Mixing
Study whether the turnout $𝑞$ is higher than $75%$ at a significance level of $1%$
$begingroup$
The famous pollster Maurice the Cat polled whether the Netherlands should stay in the European Union.
From a random sample of 1024 citizens it turns out that 250 of them are in favour of a Nexit.
From the same poll it tured out that $800$ of the $1024$ respondents would vote at a referendum about a departure of the Netherlands from the European Union.
Let $𝑞$ be the actual (still unknown) turnout at a referendum about a Nexit.
Study whether the turnout $𝑞$ is higher than $75%$ at a significance level of $1%$: formulate the relevant hypotheses, testing parameter with distribution and motivated conclusion.
I have some problem to formulate the hypothesis for this exercise, can someone help me with this?
Edit:
If I assume $H_0:q=0.75$ and $H_1:q>0.75$ with a significance level $alpha$ of $0.01$ I can find from the table the critical value that is $2.32$, now knowing that the distribution is a $Bin(1024,0.75)$, how can I proceed?
Edit 2:
$P(X>800)=Pleft(Z>frac{800-1024*0.75}{sqrt{1024*0.75*(1-0.75)}}right)=P(Z>2.309)=0.0107$ (from the standard normal table)
Edit 3:
$alpha=0.01$ correspond to a value of $2.326$ from the normal tables. $frac{n-1024*0.75}{sqrt{1024*0.75*(1-0.75)}}=2.326$ $n=800.234$, this number is bigger that $800$ so in my opinion we reject the null hypothesis, it's correct?
statistics
asked Jan 24 at 8:36
Luke MarciLuke Marci
856
$endgroup$
|
show 2 more comments
$begingroup$
The famous pollster Maurice the Cat polled whether the Netherlands should stay in the European Union.
From a random sample of 1024 citizens it turns out that 250 of them are in favour of a Nexit.
From the same poll it tured out that $800$ of the $1024$ respondents would vote at a referendum about a departure of the Netherlands from the European Union.
Let $𝑞$ be the actual (still unknown) turnout at a referendum about a Nexit.
Study whether the turnout $𝑞$ is higher than $75%$ at a significance level of $1%$: formulate the relevant hypotheses, testing parameter with distribution and motivated conclusion.
I have some problem to formulate the hypothesis for this exercise, can someone help me with this?
Edit:
If I assume $H_0:q=0.75$ and $H_1:q>0.75$ with a significance level $alpha$ of $0.01$ I can find from the table the critical value that is $2.32$, now knowing that the distribution is a $Bin(1024,0.75)$, how can I proceed?
Edit 2:
$P(X>800)=Pleft(Z>frac{800-1024*0.75}{sqrt{1024*0.75*(1-0.75)}}right)=P(Z>2.309)=0.0107$ (from the standard normal table)
Edit 3:
$alpha=0.01$ correspond to a value of $2.326$ from the normal tables. $frac{n-1024*0.75}{sqrt{1024*0.75*(1-0.75)}}=2.326$ $n=800.234$, this number is bigger that $800$ so in my opinion we reject the null hypothesis, it's correct?
statistics
asked Jan 24 at 8:36
Luke MarciLuke Marci
856
$endgroup$
1
$begingroup$
Solution using exact comp. in R: Testing $H_0: q = .75$ vs $H_a: q > .75$ and using poll results as data, one might rej at level 1% if the nr out of 1024 who say they would vote is $ge$ the 99th percentile $c$ of $mathsf{Binom}(1024, .75).$ R codeqbinom(.99, 1024, .75)returns $c=800.$ So one would (just barely) reject $H_0$ and predict that the turnout will exceed 75%. Of course, this depends on the validity of the poll and the assumption that enthusiasm for voting remains unchanged from poll to election day. // I suppose a solution by normal approx is expected in a course not using R.
$endgroup$
– BruceET
Jan 24 at 10:06
$begingroup$
@BruceET Thanks for the comment, I edited the question, now how I can I proceed?
$endgroup$
– Luke Marci
Jan 24 at 11:47
1
$begingroup$
To start, can you use normal approx to binomial to find $P(X ge 800)$ for $X sim mathsf{Binom}(1024, .75)?$ . If this clue is not enough, say what you have covered recently, what you have tried, why you're stuck. In order to provide answers relevant to you, we need to see some engagement from you.
$endgroup$
– BruceET
Jan 24 at 18:47
1
$begingroup$
Your latest result is very near 1%. If you wanted exactly $0.01$ what number would you substitute for 2.309? That would be the critical value for the normal-approximated test at 1%? Going back to the scale of binomial counts, what number, if any, would you substitute for 800 to get that critical value? (Remember you're dealing with a normal approximation to binomial,) Now maybe put it all together and bravely post an answer to your own question.
$endgroup$
– BruceET
Jan 24 at 22:45
1
$begingroup$
If your computation is correct the observed 800 (just barely) fails to be larger than your critical value 800.2. So on a very close call, you would not reject at the 1.0% level. But you could reject at the 1.1% level. If the observed value were 801 then you could reject at the 1% level. // You were asked to "study" the situation. There is strong evidence that more than 3/4 of the population will vote in a "Nexit" election, but not quite enough to reject at the 1% level. // Dealing with the discrete binomial dist'n it is not possible to test precisely at the 1% level.// Good luck; won't chat.
$endgroup$
– BruceET
Jan 25 at 9:08
|
show 2 more comments
$begingroup$
The famous pollster Maurice the Cat polled whether the Netherlands should stay in the European Union.
From a random sample of 1024 citizens it turns out that 250 of them are in favour of a Nexit.
From the same poll it tured out that $800$ of the $1024$ respondents would vote at a referendum about a departure of the Netherlands from the European Union.
Let $𝑞$ be the actual (still unknown) turnout at a referendum about a Nexit.
Study whether the turnout $𝑞$ is higher than $75%$ at a significance level of $1%$: formulate the relevant hypotheses, testing parameter with distribution and motivated conclusion.
I have some problem to formulate the hypothesis for this exercise, can someone help me with this?
Edit:
If I assume $H_0:q=0.75$ and $H_1:q>0.75$ with a significance level $alpha$ of $0.01$ I can find from the table the critical value that is $2.32$, now knowing that the distribution is a $Bin(1024,0.75)$, how can I proceed?
Edit 2:
$P(X>800)=Pleft(Z>frac{800-1024*0.75}{sqrt{1024*0.75*(1-0.75)}}right)=P(Z>2.309)=0.0107$ (from the standard normal table)
Edit 3:
$alpha=0.01$ correspond to a value of $2.326$ from the normal tables. $frac{n-1024*0.75}{sqrt{1024*0.75*(1-0.75)}}=2.326$ $n=800.234$, this number is bigger that $800$ so in my opinion we reject the null hypothesis, it's correct?
statistics
asked Jan 24 at 8:36
Luke MarciLuke Marci
856
$endgroup$
The famous pollster Maurice the Cat polled whether the Netherlands should stay in the European Union.
From a random sample of 1024 citizens it turns out that 250 of them are in favour of a Nexit.
From the same poll it tured out that $800$ of the $1024$ respondents would vote at a referendum about a departure of the Netherlands from the European Union.
Let $𝑞$ be the actual (still unknown) turnout at a referendum about a Nexit.
Study whether the turnout $𝑞$ is higher than $75%$ at a significance level of $1%$: formulate the relevant hypotheses, testing parameter with distribution and motivated conclusion.
I have some problem to formulate the hypothesis for this exercise, can someone help me with this?
Edit:
If I assume $H_0:q=0.75$ and $H_1:q>0.75$ with a significance level $alpha$ of $0.01$ I can find from the table the critical value that is $2.32$, now knowing that the distribution is a $Bin(1024,0.75)$, how can I proceed?
Edit 2:
$P(X>800)=Pleft(Z>frac{800-1024*0.75}{sqrt{1024*0.75*(1-0.75)}}right)=P(Z>2.309)=0.0107$ (from the standard normal table)
Edit 3:
$alpha=0.01$ correspond to a value of $2.326$ from the normal tables. $frac{n-1024*0.75}{sqrt{1024*0.75*(1-0.75)}}=2.326$ $n=800.234$, this number is bigger that $800$ so in my opinion we reject the null hypothesis, it's correct?
statistics
statistics
asked Jan 24 at 8:36
Luke MarciLuke Marci
856
asked Jan 24 at 8:36
Luke MarciLuke Marci
856
edited Jan 25 at 8:48
Luke Marci
asked Jan 24 at 8:36
Luke MarciLuke Marci
856
asked Jan 24 at 8:36
Luke MarciLuke Marci
856
asked Jan 24 at 8:36
Luke MarciLuke Marci
856
856
1
$begingroup$
Solution using exact comp. in R: Testing $H_0: q = .75$ vs $H_a: q > .75$ and using poll results as data, one might rej at level 1% if the nr out of 1024 who say they would vote is $ge$ the 99th percentile $c$ of $mathsf{Binom}(1024, .75).$ R codeqbinom(.99, 1024, .75)returns $c=800.$ So one would (just barely) reject $H_0$ and predict that the turnout will exceed 75%. Of course, this depends on the validity of the poll and the assumption that enthusiasm for voting remains unchanged from poll to election day. // I suppose a solution by normal approx is expected in a course not using R.
$endgroup$
– BruceET
Jan 24 at 10:06
$begingroup$
@BruceET Thanks for the comment, I edited the question, now how I can I proceed?
$endgroup$
– Luke Marci
Jan 24 at 11:47
1
$begingroup$
To start, can you use normal approx to binomial to find $P(X ge 800)$ for $X sim mathsf{Binom}(1024, .75)?$ . If this clue is not enough, say what you have covered recently, what you have tried, why you're stuck. In order to provide answers relevant to you, we need to see some engagement from you.
$endgroup$
– BruceET
Jan 24 at 18:47
1
$begingroup$
Your latest result is very near 1%. If you wanted exactly $0.01$ what number would you substitute for 2.309? That would be the critical value for the normal-approximated test at 1%? Going back to the scale of binomial counts, what number, if any, would you substitute for 800 to get that critical value? (Remember you're dealing with a normal approximation to binomial,) Now maybe put it all together and bravely post an answer to your own question.
$endgroup$
– BruceET
Jan 24 at 22:45
1
$begingroup$
If your computation is correct the observed 800 (just barely) fails to be larger than your critical value 800.2. So on a very close call, you would not reject at the 1.0% level. But you could reject at the 1.1% level. If the observed value were 801 then you could reject at the 1% level. // You were asked to "study" the situation. There is strong evidence that more than 3/4 of the population will vote in a "Nexit" election, but not quite enough to reject at the 1% level. // Dealing with the discrete binomial dist'n it is not possible to test precisely at the 1% level.// Good luck; won't chat.
$endgroup$
– BruceET
Jan 25 at 9:08
|
show 2 more comments
1
$begingroup$
Solution using exact comp. in R: Testing $H_0: q = .75$ vs $H_a: q > .75$ and using poll results as data, one might rej at level 1% if the nr out of 1024 who say they would vote is $ge$ the 99th percentile $c$ of $mathsf{Binom}(1024, .75).$ R codeqbinom(.99, 1024, .75)returns $c=800.$ So one would (just barely) reject $H_0$ and predict that the turnout will exceed 75%. Of course, this depends on the validity of the poll and the assumption that enthusiasm for voting remains unchanged from poll to election day. // I suppose a solution by normal approx is expected in a course not using R.
$endgroup$
– BruceET
Jan 24 at 10:06
$begingroup$
@BruceET Thanks for the comment, I edited the question, now how I can I proceed?
$endgroup$
– Luke Marci
Jan 24 at 11:47
1
$begingroup$
To start, can you use normal approx to binomial to find $P(X ge 800)$ for $X sim mathsf{Binom}(1024, .75)?$ . If this clue is not enough, say what you have covered recently, what you have tried, why you're stuck. In order to provide answers relevant to you, we need to see some engagement from you.
$endgroup$
– BruceET
Jan 24 at 18:47
1
$begingroup$
Your latest result is very near 1%. If you wanted exactly $0.01$ what number would you substitute for 2.309? That would be the critical value for the normal-approximated test at 1%? Going back to the scale of binomial counts, what number, if any, would you substitute for 800 to get that critical value? (Remember you're dealing with a normal approximation to binomial,) Now maybe put it all together and bravely post an answer to your own question.
$endgroup$
– BruceET
Jan 24 at 22:45
1
$begingroup$
If your computation is correct the observed 800 (just barely) fails to be larger than your critical value 800.2. So on a very close call, you would not reject at the 1.0% level. But you could reject at the 1.1% level. If the observed value were 801 then you could reject at the 1% level. // You were asked to "study" the situation. There is strong evidence that more than 3/4 of the population will vote in a "Nexit" election, but not quite enough to reject at the 1% level. // Dealing with the discrete binomial dist'n it is not possible to test precisely at the 1% level.// Good luck; won't chat.
$endgroup$
– BruceET
Jan 25 at 9:08
1
1
$begingroup$
Solution using exact comp. in R: Testing $H_0: q = .75$ vs $H_a: q > .75$ and using poll results as data, one might rej at level 1% if the nr out of 1024 who say they would vote is $ge$ the 99th percentile $c$ of $mathsf{Binom}(1024, .75).$ R code
qbinom(.99, 1024, .75) returns $c=800.$ So one would (just barely) reject $H_0$ and predict that the turnout will exceed 75%. Of course, this depends on the validity of the poll and the assumption that enthusiasm for voting remains unchanged from poll to election day. // I suppose a solution by normal approx is expected in a course not using R.$endgroup$
– BruceET
Jan 24 at 10:06
$begingroup$
Solution using exact comp. in R: Testing $H_0: q = .75$ vs $H_a: q > .75$ and using poll results as data, one might rej at level 1% if the nr out of 1024 who say they would vote is $ge$ the 99th percentile $c$ of $mathsf{Binom}(1024, .75).$ R code
qbinom(.99, 1024, .75) returns $c=800.$ So one would (just barely) reject $H_0$ and predict that the turnout will exceed 75%. Of course, this depends on the validity of the poll and the assumption that enthusiasm for voting remains unchanged from poll to election day. // I suppose a solution by normal approx is expected in a course not using R.$endgroup$
– BruceET
Jan 24 at 10:06
$begingroup$
@BruceET Thanks for the comment, I edited the question, now how I can I proceed?
$endgroup$
– Luke Marci
Jan 24 at 11:47
$begingroup$
@BruceET Thanks for the comment, I edited the question, now how I can I proceed?
$endgroup$
– Luke Marci
Jan 24 at 11:47
1
1
$begingroup$
To start, can you use normal approx to binomial to find $P(X ge 800)$ for $X sim mathsf{Binom}(1024, .75)?$ . If this clue is not enough, say what you have covered recently, what you have tried, why you're stuck. In order to provide answers relevant to you, we need to see some engagement from you.
$endgroup$
– BruceET
Jan 24 at 18:47
$begingroup$
To start, can you use normal approx to binomial to find $P(X ge 800)$ for $X sim mathsf{Binom}(1024, .75)?$ . If this clue is not enough, say what you have covered recently, what you have tried, why you're stuck. In order to provide answers relevant to you, we need to see some engagement from you.
$endgroup$
– BruceET
Jan 24 at 18:47
1
1
$begingroup$
Your latest result is very near 1%. If you wanted exactly $0.01$ what number would you substitute for 2.309? That would be the critical value for the normal-approximated test at 1%? Going back to the scale of binomial counts, what number, if any, would you substitute for 800 to get that critical value? (Remember you're dealing with a normal approximation to binomial,) Now maybe put it all together and bravely post an answer to your own question.
$endgroup$
– BruceET
Jan 24 at 22:45
$begingroup$
Your latest result is very near 1%. If you wanted exactly $0.01$ what number would you substitute for 2.309? That would be the critical value for the normal-approximated test at 1%? Going back to the scale of binomial counts, what number, if any, would you substitute for 800 to get that critical value? (Remember you're dealing with a normal approximation to binomial,) Now maybe put it all together and bravely post an answer to your own question.
$endgroup$
– BruceET
Jan 24 at 22:45
1
1
$begingroup$
If your computation is correct the observed 800 (just barely) fails to be larger than your critical value 800.2. So on a very close call, you would not reject at the 1.0% level. But you could reject at the 1.1% level. If the observed value were 801 then you could reject at the 1% level. // You were asked to "study" the situation. There is strong evidence that more than 3/4 of the population will vote in a "Nexit" election, but not quite enough to reject at the 1% level. // Dealing with the discrete binomial dist'n it is not possible to test precisely at the 1% level.// Good luck; won't chat.
$endgroup$
– BruceET
Jan 25 at 9:08
$begingroup$
If your computation is correct the observed 800 (just barely) fails to be larger than your critical value 800.2. So on a very close call, you would not reject at the 1.0% level. But you could reject at the 1.1% level. If the observed value were 801 then you could reject at the 1% level. // You were asked to "study" the situation. There is strong evidence that more than 3/4 of the population will vote in a "Nexit" election, but not quite enough to reject at the 1% level. // Dealing with the discrete binomial dist'n it is not possible to test precisely at the 1% level.// Good luck; won't chat.
$endgroup$
– BruceET
Jan 25 at 9:08
|
show 2 more comments
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$begingroup$
Solution using exact comp. in R: Testing $H_0: q = .75$ vs $H_a: q > .75$ and using poll results as data, one might rej at level 1% if the nr out of 1024 who say they would vote is $ge$ the 99th percentile $c$ of $mathsf{Binom}(1024, .75).$ R code
qbinom(.99, 1024, .75)returns $c=800.$ So one would (just barely) reject $H_0$ and predict that the turnout will exceed 75%. Of course, this depends on the validity of the poll and the assumption that enthusiasm for voting remains unchanged from poll to election day. // I suppose a solution by normal approx is expected in a course not using R.$endgroup$
– BruceET
Jan 24 at 10:06
$begingroup$
@BruceET Thanks for the comment, I edited the question, now how I can I proceed?
$endgroup$
– Luke Marci
Jan 24 at 11:47
1
$begingroup$
To start, can you use normal approx to binomial to find $P(X ge 800)$ for $X sim mathsf{Binom}(1024, .75)?$ . If this clue is not enough, say what you have covered recently, what you have tried, why you're stuck. In order to provide answers relevant to you, we need to see some engagement from you.
$endgroup$
– BruceET
Jan 24 at 18:47
1
$begingroup$
Your latest result is very near 1%. If you wanted exactly $0.01$ what number would you substitute for 2.309? That would be the critical value for the normal-approximated test at 1%? Going back to the scale of binomial counts, what number, if any, would you substitute for 800 to get that critical value? (Remember you're dealing with a normal approximation to binomial,) Now maybe put it all together and bravely post an answer to your own question.
$endgroup$
– BruceET
Jan 24 at 22:45
1
$begingroup$
If your computation is correct the observed 800 (just barely) fails to be larger than your critical value 800.2. So on a very close call, you would not reject at the 1.0% level. But you could reject at the 1.1% level. If the observed value were 801 then you could reject at the 1% level. // You were asked to "study" the situation. There is strong evidence that more than 3/4 of the population will vote in a "Nexit" election, but not quite enough to reject at the 1% level. // Dealing with the discrete binomial dist'n it is not possible to test precisely at the 1% level.// Good luck; won't chat.
$endgroup$
– BruceET
Jan 25 at 9:08