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Why do you use Z statistics instead of T for inference on categorical variables?
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If I'm comparing the difference between two proportions, why do I use the Z statistic instead of a T statistic? I thought T statistics were used when you don't know the true standard deviation in a population - why doesn't this apply for differences in standard deviations of the samples taken?
statistics statistical-inference
asked Jan 31 at 14:37
zacpolzacpol
1
$endgroup$
add a comment |
$begingroup$
If I'm comparing the difference between two proportions, why do I use the Z statistic instead of a T statistic? I thought T statistics were used when you don't know the true standard deviation in a population - why doesn't this apply for differences in standard deviations of the samples taken?
statistics statistical-inference
asked Jan 31 at 14:37
zacpolzacpol
1
$endgroup$
add a comment |
$begingroup$
If I'm comparing the difference between two proportions, why do I use the Z statistic instead of a T statistic? I thought T statistics were used when you don't know the true standard deviation in a population - why doesn't this apply for differences in standard deviations of the samples taken?
statistics statistical-inference
asked Jan 31 at 14:37
zacpolzacpol
1
$endgroup$
If I'm comparing the difference between two proportions, why do I use the Z statistic instead of a T statistic? I thought T statistics were used when you don't know the true standard deviation in a population - why doesn't this apply for differences in standard deviations of the samples taken?
statistics statistical-inference
statistics statistical-inference
asked Jan 31 at 14:37
zacpolzacpol
1
asked Jan 31 at 14:37
zacpolzacpol
1
asked Jan 31 at 14:37
zacpolzacpol
1
asked Jan 31 at 14:37
zacpolzacpol
1
asked Jan 31 at 14:37
zacpolzacpol
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
Basically it comes down to hypothesis testing. If you have a sample of $n$ Bernoulli($p$) variables $X_i$ with unknown $p$, you take the null hypothesis $p=hat{p}=overline{X}$, and then you compute tail probabilities (depending on what the alternative hypothesis is). But because you know $p$, you can exactly compute these tail probabilities using the binomial distribution. You don't need to know any other parameters to do these computations, just $p$.
The $T$ statistic strictly speaking is only correct for normally distributed data, but it arises because in the normal distribution, the mean and variance are independent parameters. That means that assuming $mu=overline{X}$ alone does not allow you to compute tail probabilities. You need to either estimating the variance or additionally include the variance in the null hypothesis (and we rarely want to do the latter). But when we estimate the variance, the error that we make changes the distribution of the test statistic so that it isn't normal anymore. Using the $t$ distribution instead corrects for this error.
answered Jan 31 at 14:58
IanIan
68.9k25392
$endgroup$
$begingroup$
Thank you - so you're saying that T is only used when you need to correct for variance in a sample? My question now is why you can assume you know p? Maybe I'm missing a part of your explanation that you can point out to me more explicitly.
$endgroup$
– zacpol
Feb 2 at 13:10
$begingroup$
for example: a paper discussed the proportion of pts who experienced tumor regrowth w/in 5 years of complete vs subtotal resection. 42/64 of the subtotal pts experienced regrowth, while only 23/65 of the total pts experienced regrowth. They reported the p value based on the Z statistic calculated by the difference in the two proportions (0.30) compared to the null (0) divided by the SE of the pooled distribution (0.088). I follow the methods, but not the assumption that you can know the true proportion of recurrence in the population. Is this based on some property of bernoulli distributions?
$endgroup$
– zacpol
Feb 2 at 13:11
$begingroup$
@zacpol It's not that you do know $p$, it's that when you're testing a hypothesis about the value of $p$, you assume a value of $p$ according to the null hypothesis and then compute probabilities based on that assumption. In the case of the binomial distribution this fully allows you to compute probabilities. In the case of the normal distribution it does not because the variance is still unknown, and then estimating the variance distorts the distribution.
$endgroup$
– Ian
Feb 2 at 14:40
add a comment |
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1 Answer
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$begingroup$
Basically it comes down to hypothesis testing. If you have a sample of $n$ Bernoulli($p$) variables $X_i$ with unknown $p$, you take the null hypothesis $p=hat{p}=overline{X}$, and then you compute tail probabilities (depending on what the alternative hypothesis is). But because you know $p$, you can exactly compute these tail probabilities using the binomial distribution. You don't need to know any other parameters to do these computations, just $p$.
The $T$ statistic strictly speaking is only correct for normally distributed data, but it arises because in the normal distribution, the mean and variance are independent parameters. That means that assuming $mu=overline{X}$ alone does not allow you to compute tail probabilities. You need to either estimating the variance or additionally include the variance in the null hypothesis (and we rarely want to do the latter). But when we estimate the variance, the error that we make changes the distribution of the test statistic so that it isn't normal anymore. Using the $t$ distribution instead corrects for this error.
answered Jan 31 at 14:58
IanIan
68.9k25392
$endgroup$
$begingroup$
Thank you - so you're saying that T is only used when you need to correct for variance in a sample? My question now is why you can assume you know p? Maybe I'm missing a part of your explanation that you can point out to me more explicitly.
$endgroup$
– zacpol
Feb 2 at 13:10
$begingroup$
for example: a paper discussed the proportion of pts who experienced tumor regrowth w/in 5 years of complete vs subtotal resection. 42/64 of the subtotal pts experienced regrowth, while only 23/65 of the total pts experienced regrowth. They reported the p value based on the Z statistic calculated by the difference in the two proportions (0.30) compared to the null (0) divided by the SE of the pooled distribution (0.088). I follow the methods, but not the assumption that you can know the true proportion of recurrence in the population. Is this based on some property of bernoulli distributions?
$endgroup$
– zacpol
Feb 2 at 13:11
$begingroup$
@zacpol It's not that you do know $p$, it's that when you're testing a hypothesis about the value of $p$, you assume a value of $p$ according to the null hypothesis and then compute probabilities based on that assumption. In the case of the binomial distribution this fully allows you to compute probabilities. In the case of the normal distribution it does not because the variance is still unknown, and then estimating the variance distorts the distribution.
$endgroup$
– Ian
Feb 2 at 14:40
add a comment |
$begingroup$
Basically it comes down to hypothesis testing. If you have a sample of $n$ Bernoulli($p$) variables $X_i$ with unknown $p$, you take the null hypothesis $p=hat{p}=overline{X}$, and then you compute tail probabilities (depending on what the alternative hypothesis is). But because you know $p$, you can exactly compute these tail probabilities using the binomial distribution. You don't need to know any other parameters to do these computations, just $p$.
The $T$ statistic strictly speaking is only correct for normally distributed data, but it arises because in the normal distribution, the mean and variance are independent parameters. That means that assuming $mu=overline{X}$ alone does not allow you to compute tail probabilities. You need to either estimating the variance or additionally include the variance in the null hypothesis (and we rarely want to do the latter). But when we estimate the variance, the error that we make changes the distribution of the test statistic so that it isn't normal anymore. Using the $t$ distribution instead corrects for this error.
answered Jan 31 at 14:58
IanIan
68.9k25392
$endgroup$
$begingroup$
Thank you - so you're saying that T is only used when you need to correct for variance in a sample? My question now is why you can assume you know p? Maybe I'm missing a part of your explanation that you can point out to me more explicitly.
$endgroup$
– zacpol
Feb 2 at 13:10
$begingroup$
for example: a paper discussed the proportion of pts who experienced tumor regrowth w/in 5 years of complete vs subtotal resection. 42/64 of the subtotal pts experienced regrowth, while only 23/65 of the total pts experienced regrowth. They reported the p value based on the Z statistic calculated by the difference in the two proportions (0.30) compared to the null (0) divided by the SE of the pooled distribution (0.088). I follow the methods, but not the assumption that you can know the true proportion of recurrence in the population. Is this based on some property of bernoulli distributions?
$endgroup$
– zacpol
Feb 2 at 13:11
$begingroup$
@zacpol It's not that you do know $p$, it's that when you're testing a hypothesis about the value of $p$, you assume a value of $p$ according to the null hypothesis and then compute probabilities based on that assumption. In the case of the binomial distribution this fully allows you to compute probabilities. In the case of the normal distribution it does not because the variance is still unknown, and then estimating the variance distorts the distribution.
$endgroup$
– Ian
Feb 2 at 14:40
add a comment |
$begingroup$
Basically it comes down to hypothesis testing. If you have a sample of $n$ Bernoulli($p$) variables $X_i$ with unknown $p$, you take the null hypothesis $p=hat{p}=overline{X}$, and then you compute tail probabilities (depending on what the alternative hypothesis is). But because you know $p$, you can exactly compute these tail probabilities using the binomial distribution. You don't need to know any other parameters to do these computations, just $p$.
The $T$ statistic strictly speaking is only correct for normally distributed data, but it arises because in the normal distribution, the mean and variance are independent parameters. That means that assuming $mu=overline{X}$ alone does not allow you to compute tail probabilities. You need to either estimating the variance or additionally include the variance in the null hypothesis (and we rarely want to do the latter). But when we estimate the variance, the error that we make changes the distribution of the test statistic so that it isn't normal anymore. Using the $t$ distribution instead corrects for this error.
answered Jan 31 at 14:58
IanIan
68.9k25392
$endgroup$
Basically it comes down to hypothesis testing. If you have a sample of $n$ Bernoulli($p$) variables $X_i$ with unknown $p$, you take the null hypothesis $p=hat{p}=overline{X}$, and then you compute tail probabilities (depending on what the alternative hypothesis is). But because you know $p$, you can exactly compute these tail probabilities using the binomial distribution. You don't need to know any other parameters to do these computations, just $p$.
The $T$ statistic strictly speaking is only correct for normally distributed data, but it arises because in the normal distribution, the mean and variance are independent parameters. That means that assuming $mu=overline{X}$ alone does not allow you to compute tail probabilities. You need to either estimating the variance or additionally include the variance in the null hypothesis (and we rarely want to do the latter). But when we estimate the variance, the error that we make changes the distribution of the test statistic so that it isn't normal anymore. Using the $t$ distribution instead corrects for this error.
answered Jan 31 at 14:58
IanIan
68.9k25392
answered Jan 31 at 14:58
IanIan
68.9k25392
answered Jan 31 at 14:58
IanIan
68.9k25392
answered Jan 31 at 14:58
IanIan
68.9k25392
68.9k25392
$begingroup$
Thank you - so you're saying that T is only used when you need to correct for variance in a sample? My question now is why you can assume you know p? Maybe I'm missing a part of your explanation that you can point out to me more explicitly.
$endgroup$
– zacpol
Feb 2 at 13:10
$begingroup$
for example: a paper discussed the proportion of pts who experienced tumor regrowth w/in 5 years of complete vs subtotal resection. 42/64 of the subtotal pts experienced regrowth, while only 23/65 of the total pts experienced regrowth. They reported the p value based on the Z statistic calculated by the difference in the two proportions (0.30) compared to the null (0) divided by the SE of the pooled distribution (0.088). I follow the methods, but not the assumption that you can know the true proportion of recurrence in the population. Is this based on some property of bernoulli distributions?
$endgroup$
– zacpol
Feb 2 at 13:11
$begingroup$
@zacpol It's not that you do know $p$, it's that when you're testing a hypothesis about the value of $p$, you assume a value of $p$ according to the null hypothesis and then compute probabilities based on that assumption. In the case of the binomial distribution this fully allows you to compute probabilities. In the case of the normal distribution it does not because the variance is still unknown, and then estimating the variance distorts the distribution.
$endgroup$
– Ian
Feb 2 at 14:40
add a comment |
$begingroup$
Thank you - so you're saying that T is only used when you need to correct for variance in a sample? My question now is why you can assume you know p? Maybe I'm missing a part of your explanation that you can point out to me more explicitly.
$endgroup$
– zacpol
Feb 2 at 13:10
$begingroup$
for example: a paper discussed the proportion of pts who experienced tumor regrowth w/in 5 years of complete vs subtotal resection. 42/64 of the subtotal pts experienced regrowth, while only 23/65 of the total pts experienced regrowth. They reported the p value based on the Z statistic calculated by the difference in the two proportions (0.30) compared to the null (0) divided by the SE of the pooled distribution (0.088). I follow the methods, but not the assumption that you can know the true proportion of recurrence in the population. Is this based on some property of bernoulli distributions?
$endgroup$
– zacpol
Feb 2 at 13:11
$begingroup$
@zacpol It's not that you do know $p$, it's that when you're testing a hypothesis about the value of $p$, you assume a value of $p$ according to the null hypothesis and then compute probabilities based on that assumption. In the case of the binomial distribution this fully allows you to compute probabilities. In the case of the normal distribution it does not because the variance is still unknown, and then estimating the variance distorts the distribution.
$endgroup$
– Ian
Feb 2 at 14:40
$begingroup$
Thank you - so you're saying that T is only used when you need to correct for variance in a sample? My question now is why you can assume you know p? Maybe I'm missing a part of your explanation that you can point out to me more explicitly.
$endgroup$
– zacpol
Feb 2 at 13:10
$begingroup$
Thank you - so you're saying that T is only used when you need to correct for variance in a sample? My question now is why you can assume you know p? Maybe I'm missing a part of your explanation that you can point out to me more explicitly.
$endgroup$
– zacpol
Feb 2 at 13:10
$begingroup$
for example: a paper discussed the proportion of pts who experienced tumor regrowth w/in 5 years of complete vs subtotal resection. 42/64 of the subtotal pts experienced regrowth, while only 23/65 of the total pts experienced regrowth. They reported the p value based on the Z statistic calculated by the difference in the two proportions (0.30) compared to the null (0) divided by the SE of the pooled distribution (0.088). I follow the methods, but not the assumption that you can know the true proportion of recurrence in the population. Is this based on some property of bernoulli distributions?
$endgroup$
– zacpol
Feb 2 at 13:11
$begingroup$
for example: a paper discussed the proportion of pts who experienced tumor regrowth w/in 5 years of complete vs subtotal resection. 42/64 of the subtotal pts experienced regrowth, while only 23/65 of the total pts experienced regrowth. They reported the p value based on the Z statistic calculated by the difference in the two proportions (0.30) compared to the null (0) divided by the SE of the pooled distribution (0.088). I follow the methods, but not the assumption that you can know the true proportion of recurrence in the population. Is this based on some property of bernoulli distributions?
$endgroup$
– zacpol
Feb 2 at 13:11
$begingroup$
@zacpol It's not that you do know $p$, it's that when you're testing a hypothesis about the value of $p$, you assume a value of $p$ according to the null hypothesis and then compute probabilities based on that assumption. In the case of the binomial distribution this fully allows you to compute probabilities. In the case of the normal distribution it does not because the variance is still unknown, and then estimating the variance distorts the distribution.
$endgroup$
– Ian
Feb 2 at 14:40
$begingroup$
@zacpol It's not that you do know $p$, it's that when you're testing a hypothesis about the value of $p$, you assume a value of $p$ according to the null hypothesis and then compute probabilities based on that assumption. In the case of the binomial distribution this fully allows you to compute probabilities. In the case of the normal distribution it does not because the variance is still unknown, and then estimating the variance distorts the distribution.
$endgroup$
– Ian
Feb 2 at 14:40
add a comment |
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