n^{i}}{i!}$$ $$mathbb{P}left(X<mright)=sum_{i=0}^{m}left(begin{array}{c} n\ i end{array}right)p^{i}left(1-pright)^{n-i} approxsum_{i=0}^{m}frac{n^{k}}{i!}p^{i}left(1-pright)^{n-i}$$ Should be a little bit easier than binomial to compute. A normal won't be bad as long as you aren't too too far in the tail--just keep in mind that if you have something like $p=10^{-6}$, you're probably underestimating the p value by using a normal approximation.
    $endgroup$
    – Alex
    Dec 12 '18 at 19:49












  • $begingroup$
    Also, to answer the question of when the normal approximation is good, it will be good with a sample size of 10,000 up to very deep into the tails, and it is more accurate when $theta$ is closer to 0.5 as the distribution will be more symmetric.
    $endgroup$
    – Alex
    Dec 12 '18 at 23:20


















  • $begingroup$
    What if I have, let's say 10000 people? What is an appropriate condition to apply normal approximation?
    $endgroup$
    – Zakkery
    Dec 12 '18 at 19:20










  • $begingroup$
    A stirling approximation would be better, $$left(begin{array}{c} n\ i end{array}right)approxfrac{n^{i}}{i!}$$ $$mathbb{P}left(X<mright)=sum_{i=0}^{m}left(begin{array}{c} n\ i end{array}right)p^{i}left(1-pright)^{n-i} approxsum_{i=0}^{m}frac{n^{k}}{i!}p^{i}left(1-pright)^{n-i}$$ Should be a little bit easier than binomial to compute. A normal won't be bad as long as you aren't too too far in the tail--just keep in mind that if you have something like $p=10^{-6}$, you're probably underestimating the p value by using a normal approximation.
    $endgroup$
    – Alex
    Dec 12 '18 at 19:49












  • $begingroup$
    Also, to answer the question of when the normal approximation is good, it will be good with a sample size of 10,000 up to very deep into the tails, and it is more accurate when $theta$ is closer to 0.5 as the distribution will be more symmetric.
    $endgroup$
    – Alex
    Dec 12 '18 at 23:20
















$begingroup$
What if I have, let's say 10000 people? What is an appropriate condition to apply normal approximation?
$endgroup$
– Zakkery
Dec 12 '18 at 19:20




$begingroup$
What if I have, let's say 10000 people? What is an appropriate condition to apply normal approximation?
$endgroup$
– Zakkery
Dec 12 '18 at 19:20












$begingroup$
A stirling approximation would be better, $$left(begin{array}{c} n\ i end{array}right)approxfrac{n^{i}}{i!}$$ $$mathbb{P}left(X<mright)=sum_{i=0}^{m}left(begin{array}{c} n\ i end{array}right)p^{i}left(1-pright)^{n-i} approxsum_{i=0}^{m}frac{n^{k}}{i!}p^{i}left(1-pright)^{n-i}$$ Should be a little bit easier than binomial to compute. A normal won't be bad as long as you aren't too too far in the tail--just keep in mind that if you have something like $p=10^{-6}$, you're probably underestimating the p value by using a normal approximation.
$endgroup$
– Alex
Dec 12 '18 at 19:49






$begingroup$
A stirling approximation would be better, $$left(begin{array}{c} n\ i end{array}right)approxfrac{n^{i}}{i!}$$ $$mathbb{P}left(X<mright)=sum_{i=0}^{m}left(begin{array}{c} n\ i end{array}right)p^{i}left(1-pright)^{n-i} approxsum_{i=0}^{m}frac{n^{k}}{i!}p^{i}left(1-pright)^{n-i}$$ Should be a little bit easier than binomial to compute. A normal won't be bad as long as you aren't too too far in the tail--just keep in mind that if you have something like $p=10^{-6}$, you're probably underestimating the p value by using a normal approximation.
$endgroup$
– Alex
Dec 12 '18 at 19:49














$begingroup$
Also, to answer the question of when the normal approximation is good, it will be good with a sample size of 10,000 up to very deep into the tails, and it is more accurate when $theta$ is closer to 0.5 as the distribution will be more symmetric.
$endgroup$
– Alex
Dec 12 '18 at 23:20




$begingroup$
Also, to answer the question of when the normal approximation is good, it will be good with a sample size of 10,000 up to very deep into the tails, and it is more accurate when $theta$ is closer to 0.5 as the distribution will be more symmetric.
$endgroup$
– Alex
Dec 12 '18 at 23:20


















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