calculate sample size for specified type II error probability, comparing 2 proportions
$begingroup$
This is from Devore Probability and Statistics for Engineering and Sciences 9th edition, chapter 9, section 9.4, exercise #53 on p.398.
The problem involves a hypothesis test comparing two proportions. Part b of the problem states:
"If the true percentages for the two treatments were 15% and 20%, respectively, what sample sizes $(m=n)$ would be necessary to detect such a difference?"
I understand that it is asking us to raise the probability of rejecting the null hypothesis given a specific alternative hypothesis to $1-beta$ (i.e. reduce the probability of type II error by making the sample size large enough). This wording of the question is vastly different from that in definitions, theorems, and examples and can easily cause confusion for students, but that is really a separate issue from my question here.
Equation 9.7 on p.395 provides the required formula:
With $p_1$ and $p_2$ ($q_i=1-p_i$) being the alternative proportions and $z_p$ being the $p^{th}$ percentile of the standard normal, the sample size is
$$n=frac{left(z_{1-alpha/2}sqrt{(p_1+p_2)(q_1+q_2)/2}+z_{1-beta}sqrt{p_1q_1+p_2q_2}right)^2}{(p_1-p_2)^2}.$$
I will assume $p_1>p_2$ in the alternative (it doesn't matter). I will also thus ignore the left tail since it will have extremely small probability under the alternative hypothesis.
We have that:
$$hat{p}_isim Nleft(p_i,sqrt{frac{p_iq_i}{n}}right)$$
$$hat{p}_1-hat p_2sim Nleft(p_1-p_2,sqrt{frac{p_1q_1+p_2q_2}{n}}right)$$
Thus under the null hypothesis, $hat{p}_1-hat p_2sim Nleft(0,sqrt{frac{2pq}{n}}right)$ where we assume that $p_1=p_2=p$. However the null hypothesis only specifies the difference $p_1-p_2=0$ but does not posit any specific value for $p$.
My question is: what justifies setting $p=frac12(p_1+p_2)$ (as used in the sample size calculation formula above)? In other words why are we using the alternative proportions to calculate the percentile under the null hypothesis distribution?
It seems that we might want to maximize this $n$ by using $p=frac12$. I guess there is a built in assumption that the alternative proportions will naturally be somewhat close to our null proportions, or maybe that we can set our null proportions to whatever we want.
There can be a wide range of sample sizes depending on what we input for $p$. For example, with $p_1=0.2$ and $p_2=0.15$, letting $p$ range from 0 to 0.5 gives $n$ ranging from 479 to 1719 (requiring also that $np>10$ and $n(1-p)>10$ otherwise the lower bound on $n$ is 189).
statistics hypothesis-testing
asked Dec 10 '18 at 19:57
jdodsjdods
3,67311234
$endgroup$
add a comment |
$begingroup$
This is from Devore Probability and Statistics for Engineering and Sciences 9th edition, chapter 9, section 9.4, exercise #53 on p.398.
The problem involves a hypothesis test comparing two proportions. Part b of the problem states:
"If the true percentages for the two treatments were 15% and 20%, respectively, what sample sizes $(m=n)$ would be necessary to detect such a difference?"
I understand that it is asking us to raise the probability of rejecting the null hypothesis given a specific alternative hypothesis to $1-beta$ (i.e. reduce the probability of type II error by making the sample size large enough). This wording of the question is vastly different from that in definitions, theorems, and examples and can easily cause confusion for students, but that is really a separate issue from my question here.
Equation 9.7 on p.395 provides the required formula:
With $p_1$ and $p_2$ ($q_i=1-p_i$) being the alternative proportions and $z_p$ being the $p^{th}$ percentile of the standard normal, the sample size is
$$n=frac{left(z_{1-alpha/2}sqrt{(p_1+p_2)(q_1+q_2)/2}+z_{1-beta}sqrt{p_1q_1+p_2q_2}right)^2}{(p_1-p_2)^2}.$$
I will assume $p_1>p_2$ in the alternative (it doesn't matter). I will also thus ignore the left tail since it will have extremely small probability under the alternative hypothesis.
We have that:
$$hat{p}_isim Nleft(p_i,sqrt{frac{p_iq_i}{n}}right)$$
$$hat{p}_1-hat p_2sim Nleft(p_1-p_2,sqrt{frac{p_1q_1+p_2q_2}{n}}right)$$
Thus under the null hypothesis, $hat{p}_1-hat p_2sim Nleft(0,sqrt{frac{2pq}{n}}right)$ where we assume that $p_1=p_2=p$. However the null hypothesis only specifies the difference $p_1-p_2=0$ but does not posit any specific value for $p$.
My question is: what justifies setting $p=frac12(p_1+p_2)$ (as used in the sample size calculation formula above)? In other words why are we using the alternative proportions to calculate the percentile under the null hypothesis distribution?
It seems that we might want to maximize this $n$ by using $p=frac12$. I guess there is a built in assumption that the alternative proportions will naturally be somewhat close to our null proportions, or maybe that we can set our null proportions to whatever we want.
There can be a wide range of sample sizes depending on what we input for $p$. For example, with $p_1=0.2$ and $p_2=0.15$, letting $p$ range from 0 to 0.5 gives $n$ ranging from 479 to 1719 (requiring also that $np>10$ and $n(1-p)>10$ otherwise the lower bound on $n$ is 189).
statistics hypothesis-testing
asked Dec 10 '18 at 19:57
jdodsjdods
3,67311234
$endgroup$
add a comment |
$begingroup$
This is from Devore Probability and Statistics for Engineering and Sciences 9th edition, chapter 9, section 9.4, exercise #53 on p.398.
The problem involves a hypothesis test comparing two proportions. Part b of the problem states:
"If the true percentages for the two treatments were 15% and 20%, respectively, what sample sizes $(m=n)$ would be necessary to detect such a difference?"
I understand that it is asking us to raise the probability of rejecting the null hypothesis given a specific alternative hypothesis to $1-beta$ (i.e. reduce the probability of type II error by making the sample size large enough). This wording of the question is vastly different from that in definitions, theorems, and examples and can easily cause confusion for students, but that is really a separate issue from my question here.
Equation 9.7 on p.395 provides the required formula:
With $p_1$ and $p_2$ ($q_i=1-p_i$) being the alternative proportions and $z_p$ being the $p^{th}$ percentile of the standard normal, the sample size is
$$n=frac{left(z_{1-alpha/2}sqrt{(p_1+p_2)(q_1+q_2)/2}+z_{1-beta}sqrt{p_1q_1+p_2q_2}right)^2}{(p_1-p_2)^2}.$$
I will assume $p_1>p_2$ in the alternative (it doesn't matter). I will also thus ignore the left tail since it will have extremely small probability under the alternative hypothesis.
We have that:
$$hat{p}_isim Nleft(p_i,sqrt{frac{p_iq_i}{n}}right)$$
$$hat{p}_1-hat p_2sim Nleft(p_1-p_2,sqrt{frac{p_1q_1+p_2q_2}{n}}right)$$
Thus under the null hypothesis, $hat{p}_1-hat p_2sim Nleft(0,sqrt{frac{2pq}{n}}right)$ where we assume that $p_1=p_2=p$. However the null hypothesis only specifies the difference $p_1-p_2=0$ but does not posit any specific value for $p$.
My question is: what justifies setting $p=frac12(p_1+p_2)$ (as used in the sample size calculation formula above)? In other words why are we using the alternative proportions to calculate the percentile under the null hypothesis distribution?
It seems that we might want to maximize this $n$ by using $p=frac12$. I guess there is a built in assumption that the alternative proportions will naturally be somewhat close to our null proportions, or maybe that we can set our null proportions to whatever we want.
There can be a wide range of sample sizes depending on what we input for $p$. For example, with $p_1=0.2$ and $p_2=0.15$, letting $p$ range from 0 to 0.5 gives $n$ ranging from 479 to 1719 (requiring also that $np>10$ and $n(1-p)>10$ otherwise the lower bound on $n$ is 189).
statistics hypothesis-testing
asked Dec 10 '18 at 19:57
jdodsjdods
3,67311234
$endgroup$
This is from Devore Probability and Statistics for Engineering and Sciences 9th edition, chapter 9, section 9.4, exercise #53 on p.398.
The problem involves a hypothesis test comparing two proportions. Part b of the problem states:
"If the true percentages for the two treatments were 15% and 20%, respectively, what sample sizes $(m=n)$ would be necessary to detect such a difference?"
I understand that it is asking us to raise the probability of rejecting the null hypothesis given a specific alternative hypothesis to $1-beta$ (i.e. reduce the probability of type II error by making the sample size large enough). This wording of the question is vastly different from that in definitions, theorems, and examples and can easily cause confusion for students, but that is really a separate issue from my question here.
Equation 9.7 on p.395 provides the required formula:
With $p_1$ and $p_2$ ($q_i=1-p_i$) being the alternative proportions and $z_p$ being the $p^{th}$ percentile of the standard normal, the sample size is
$$n=frac{left(z_{1-alpha/2}sqrt{(p_1+p_2)(q_1+q_2)/2}+z_{1-beta}sqrt{p_1q_1+p_2q_2}right)^2}{(p_1-p_2)^2}.$$
I will assume $p_1>p_2$ in the alternative (it doesn't matter). I will also thus ignore the left tail since it will have extremely small probability under the alternative hypothesis.
We have that:
$$hat{p}_isim Nleft(p_i,sqrt{frac{p_iq_i}{n}}right)$$
$$hat{p}_1-hat p_2sim Nleft(p_1-p_2,sqrt{frac{p_1q_1+p_2q_2}{n}}right)$$
Thus under the null hypothesis, $hat{p}_1-hat p_2sim Nleft(0,sqrt{frac{2pq}{n}}right)$ where we assume that $p_1=p_2=p$. However the null hypothesis only specifies the difference $p_1-p_2=0$ but does not posit any specific value for $p$.
My question is: what justifies setting $p=frac12(p_1+p_2)$ (as used in the sample size calculation formula above)? In other words why are we using the alternative proportions to calculate the percentile under the null hypothesis distribution?
It seems that we might want to maximize this $n$ by using $p=frac12$. I guess there is a built in assumption that the alternative proportions will naturally be somewhat close to our null proportions, or maybe that we can set our null proportions to whatever we want.
There can be a wide range of sample sizes depending on what we input for $p$. For example, with $p_1=0.2$ and $p_2=0.15$, letting $p$ range from 0 to 0.5 gives $n$ ranging from 479 to 1719 (requiring also that $np>10$ and $n(1-p)>10$ otherwise the lower bound on $n$ is 189).
statistics hypothesis-testing
statistics hypothesis-testing
asked Dec 10 '18 at 19:57
jdodsjdods
3,67311234
asked Dec 10 '18 at 19:57
jdodsjdods
3,67311234
asked Dec 10 '18 at 19:57
jdodsjdods
3,67311234
asked Dec 10 '18 at 19:57
jdodsjdods
3,67311234
asked Dec 10 '18 at 19:57
jdodsjdods
3,67311234
3,67311234
add a comment |
add a comment |
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