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Calculating the Standard Error for a one sample T-test: σ/sqrt(n) or s/sqrt(n)?
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Consider a case when you are looking to test whether a small sample deviates significant from the population (normally distributed) whence it is drawn. Both the population standard deviation (σ) and mean (μ) are known, as are the sample standard deviation (s) and mean (Xbar).
Because the sample is small, you will need to use a t-test. However, my question is, do we calculate the standard error as s/sqrt(n) or σ/sqrt(n)? With a z-test, we usually use σ if we know it, but I'm wondering if the same applies for a t-test.
statistics statistical-inference hypothesis-testing
asked Oct 18 '15 at 22:41
cybervision
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bumped to the homepage by Community♦ 2 days ago
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Consider a case when you are looking to test whether a small sample deviates significant from the population (normally distributed) whence it is drawn. Both the population standard deviation (σ) and mean (μ) are known, as are the sample standard deviation (s) and mean (Xbar).
Because the sample is small, you will need to use a t-test. However, my question is, do we calculate the standard error as s/sqrt(n) or σ/sqrt(n)? With a z-test, we usually use σ if we know it, but I'm wondering if the same applies for a t-test.
statistics statistical-inference hypothesis-testing
asked Oct 18 '15 at 22:41
cybervision
11
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider a case when you are looking to test whether a small sample deviates significant from the population (normally distributed) whence it is drawn. Both the population standard deviation (σ) and mean (μ) are known, as are the sample standard deviation (s) and mean (Xbar).
Because the sample is small, you will need to use a t-test. However, my question is, do we calculate the standard error as s/sqrt(n) or σ/sqrt(n)? With a z-test, we usually use σ if we know it, but I'm wondering if the same applies for a t-test.
statistics statistical-inference hypothesis-testing
asked Oct 18 '15 at 22:41
cybervision
11
Consider a case when you are looking to test whether a small sample deviates significant from the population (normally distributed) whence it is drawn. Both the population standard deviation (σ) and mean (μ) are known, as are the sample standard deviation (s) and mean (Xbar).
Because the sample is small, you will need to use a t-test. However, my question is, do we calculate the standard error as s/sqrt(n) or σ/sqrt(n)? With a z-test, we usually use σ if we know it, but I'm wondering if the same applies for a t-test.
statistics statistical-inference hypothesis-testing
statistics statistical-inference hypothesis-testing
asked Oct 18 '15 at 22:41
cybervision
11
asked Oct 18 '15 at 22:41
cybervision
11
asked Oct 18 '15 at 22:41
cybervision
11
asked Oct 18 '15 at 22:41
cybervision
11
asked Oct 18 '15 at 22:41
cybervision
11
11
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
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Strictly, speaking, if both $mu$ and $sigma$ are known,
you have no reason to do a test.
I think you must mean that you want to test the null hypothesis
$H_0: mu = mu_0$ against the alternative $H_a: mu ne mu_0$
and that $mu_0$ is a specified number.
z-test. If the numerical value of $sigma$ is known, there is no need
to estimate it using the sample standard deviation $S$.
In that case you would have a z-test, with test statistic
$$Z = frac{bar X - mu_0}{sigma/sqrt{n}},$$
where $Z$ has a standard normal distribution if $H_0$ is
true. Then you would reject $H_0$ at the 5% level of
significance if $|Z| > 1.96.$
t test. If the numerical value of $sigma$ is not known, then you
would use a t-test, with test statistic
$$T = frac{bar X - mu_0}{S/sqrt{n}},$$
where $T$ has Student's t distribution with $n - 1$
degrees of freedom if $H_0$ is true.
Then you would reject $H_0$ at the 5% level of significance
if $|T| > t^*,$ where $t^*$ (obtained from tables)
cuts 2.5% of the area from the upper tail of Student's
t distribution with $n - 1$ degrees of freedom.
Distinction between z-test and t-test. For $n > 30,$ you will find that the tabled value $t^*$
is just a bit larger than 1.96. This leads some authors
to say you should use a t test only if $n$ is small.
However, if you use software, you will find that whenever
you do a z test, you will be asked for the numerical
value of $sigma.$ Also, the "rule of 30" really only
works for testing at the 5% level of significance. [At the
1% level, it would be the (seldom mentioned) "rule of 120."
And if you're looking at P-values, no such rule suffices.]
The best rule for z-test vs. t-test is very simple:
If the numerical value of $sigma$ is known, then use
a z-test. If $sigma$ is not known, then it is estimated
by $S$ and you will use a t-test. The distinction
has to do purely with whether $sigma$ is known; it
really has nothing to do with sample size.
answered Oct 19 '15 at 0:13
BruceET
34.7k71440
add a comment |
1 Answer
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1 Answer
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oldest
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up vote
0
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Strictly, speaking, if both $mu$ and $sigma$ are known,
you have no reason to do a test.
I think you must mean that you want to test the null hypothesis
$H_0: mu = mu_0$ against the alternative $H_a: mu ne mu_0$
and that $mu_0$ is a specified number.
z-test. If the numerical value of $sigma$ is known, there is no need
to estimate it using the sample standard deviation $S$.
In that case you would have a z-test, with test statistic
$$Z = frac{bar X - mu_0}{sigma/sqrt{n}},$$
where $Z$ has a standard normal distribution if $H_0$ is
true. Then you would reject $H_0$ at the 5% level of
significance if $|Z| > 1.96.$
t test. If the numerical value of $sigma$ is not known, then you
would use a t-test, with test statistic
$$T = frac{bar X - mu_0}{S/sqrt{n}},$$
where $T$ has Student's t distribution with $n - 1$
degrees of freedom if $H_0$ is true.
Then you would reject $H_0$ at the 5% level of significance
if $|T| > t^*,$ where $t^*$ (obtained from tables)
cuts 2.5% of the area from the upper tail of Student's
t distribution with $n - 1$ degrees of freedom.
Distinction between z-test and t-test. For $n > 30,$ you will find that the tabled value $t^*$
is just a bit larger than 1.96. This leads some authors
to say you should use a t test only if $n$ is small.
However, if you use software, you will find that whenever
you do a z test, you will be asked for the numerical
value of $sigma.$ Also, the "rule of 30" really only
works for testing at the 5% level of significance. [At the
1% level, it would be the (seldom mentioned) "rule of 120."
And if you're looking at P-values, no such rule suffices.]
The best rule for z-test vs. t-test is very simple:
If the numerical value of $sigma$ is known, then use
a z-test. If $sigma$ is not known, then it is estimated
by $S$ and you will use a t-test. The distinction
has to do purely with whether $sigma$ is known; it
really has nothing to do with sample size.
answered Oct 19 '15 at 0:13
BruceET
34.7k71440
add a comment |
up vote
0
down vote
Strictly, speaking, if both $mu$ and $sigma$ are known,
you have no reason to do a test.
I think you must mean that you want to test the null hypothesis
$H_0: mu = mu_0$ against the alternative $H_a: mu ne mu_0$
and that $mu_0$ is a specified number.
z-test. If the numerical value of $sigma$ is known, there is no need
to estimate it using the sample standard deviation $S$.
In that case you would have a z-test, with test statistic
$$Z = frac{bar X - mu_0}{sigma/sqrt{n}},$$
where $Z$ has a standard normal distribution if $H_0$ is
true. Then you would reject $H_0$ at the 5% level of
significance if $|Z| > 1.96.$
t test. If the numerical value of $sigma$ is not known, then you
would use a t-test, with test statistic
$$T = frac{bar X - mu_0}{S/sqrt{n}},$$
where $T$ has Student's t distribution with $n - 1$
degrees of freedom if $H_0$ is true.
Then you would reject $H_0$ at the 5% level of significance
if $|T| > t^*,$ where $t^*$ (obtained from tables)
cuts 2.5% of the area from the upper tail of Student's
t distribution with $n - 1$ degrees of freedom.
Distinction between z-test and t-test. For $n > 30,$ you will find that the tabled value $t^*$
is just a bit larger than 1.96. This leads some authors
to say you should use a t test only if $n$ is small.
However, if you use software, you will find that whenever
you do a z test, you will be asked for the numerical
value of $sigma.$ Also, the "rule of 30" really only
works for testing at the 5% level of significance. [At the
1% level, it would be the (seldom mentioned) "rule of 120."
And if you're looking at P-values, no such rule suffices.]
The best rule for z-test vs. t-test is very simple:
If the numerical value of $sigma$ is known, then use
a z-test. If $sigma$ is not known, then it is estimated
by $S$ and you will use a t-test. The distinction
has to do purely with whether $sigma$ is known; it
really has nothing to do with sample size.
answered Oct 19 '15 at 0:13
BruceET
34.7k71440
add a comment |
up vote
0
down vote
up vote
0
down vote
Strictly, speaking, if both $mu$ and $sigma$ are known,
you have no reason to do a test.
I think you must mean that you want to test the null hypothesis
$H_0: mu = mu_0$ against the alternative $H_a: mu ne mu_0$
and that $mu_0$ is a specified number.
z-test. If the numerical value of $sigma$ is known, there is no need
to estimate it using the sample standard deviation $S$.
In that case you would have a z-test, with test statistic
$$Z = frac{bar X - mu_0}{sigma/sqrt{n}},$$
where $Z$ has a standard normal distribution if $H_0$ is
true. Then you would reject $H_0$ at the 5% level of
significance if $|Z| > 1.96.$
t test. If the numerical value of $sigma$ is not known, then you
would use a t-test, with test statistic
$$T = frac{bar X - mu_0}{S/sqrt{n}},$$
where $T$ has Student's t distribution with $n - 1$
degrees of freedom if $H_0$ is true.
Then you would reject $H_0$ at the 5% level of significance
if $|T| > t^*,$ where $t^*$ (obtained from tables)
cuts 2.5% of the area from the upper tail of Student's
t distribution with $n - 1$ degrees of freedom.
Distinction between z-test and t-test. For $n > 30,$ you will find that the tabled value $t^*$
is just a bit larger than 1.96. This leads some authors
to say you should use a t test only if $n$ is small.
However, if you use software, you will find that whenever
you do a z test, you will be asked for the numerical
value of $sigma.$ Also, the "rule of 30" really only
works for testing at the 5% level of significance. [At the
1% level, it would be the (seldom mentioned) "rule of 120."
And if you're looking at P-values, no such rule suffices.]
The best rule for z-test vs. t-test is very simple:
If the numerical value of $sigma$ is known, then use
a z-test. If $sigma$ is not known, then it is estimated
by $S$ and you will use a t-test. The distinction
has to do purely with whether $sigma$ is known; it
really has nothing to do with sample size.
answered Oct 19 '15 at 0:13
BruceET
34.7k71440
Strictly, speaking, if both $mu$ and $sigma$ are known,
you have no reason to do a test.
I think you must mean that you want to test the null hypothesis
$H_0: mu = mu_0$ against the alternative $H_a: mu ne mu_0$
and that $mu_0$ is a specified number.
z-test. If the numerical value of $sigma$ is known, there is no need
to estimate it using the sample standard deviation $S$.
In that case you would have a z-test, with test statistic
$$Z = frac{bar X - mu_0}{sigma/sqrt{n}},$$
where $Z$ has a standard normal distribution if $H_0$ is
true. Then you would reject $H_0$ at the 5% level of
significance if $|Z| > 1.96.$
t test. If the numerical value of $sigma$ is not known, then you
would use a t-test, with test statistic
$$T = frac{bar X - mu_0}{S/sqrt{n}},$$
where $T$ has Student's t distribution with $n - 1$
degrees of freedom if $H_0$ is true.
Then you would reject $H_0$ at the 5% level of significance
if $|T| > t^*,$ where $t^*$ (obtained from tables)
cuts 2.5% of the area from the upper tail of Student's
t distribution with $n - 1$ degrees of freedom.
Distinction between z-test and t-test. For $n > 30,$ you will find that the tabled value $t^*$
is just a bit larger than 1.96. This leads some authors
to say you should use a t test only if $n$ is small.
However, if you use software, you will find that whenever
you do a z test, you will be asked for the numerical
value of $sigma.$ Also, the "rule of 30" really only
works for testing at the 5% level of significance. [At the
1% level, it would be the (seldom mentioned) "rule of 120."
And if you're looking at P-values, no such rule suffices.]
The best rule for z-test vs. t-test is very simple:
If the numerical value of $sigma$ is known, then use
a z-test. If $sigma$ is not known, then it is estimated
by $S$ and you will use a t-test. The distinction
has to do purely with whether $sigma$ is known; it
really has nothing to do with sample size.
answered Oct 19 '15 at 0:13
BruceET
34.7k71440
edited Oct 19 '15 at 0:27
answered Oct 19 '15 at 0:13
BruceET
34.7k71440
answered Oct 19 '15 at 0:13
BruceET
34.7k71440
answered Oct 19 '15 at 0:13
BruceET
34.7k71440
34.7k71440
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