Mixing
Find the 99% confidence interval (Interval and test for proportion)
Clash Royale CLAN TAG#URR8PPP
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I have the following question and I am getting different results from a friend (I think he forgot to halve $ alpha $). Problem is:
A random sample of 100 students from a large school was taken. It was found 38 went on a trip last month, 62 had not. Obtain a 99% confidence interval for the proportion of students who went on a trip last month.
Solution:
$$ N=100, X=38$$
$$hatp = XoverN$$
$$ hatp pm z_alpha/2 sqrthatp(1-hatp)overN$$
I then take $z_alpha/2$ from the t-table corresponding to $alpha = 0.005$ and $n=infty$ which is 2.576. Is this correct?
Therefore, I get the interval of (0.255, 0.505).
statistics statistical-inference confidence-interval
asked 5 hours ago
s5s
18516
add a comment |Â
up vote
2
down vote
favorite
I have the following question and I am getting different results from a friend (I think he forgot to halve $ alpha $). Problem is:
A random sample of 100 students from a large school was taken. It was found 38 went on a trip last month, 62 had not. Obtain a 99% confidence interval for the proportion of students who went on a trip last month.
Solution:
$$ N=100, X=38$$
$$hatp = XoverN$$
$$ hatp pm z_alpha/2 sqrthatp(1-hatp)overN$$
I then take $z_alpha/2$ from the t-table corresponding to $alpha = 0.005$ and $n=infty$ which is 2.576. Is this correct?
Therefore, I get the interval of (0.255, 0.505).
statistics statistical-inference confidence-interval
asked 5 hours ago
s5s
18516
What is the result of your friend?
– callculus
4 hours ago
After I´ve made my own calculation I got the same result like you. Therefore it is correct :) The bounds are $0.38pm 2.576cdot sqrtfrac0.62cdot 0.38100$
– callculus
4 hours ago
The results are here
– callculus
4 hours ago
@callculus he didn't halve $alpha$ so he used $z=1.96$
– s5s
4 hours ago
That´s indeed not right, since it is a two sided interval. But you are right.
– callculus
4 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the following question and I am getting different results from a friend (I think he forgot to halve $ alpha $). Problem is:
A random sample of 100 students from a large school was taken. It was found 38 went on a trip last month, 62 had not. Obtain a 99% confidence interval for the proportion of students who went on a trip last month.
Solution:
$$ N=100, X=38$$
$$hatp = XoverN$$
$$ hatp pm z_alpha/2 sqrthatp(1-hatp)overN$$
I then take $z_alpha/2$ from the t-table corresponding to $alpha = 0.005$ and $n=infty$ which is 2.576. Is this correct?
Therefore, I get the interval of (0.255, 0.505).
statistics statistical-inference confidence-interval
asked 5 hours ago
s5s
18516
I have the following question and I am getting different results from a friend (I think he forgot to halve $ alpha $). Problem is:
A random sample of 100 students from a large school was taken. It was found 38 went on a trip last month, 62 had not. Obtain a 99% confidence interval for the proportion of students who went on a trip last month.
Solution:
$$ N=100, X=38$$
$$hatp = XoverN$$
$$ hatp pm z_alpha/2 sqrthatp(1-hatp)overN$$
I then take $z_alpha/2$ from the t-table corresponding to $alpha = 0.005$ and $n=infty$ which is 2.576. Is this correct?
Therefore, I get the interval of (0.255, 0.505).
statistics statistical-inference confidence-interval
asked 5 hours ago
s5s
18516
edited 3 hours ago
asked 5 hours ago
s5s
18516
asked 5 hours ago
s5s
18516
asked 5 hours ago
s5s
18516
18516
What is the result of your friend?
– callculus
4 hours ago
After I´ve made my own calculation I got the same result like you. Therefore it is correct :) The bounds are $0.38pm 2.576cdot sqrtfrac0.62cdot 0.38100$
– callculus
4 hours ago
The results are here
– callculus
4 hours ago
@callculus he didn't halve $alpha$ so he used $z=1.96$
– s5s
4 hours ago
That´s indeed not right, since it is a two sided interval. But you are right.
– callculus
4 hours ago
add a comment |Â
What is the result of your friend?
– callculus
4 hours ago
After I´ve made my own calculation I got the same result like you. Therefore it is correct :) The bounds are $0.38pm 2.576cdot sqrtfrac0.62cdot 0.38100$
– callculus
4 hours ago
The results are here
– callculus
4 hours ago
@callculus he didn't halve $alpha$ so he used $z=1.96$
– s5s
4 hours ago
That´s indeed not right, since it is a two sided interval. But you are right.
– callculus
4 hours ago
What is the result of your friend?
– callculus
4 hours ago
What is the result of your friend?
– callculus
4 hours ago
After I´ve made my own calculation I got the same result like you. Therefore it is correct :) The bounds are $0.38pm 2.576cdot sqrtfrac0.62cdot 0.38100$
– callculus
4 hours ago
After I´ve made my own calculation I got the same result like you. Therefore it is correct :) The bounds are $0.38pm 2.576cdot sqrtfrac0.62cdot 0.38100$
– callculus
4 hours ago
The results are here
– callculus
4 hours ago
The results are here
– callculus
4 hours ago
@callculus he didn't halve $alpha$ so he used $z=1.96$
– s5s
4 hours ago
@callculus he didn't halve $alpha$ so he used $z=1.96$
– s5s
4 hours ago
That´s indeed not right, since it is a two sided interval. But you are right.
– callculus
4 hours ago
That´s indeed not right, since it is a two sided interval. But you are right.
– callculus
4 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
Your computation is correct for the type of confidence interval you are using.
However, this kind of confidence interval is known not to provide the
promised level of confidence, in your case 99%.
An improved "Agresti-Coull" 99% confidence interval uses $check p = (X+.5c^2)/(n+c^2),$
where $X$ is the number of successes in $n$ trials, and $c = 2.576.$
Then the CI is
$$ check p pm csqrtfraccheck p(1- check p)n+c^2.$$
Then, for your data, $check p = 0.3875$ and the 99% CI is
$( 0.266, .509).$
For very large $n$ (say $n ge 1000$) the difference between the two types of CIs disappears. For more on binomial confidence intervals see this Q & A and its References.
answered 4 hours ago
BruceET
33k61439
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Your computation is correct for the type of confidence interval you are using.
However, this kind of confidence interval is known not to provide the
promised level of confidence, in your case 99%.
An improved "Agresti-Coull" 99% confidence interval uses $check p = (X+.5c^2)/(n+c^2),$
where $X$ is the number of successes in $n$ trials, and $c = 2.576.$
Then the CI is
$$ check p pm csqrtfraccheck p(1- check p)n+c^2.$$
Then, for your data, $check p = 0.3875$ and the 99% CI is
$( 0.266, .509).$
For very large $n$ (say $n ge 1000$) the difference between the two types of CIs disappears. For more on binomial confidence intervals see this Q & A and its References.
answered 4 hours ago
BruceET
33k61439
add a comment |Â
up vote
2
down vote
Your computation is correct for the type of confidence interval you are using.
However, this kind of confidence interval is known not to provide the
promised level of confidence, in your case 99%.
An improved "Agresti-Coull" 99% confidence interval uses $check p = (X+.5c^2)/(n+c^2),$
where $X$ is the number of successes in $n$ trials, and $c = 2.576.$
Then the CI is
$$ check p pm csqrtfraccheck p(1- check p)n+c^2.$$
Then, for your data, $check p = 0.3875$ and the 99% CI is
$( 0.266, .509).$
For very large $n$ (say $n ge 1000$) the difference between the two types of CIs disappears. For more on binomial confidence intervals see this Q & A and its References.
answered 4 hours ago
BruceET
33k61439
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your computation is correct for the type of confidence interval you are using.
However, this kind of confidence interval is known not to provide the
promised level of confidence, in your case 99%.
An improved "Agresti-Coull" 99% confidence interval uses $check p = (X+.5c^2)/(n+c^2),$
where $X$ is the number of successes in $n$ trials, and $c = 2.576.$
Then the CI is
$$ check p pm csqrtfraccheck p(1- check p)n+c^2.$$
Then, for your data, $check p = 0.3875$ and the 99% CI is
$( 0.266, .509).$
For very large $n$ (say $n ge 1000$) the difference between the two types of CIs disappears. For more on binomial confidence intervals see this Q & A and its References.
answered 4 hours ago
BruceET
33k61439
Your computation is correct for the type of confidence interval you are using.
However, this kind of confidence interval is known not to provide the
promised level of confidence, in your case 99%.
An improved "Agresti-Coull" 99% confidence interval uses $check p = (X+.5c^2)/(n+c^2),$
where $X$ is the number of successes in $n$ trials, and $c = 2.576.$
Then the CI is
$$ check p pm csqrtfraccheck p(1- check p)n+c^2.$$
Then, for your data, $check p = 0.3875$ and the 99% CI is
$( 0.266, .509).$
For very large $n$ (say $n ge 1000$) the difference between the two types of CIs disappears. For more on binomial confidence intervals see this Q & A and its References.
answered 4 hours ago
BruceET
33k61439
edited 4 hours ago
answered 4 hours ago
BruceET
33k61439
answered 4 hours ago
BruceET
33k61439
answered 4 hours ago
BruceET
33k61439
33k61439
add a comment |Â
add a comment |Â
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What is the result of your friend?
– callculus
4 hours ago
After I´ve made my own calculation I got the same result like you. Therefore it is correct :) The bounds are $0.38pm 2.576cdot sqrtfrac0.62cdot 0.38100$
– callculus
4 hours ago
The results are here
– callculus
4 hours ago
@callculus he didn't halve $alpha$ so he used $z=1.96$
– s5s
4 hours ago
That´s indeed not right, since it is a two sided interval. But you are right.
– callculus
4 hours ago