Mixing
Statistics : Hypothesis Testing [closed]
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I'm back with another question I can't solve. For this question, I was able to get part of the answer; nk$^{n-1}$(1-k), but I can't seem to see where you'd get the k$^n$ part.
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Answers
binomial-distribution hypothesis-testing
asked Jan 9 at 19:43
Jia Xuan NgJia Xuan Ng
101
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closed as off-topic by heropup, verret, Math1000, Eevee Trainer, Leucippus Jan 10 at 3:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – heropup, Math1000, Eevee Trainer, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
I'm back with another question I can't solve. For this question, I was able to get part of the answer; nk$^{n-1}$(1-k), but I can't seem to see where you'd get the k$^n$ part.
Questions
Answers
binomial-distribution hypothesis-testing
asked Jan 9 at 19:43
Jia Xuan NgJia Xuan Ng
101
$endgroup$
closed as off-topic by heropup, verret, Math1000, Eevee Trainer, Leucippus Jan 10 at 3:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – heropup, Math1000, Eevee Trainer, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm back with another question I can't solve. For this question, I was able to get part of the answer; nk$^{n-1}$(1-k), but I can't seem to see where you'd get the k$^n$ part.
Questions
Answers
binomial-distribution hypothesis-testing
asked Jan 9 at 19:43
Jia Xuan NgJia Xuan Ng
101
$endgroup$
I'm back with another question I can't solve. For this question, I was able to get part of the answer; nk$^{n-1}$(1-k), but I can't seem to see where you'd get the k$^n$ part.
Questions
Answers
binomial-distribution hypothesis-testing
binomial-distribution hypothesis-testing
asked Jan 9 at 19:43
Jia Xuan NgJia Xuan Ng
101
asked Jan 9 at 19:43
Jia Xuan NgJia Xuan Ng
101
asked Jan 9 at 19:43
Jia Xuan NgJia Xuan Ng
101
asked Jan 9 at 19:43
Jia Xuan NgJia Xuan Ng
101
asked Jan 9 at 19:43
Jia Xuan NgJia Xuan Ng
101
101
closed as off-topic by heropup, verret, Math1000, Eevee Trainer, Leucippus Jan 10 at 3:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – heropup, Math1000, Eevee Trainer, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by heropup, verret, Math1000, Eevee Trainer, Leucippus Jan 10 at 3:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – heropup, Math1000, Eevee Trainer, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If the alternative hypothesis is $H_1: p>k$ then it is an right tailed test, where $c$ is the critical value. If the critical value is part of the critical region, then the critical region is ${c,c+1,ldots, n}$.
In your case $c=n-1$. Therefore the critical region is ${n-1, n}$. The largest element is always $n$. Now you sum up these probabilities:
$$sum_{x=n-1}^{n} binom{n}{x}cdot k^xcdot (1-k)^{n-x}$$
$$=binom{n}{n-1}cdot k^{n-1}cdot (1-k)^{1}+binom{n}{n}cdot k^{n}cdot (1-k)^{0}=ncdot k^{n-1}cdot (1-k)+k^n$$
The source is german wiki (translated).
$alpha$ is the significance level.
answered Jan 9 at 20:59
callculuscallculus
18k31427
$endgroup$
$begingroup$
I'm still confused on the k$^n$ part, do you do another binomial expansion but use n instead of n-1,and then add the two together?
$endgroup$
– Jia Xuan Ng
Jan 9 at 22:57
$begingroup$
@JiaXuanNg It has less to do withbinomial expansionthan the probabilities $P(X=n-1)$ and $P(X=n)$, where $Xsim Bin(n, k)$. I´ve calculated these two probabilities since $n-1$ and $n$ are the elements of the critical region. And yes, you have to add together since we want the probability for the (whole) critical region. Please, give a reply if you have additional questions or not.
$endgroup$
– callculus
Jan 10 at 0:54
$begingroup$
Sorry, was asleep. I'm starting to get it now, thank you very much. So if the critical region was n+1,would n also be part of it? Or is it n+2?
$endgroup$
– Jia Xuan Ng
Jan 10 at 7:58
$begingroup$
@JiaXuanNg No problem. I appreciate to get a reply, but it musn´t be immediately. No, the critical region goes from the critical value to n. See my edit. At the table the yellow marked row refers to your case. If the lower bound is $c=n-1$ then the probablity of the critical region is $sum_{i=c}^n B(i| p_0,n)=sumlimits_{i=n-1}^n B(i| k,n)$
$endgroup$
– callculus
Jan 10 at 21:06
$begingroup$
Ooh, I think I understand it now. I'm going to practice doing this types of questions. Thank you again!
$endgroup$
– Jia Xuan Ng
Jan 11 at 7:53
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the alternative hypothesis is $H_1: p>k$ then it is an right tailed test, where $c$ is the critical value. If the critical value is part of the critical region, then the critical region is ${c,c+1,ldots, n}$.
In your case $c=n-1$. Therefore the critical region is ${n-1, n}$. The largest element is always $n$. Now you sum up these probabilities:
$$sum_{x=n-1}^{n} binom{n}{x}cdot k^xcdot (1-k)^{n-x}$$
$$=binom{n}{n-1}cdot k^{n-1}cdot (1-k)^{1}+binom{n}{n}cdot k^{n}cdot (1-k)^{0}=ncdot k^{n-1}cdot (1-k)+k^n$$
The source is german wiki (translated).
$alpha$ is the significance level.
answered Jan 9 at 20:59
callculuscallculus
18k31427
$endgroup$
$begingroup$
I'm still confused on the k$^n$ part, do you do another binomial expansion but use n instead of n-1,and then add the two together?
$endgroup$
– Jia Xuan Ng
Jan 9 at 22:57
$begingroup$
@JiaXuanNg It has less to do withbinomial expansionthan the probabilities $P(X=n-1)$ and $P(X=n)$, where $Xsim Bin(n, k)$. I´ve calculated these two probabilities since $n-1$ and $n$ are the elements of the critical region. And yes, you have to add together since we want the probability for the (whole) critical region. Please, give a reply if you have additional questions or not.
$endgroup$
– callculus
Jan 10 at 0:54
$begingroup$
Sorry, was asleep. I'm starting to get it now, thank you very much. So if the critical region was n+1,would n also be part of it? Or is it n+2?
$endgroup$
– Jia Xuan Ng
Jan 10 at 7:58
$begingroup$
@JiaXuanNg No problem. I appreciate to get a reply, but it musn´t be immediately. No, the critical region goes from the critical value to n. See my edit. At the table the yellow marked row refers to your case. If the lower bound is $c=n-1$ then the probablity of the critical region is $sum_{i=c}^n B(i| p_0,n)=sumlimits_{i=n-1}^n B(i| k,n)$
$endgroup$
– callculus
Jan 10 at 21:06
$begingroup$
Ooh, I think I understand it now. I'm going to practice doing this types of questions. Thank you again!
$endgroup$
– Jia Xuan Ng
Jan 11 at 7:53
|
show 2 more comments
$begingroup$
If the alternative hypothesis is $H_1: p>k$ then it is an right tailed test, where $c$ is the critical value. If the critical value is part of the critical region, then the critical region is ${c,c+1,ldots, n}$.
In your case $c=n-1$. Therefore the critical region is ${n-1, n}$. The largest element is always $n$. Now you sum up these probabilities:
$$sum_{x=n-1}^{n} binom{n}{x}cdot k^xcdot (1-k)^{n-x}$$
$$=binom{n}{n-1}cdot k^{n-1}cdot (1-k)^{1}+binom{n}{n}cdot k^{n}cdot (1-k)^{0}=ncdot k^{n-1}cdot (1-k)+k^n$$
The source is german wiki (translated).
$alpha$ is the significance level.
answered Jan 9 at 20:59
callculuscallculus
18k31427
$endgroup$
$begingroup$
I'm still confused on the k$^n$ part, do you do another binomial expansion but use n instead of n-1,and then add the two together?
$endgroup$
– Jia Xuan Ng
Jan 9 at 22:57
$begingroup$
@JiaXuanNg It has less to do withbinomial expansionthan the probabilities $P(X=n-1)$ and $P(X=n)$, where $Xsim Bin(n, k)$. I´ve calculated these two probabilities since $n-1$ and $n$ are the elements of the critical region. And yes, you have to add together since we want the probability for the (whole) critical region. Please, give a reply if you have additional questions or not.
$endgroup$
– callculus
Jan 10 at 0:54
$begingroup$
Sorry, was asleep. I'm starting to get it now, thank you very much. So if the critical region was n+1,would n also be part of it? Or is it n+2?
$endgroup$
– Jia Xuan Ng
Jan 10 at 7:58
$begingroup$
@JiaXuanNg No problem. I appreciate to get a reply, but it musn´t be immediately. No, the critical region goes from the critical value to n. See my edit. At the table the yellow marked row refers to your case. If the lower bound is $c=n-1$ then the probablity of the critical region is $sum_{i=c}^n B(i| p_0,n)=sumlimits_{i=n-1}^n B(i| k,n)$
$endgroup$
– callculus
Jan 10 at 21:06
$begingroup$
Ooh, I think I understand it now. I'm going to practice doing this types of questions. Thank you again!
$endgroup$
– Jia Xuan Ng
Jan 11 at 7:53
|
show 2 more comments
$begingroup$
If the alternative hypothesis is $H_1: p>k$ then it is an right tailed test, where $c$ is the critical value. If the critical value is part of the critical region, then the critical region is ${c,c+1,ldots, n}$.
In your case $c=n-1$. Therefore the critical region is ${n-1, n}$. The largest element is always $n$. Now you sum up these probabilities:
$$sum_{x=n-1}^{n} binom{n}{x}cdot k^xcdot (1-k)^{n-x}$$
$$=binom{n}{n-1}cdot k^{n-1}cdot (1-k)^{1}+binom{n}{n}cdot k^{n}cdot (1-k)^{0}=ncdot k^{n-1}cdot (1-k)+k^n$$
The source is german wiki (translated).
$alpha$ is the significance level.
answered Jan 9 at 20:59
callculuscallculus
18k31427
$endgroup$
If the alternative hypothesis is $H_1: p>k$ then it is an right tailed test, where $c$ is the critical value. If the critical value is part of the critical region, then the critical region is ${c,c+1,ldots, n}$.
In your case $c=n-1$. Therefore the critical region is ${n-1, n}$. The largest element is always $n$. Now you sum up these probabilities:
$$sum_{x=n-1}^{n} binom{n}{x}cdot k^xcdot (1-k)^{n-x}$$
$$=binom{n}{n-1}cdot k^{n-1}cdot (1-k)^{1}+binom{n}{n}cdot k^{n}cdot (1-k)^{0}=ncdot k^{n-1}cdot (1-k)+k^n$$
The source is german wiki (translated).
$alpha$ is the significance level.
answered Jan 9 at 20:59
callculuscallculus
18k31427
edited Jan 10 at 21:04
answered Jan 9 at 20:59
callculuscallculus
18k31427
answered Jan 9 at 20:59
callculuscallculus
18k31427
answered Jan 9 at 20:59
callculuscallculus
18k31427
18k31427
$begingroup$
I'm still confused on the k$^n$ part, do you do another binomial expansion but use n instead of n-1,and then add the two together?
$endgroup$
– Jia Xuan Ng
Jan 9 at 22:57
$begingroup$
@JiaXuanNg It has less to do withbinomial expansionthan the probabilities $P(X=n-1)$ and $P(X=n)$, where $Xsim Bin(n, k)$. I´ve calculated these two probabilities since $n-1$ and $n$ are the elements of the critical region. And yes, you have to add together since we want the probability for the (whole) critical region. Please, give a reply if you have additional questions or not.
$endgroup$
– callculus
Jan 10 at 0:54
$begingroup$
Sorry, was asleep. I'm starting to get it now, thank you very much. So if the critical region was n+1,would n also be part of it? Or is it n+2?
$endgroup$
– Jia Xuan Ng
Jan 10 at 7:58
$begingroup$
@JiaXuanNg No problem. I appreciate to get a reply, but it musn´t be immediately. No, the critical region goes from the critical value to n. See my edit. At the table the yellow marked row refers to your case. If the lower bound is $c=n-1$ then the probablity of the critical region is $sum_{i=c}^n B(i| p_0,n)=sumlimits_{i=n-1}^n B(i| k,n)$
$endgroup$
– callculus
Jan 10 at 21:06
$begingroup$
Ooh, I think I understand it now. I'm going to practice doing this types of questions. Thank you again!
$endgroup$
– Jia Xuan Ng
Jan 11 at 7:53
|
show 2 more comments
$begingroup$
I'm still confused on the k$^n$ part, do you do another binomial expansion but use n instead of n-1,and then add the two together?
$endgroup$
– Jia Xuan Ng
Jan 9 at 22:57
$begingroup$
@JiaXuanNg It has less to do withbinomial expansionthan the probabilities $P(X=n-1)$ and $P(X=n)$, where $Xsim Bin(n, k)$. I´ve calculated these two probabilities since $n-1$ and $n$ are the elements of the critical region. And yes, you have to add together since we want the probability for the (whole) critical region. Please, give a reply if you have additional questions or not.
$endgroup$
– callculus
Jan 10 at 0:54
$begingroup$
Sorry, was asleep. I'm starting to get it now, thank you very much. So if the critical region was n+1,would n also be part of it? Or is it n+2?
$endgroup$
– Jia Xuan Ng
Jan 10 at 7:58
$begingroup$
@JiaXuanNg No problem. I appreciate to get a reply, but it musn´t be immediately. No, the critical region goes from the critical value to n. See my edit. At the table the yellow marked row refers to your case. If the lower bound is $c=n-1$ then the probablity of the critical region is $sum_{i=c}^n B(i| p_0,n)=sumlimits_{i=n-1}^n B(i| k,n)$
$endgroup$
– callculus
Jan 10 at 21:06
$begingroup$
Ooh, I think I understand it now. I'm going to practice doing this types of questions. Thank you again!
$endgroup$
– Jia Xuan Ng
Jan 11 at 7:53
$begingroup$
I'm still confused on the k$^n$ part, do you do another binomial expansion but use n instead of n-1,and then add the two together?
$endgroup$
– Jia Xuan Ng
Jan 9 at 22:57
$begingroup$
I'm still confused on the k$^n$ part, do you do another binomial expansion but use n instead of n-1,and then add the two together?
$endgroup$
– Jia Xuan Ng
Jan 9 at 22:57
$begingroup$
@JiaXuanNg It has less to do with
binomial expansion than the probabilities $P(X=n-1)$ and $P(X=n)$, where $Xsim Bin(n, k)$. I´ve calculated these two probabilities since $n-1$ and $n$ are the elements of the critical region. And yes, you have to add together since we want the probability for the (whole) critical region. Please, give a reply if you have additional questions or not.$endgroup$
– callculus
Jan 10 at 0:54
$begingroup$
@JiaXuanNg It has less to do with
binomial expansion than the probabilities $P(X=n-1)$ and $P(X=n)$, where $Xsim Bin(n, k)$. I´ve calculated these two probabilities since $n-1$ and $n$ are the elements of the critical region. And yes, you have to add together since we want the probability for the (whole) critical region. Please, give a reply if you have additional questions or not.$endgroup$
– callculus
Jan 10 at 0:54
$begingroup$
Sorry, was asleep. I'm starting to get it now, thank you very much. So if the critical region was n+1,would n also be part of it? Or is it n+2?
$endgroup$
– Jia Xuan Ng
Jan 10 at 7:58
$begingroup$
Sorry, was asleep. I'm starting to get it now, thank you very much. So if the critical region was n+1,would n also be part of it? Or is it n+2?
$endgroup$
– Jia Xuan Ng
Jan 10 at 7:58
$begingroup$
@JiaXuanNg No problem. I appreciate to get a reply, but it musn´t be immediately. No, the critical region goes from the critical value to n. See my edit. At the table the yellow marked row refers to your case. If the lower bound is $c=n-1$ then the probablity of the critical region is $sum_{i=c}^n B(i| p_0,n)=sumlimits_{i=n-1}^n B(i| k,n)$
$endgroup$
– callculus
Jan 10 at 21:06
$begingroup$
@JiaXuanNg No problem. I appreciate to get a reply, but it musn´t be immediately. No, the critical region goes from the critical value to n. See my edit. At the table the yellow marked row refers to your case. If the lower bound is $c=n-1$ then the probablity of the critical region is $sum_{i=c}^n B(i| p_0,n)=sumlimits_{i=n-1}^n B(i| k,n)$
$endgroup$
– callculus
Jan 10 at 21:06
$begingroup$
Ooh, I think I understand it now. I'm going to practice doing this types of questions. Thank you again!
$endgroup$
– Jia Xuan Ng
Jan 11 at 7:53
$begingroup$
Ooh, I think I understand it now. I'm going to practice doing this types of questions. Thank you again!
$endgroup$
– Jia Xuan Ng
Jan 11 at 7:53
|
show 2 more comments
