Determine whether the following $H_0$ can be accepted or not using hypothesis testing
$begingroup$
$H_0 : p=0.5$
$H_1 : p > 0.5 $
where p is the probability of heads from a coin flip.
Let $W_1$ be the number of heads from 10 coin flips and and $W_2$ be the number of heads from 1000 coin flips.
Accept $H_0$ if $W_1 leq c_1 $ and $W_2 leq c_2$, otherwise accept $H_1$. Determine the value of $c_1$ and $c_2$, or in other words the number of heads, using a level 0.05 test so that we may accept $H_0$.
I know that coin flips follow a binomial distribution, however I am not sure how to determine the two values using that. So far I've only solved problems with normal distributed random events. I am aware of CLT but is n=10 large enough to use it? For n=1000 it seems more reasonable. So do I have to use the fact that they are binomial distributed or is there some formula or trick that I am unaware of? Any help is appreciated!
statistics binomial-distribution hypothesis-testing
asked Nov 29 '18 at 8:14
CruZCruZ
527
$endgroup$
add a comment |
$begingroup$
$H_0 : p=0.5$
$H_1 : p > 0.5 $
where p is the probability of heads from a coin flip.
Let $W_1$ be the number of heads from 10 coin flips and and $W_2$ be the number of heads from 1000 coin flips.
Accept $H_0$ if $W_1 leq c_1 $ and $W_2 leq c_2$, otherwise accept $H_1$. Determine the value of $c_1$ and $c_2$, or in other words the number of heads, using a level 0.05 test so that we may accept $H_0$.
I know that coin flips follow a binomial distribution, however I am not sure how to determine the two values using that. So far I've only solved problems with normal distributed random events. I am aware of CLT but is n=10 large enough to use it? For n=1000 it seems more reasonable. So do I have to use the fact that they are binomial distributed or is there some formula or trick that I am unaware of? Any help is appreciated!
statistics binomial-distribution hypothesis-testing
asked Nov 29 '18 at 8:14
CruZCruZ
527
$endgroup$
add a comment |
$begingroup$
$H_0 : p=0.5$
$H_1 : p > 0.5 $
where p is the probability of heads from a coin flip.
Let $W_1$ be the number of heads from 10 coin flips and and $W_2$ be the number of heads from 1000 coin flips.
Accept $H_0$ if $W_1 leq c_1 $ and $W_2 leq c_2$, otherwise accept $H_1$. Determine the value of $c_1$ and $c_2$, or in other words the number of heads, using a level 0.05 test so that we may accept $H_0$.
I know that coin flips follow a binomial distribution, however I am not sure how to determine the two values using that. So far I've only solved problems with normal distributed random events. I am aware of CLT but is n=10 large enough to use it? For n=1000 it seems more reasonable. So do I have to use the fact that they are binomial distributed or is there some formula or trick that I am unaware of? Any help is appreciated!
statistics binomial-distribution hypothesis-testing
asked Nov 29 '18 at 8:14
CruZCruZ
527
$endgroup$
$H_0 : p=0.5$
$H_1 : p > 0.5 $
where p is the probability of heads from a coin flip.
Let $W_1$ be the number of heads from 10 coin flips and and $W_2$ be the number of heads from 1000 coin flips.
Accept $H_0$ if $W_1 leq c_1 $ and $W_2 leq c_2$, otherwise accept $H_1$. Determine the value of $c_1$ and $c_2$, or in other words the number of heads, using a level 0.05 test so that we may accept $H_0$.
I know that coin flips follow a binomial distribution, however I am not sure how to determine the two values using that. So far I've only solved problems with normal distributed random events. I am aware of CLT but is n=10 large enough to use it? For n=1000 it seems more reasonable. So do I have to use the fact that they are binomial distributed or is there some formula or trick that I am unaware of? Any help is appreciated!
statistics binomial-distribution hypothesis-testing
statistics binomial-distribution hypothesis-testing
asked Nov 29 '18 at 8:14
CruZCruZ
527
asked Nov 29 '18 at 8:14
CruZCruZ
527
edited Nov 29 '18 at 8:31
CruZ
asked Nov 29 '18 at 8:14
CruZCruZ
527
asked Nov 29 '18 at 8:14
CruZCruZ
527
asked Nov 29 '18 at 8:14
CruZCruZ
527
527
add a comment |
add a comment |
1 Answer
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$begingroup$
For $n=1000$ it is certainly reasonable to just use a normal approximation. Be careful to use a one-tailed test.
For $n$ too small to use CLT (10 is on the small side), you can go back to the basic definition of a binomial distribution to write
$$alpha=Pr(W_1>c_1,|,H_0)=sum_{i=c_1+1}^{10}binom{10}{i}left(frac{1}{2}right)^{10}text{,}$$
and solve for the smallest value of $c_1$ that leaves this probability below 0.05.
This is known as the (binomial) exact test since no approximations are used. Some authors take issue with saying the test is "at significance level 0.05" when the actual level is much lower. Of course, with 10 flips it is impossible to get that close to 0.05.
Note that we usually do not talk about "accepting" $H_0$ so much as failing to reject it, because not (enough) evidence that $H_0$ is false is not the same as evidence that $H_0$ is in fact true, in this frequentist setting.
answered Nov 29 '18 at 10:40
obscuransobscurans
1,152311
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1 Answer
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$begingroup$
For $n=1000$ it is certainly reasonable to just use a normal approximation. Be careful to use a one-tailed test.
For $n$ too small to use CLT (10 is on the small side), you can go back to the basic definition of a binomial distribution to write
$$alpha=Pr(W_1>c_1,|,H_0)=sum_{i=c_1+1}^{10}binom{10}{i}left(frac{1}{2}right)^{10}text{,}$$
and solve for the smallest value of $c_1$ that leaves this probability below 0.05.
This is known as the (binomial) exact test since no approximations are used. Some authors take issue with saying the test is "at significance level 0.05" when the actual level is much lower. Of course, with 10 flips it is impossible to get that close to 0.05.
Note that we usually do not talk about "accepting" $H_0$ so much as failing to reject it, because not (enough) evidence that $H_0$ is false is not the same as evidence that $H_0$ is in fact true, in this frequentist setting.
answered Nov 29 '18 at 10:40
obscuransobscurans
1,152311
$endgroup$
add a comment |
$begingroup$
For $n=1000$ it is certainly reasonable to just use a normal approximation. Be careful to use a one-tailed test.
For $n$ too small to use CLT (10 is on the small side), you can go back to the basic definition of a binomial distribution to write
$$alpha=Pr(W_1>c_1,|,H_0)=sum_{i=c_1+1}^{10}binom{10}{i}left(frac{1}{2}right)^{10}text{,}$$
and solve for the smallest value of $c_1$ that leaves this probability below 0.05.
This is known as the (binomial) exact test since no approximations are used. Some authors take issue with saying the test is "at significance level 0.05" when the actual level is much lower. Of course, with 10 flips it is impossible to get that close to 0.05.
Note that we usually do not talk about "accepting" $H_0$ so much as failing to reject it, because not (enough) evidence that $H_0$ is false is not the same as evidence that $H_0$ is in fact true, in this frequentist setting.
answered Nov 29 '18 at 10:40
obscuransobscurans
1,152311
$endgroup$
add a comment |
$begingroup$
For $n=1000$ it is certainly reasonable to just use a normal approximation. Be careful to use a one-tailed test.
For $n$ too small to use CLT (10 is on the small side), you can go back to the basic definition of a binomial distribution to write
$$alpha=Pr(W_1>c_1,|,H_0)=sum_{i=c_1+1}^{10}binom{10}{i}left(frac{1}{2}right)^{10}text{,}$$
and solve for the smallest value of $c_1$ that leaves this probability below 0.05.
This is known as the (binomial) exact test since no approximations are used. Some authors take issue with saying the test is "at significance level 0.05" when the actual level is much lower. Of course, with 10 flips it is impossible to get that close to 0.05.
Note that we usually do not talk about "accepting" $H_0$ so much as failing to reject it, because not (enough) evidence that $H_0$ is false is not the same as evidence that $H_0$ is in fact true, in this frequentist setting.
answered Nov 29 '18 at 10:40
obscuransobscurans
1,152311
$endgroup$
For $n=1000$ it is certainly reasonable to just use a normal approximation. Be careful to use a one-tailed test.
For $n$ too small to use CLT (10 is on the small side), you can go back to the basic definition of a binomial distribution to write
$$alpha=Pr(W_1>c_1,|,H_0)=sum_{i=c_1+1}^{10}binom{10}{i}left(frac{1}{2}right)^{10}text{,}$$
and solve for the smallest value of $c_1$ that leaves this probability below 0.05.
This is known as the (binomial) exact test since no approximations are used. Some authors take issue with saying the test is "at significance level 0.05" when the actual level is much lower. Of course, with 10 flips it is impossible to get that close to 0.05.
Note that we usually do not talk about "accepting" $H_0$ so much as failing to reject it, because not (enough) evidence that $H_0$ is false is not the same as evidence that $H_0$ is in fact true, in this frequentist setting.
answered Nov 29 '18 at 10:40
obscuransobscurans
1,152311
edited Nov 29 '18 at 10:45
answered Nov 29 '18 at 10:40
obscuransobscurans
1,152311
answered Nov 29 '18 at 10:40
obscuransobscurans
1,152311
answered Nov 29 '18 at 10:40
obscuransobscurans
1,152311
1,152311
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