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Approximation using the t-distribution or the z-distribution
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When we approximate a parameter of a population(let’s say μ) and the size of our sample is small( for example:n=9)
Given the standard deviation of the popluation( not the sample). Do we use the Z-distribution(normal) or the T-distribution to approximate the parameter?
I am stuck here since I know that we use the t-distribution for a small sample given the standard deviation of the sample, and we use the Z-test with a large sample since we can approximate its distribution to be normal by the central limit theorem(with either, the standard deviation of the population or of the sample given).
According to the solution of a certain problem, the z-distribution is used in the latter case.Can you please explain why?
probability probability-distributions
asked Jul 28 at 14:16
Roy Rizk
887
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up vote
1
down vote
favorite
When we approximate a parameter of a population(let’s say μ) and the size of our sample is small( for example:n=9)
Given the standard deviation of the popluation( not the sample). Do we use the Z-distribution(normal) or the T-distribution to approximate the parameter?
I am stuck here since I know that we use the t-distribution for a small sample given the standard deviation of the sample, and we use the Z-test with a large sample since we can approximate its distribution to be normal by the central limit theorem(with either, the standard deviation of the population or of the sample given).
According to the solution of a certain problem, the z-distribution is used in the latter case.Can you please explain why?
probability probability-distributions
asked Jul 28 at 14:16
Roy Rizk
887
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
When we approximate a parameter of a population(let’s say μ) and the size of our sample is small( for example:n=9)
Given the standard deviation of the popluation( not the sample). Do we use the Z-distribution(normal) or the T-distribution to approximate the parameter?
I am stuck here since I know that we use the t-distribution for a small sample given the standard deviation of the sample, and we use the Z-test with a large sample since we can approximate its distribution to be normal by the central limit theorem(with either, the standard deviation of the population or of the sample given).
According to the solution of a certain problem, the z-distribution is used in the latter case.Can you please explain why?
probability probability-distributions
asked Jul 28 at 14:16
Roy Rizk
887
When we approximate a parameter of a population(let’s say μ) and the size of our sample is small( for example:n=9)
Given the standard deviation of the popluation( not the sample). Do we use the Z-distribution(normal) or the T-distribution to approximate the parameter?
I am stuck here since I know that we use the t-distribution for a small sample given the standard deviation of the sample, and we use the Z-test with a large sample since we can approximate its distribution to be normal by the central limit theorem(with either, the standard deviation of the population or of the sample given).
According to the solution of a certain problem, the z-distribution is used in the latter case.Can you please explain why?
probability probability-distributions
asked Jul 28 at 14:16
Roy Rizk
887
asked Jul 28 at 14:16
Roy Rizk
887
asked Jul 28 at 14:16
Roy Rizk
887
asked Jul 28 at 14:16
Roy Rizk
887
887
add a comment |Â
add a comment |Â
1 Answer
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Data must be normal (or nearly normal) in order to use either z or t test (or confidence interval).
Then the choice between z and t is easy:
- If the population standard deviation $sigma$ is known, then use z methods (test or CI).
- If the population standard deviation $sigma$ is unknown and estimated by
the sample standard deviation $S,$ then use t methods (test or CI).
The sample size $n$ has nothing to do with this choice between z and t.
Note: If (i) the significance level of a test is $alpha = 0.05$ or a confidence interval has confidence level 95%,
(ii) the sample standard deviation is unknown, and (iii) the sample size is $n ge 30,$
then some textbooks point out that you can use
a cut-off value from z tables instead of t tables as an approximation.
For example, if $n = 35,$ the exact 95% CI is of the form
$bar X pm 2.032 S/sqrt35,$ where 2.032 cuts area 2.5% from the upper
tail of $mathsfT(df = 34).$ But for some authors it is good enough
to use the CI $bar X pm 1.96 S/sqrt35,$ where 1.96 cuts area 2.5% from
the upper tail of the standard normal distribution.
Properly used, the approximation usually causes no harm. (In the example, 2.032 and 1.96 both round to 2.0.)
The only real problem with this is that some people have trouble remembering
all three conditions (i), (ii), and (iii). Then they get to thinking that t methods
are only good for small samples. Or they use z methods for anything where $n ge 30.$
answered Jul 29 at 3:54
BruceET
33.1k61440
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Data must be normal (or nearly normal) in order to use either z or t test (or confidence interval).
Then the choice between z and t is easy:
- If the population standard deviation $sigma$ is known, then use z methods (test or CI).
- If the population standard deviation $sigma$ is unknown and estimated by
the sample standard deviation $S,$ then use t methods (test or CI).
The sample size $n$ has nothing to do with this choice between z and t.
Note: If (i) the significance level of a test is $alpha = 0.05$ or a confidence interval has confidence level 95%,
(ii) the sample standard deviation is unknown, and (iii) the sample size is $n ge 30,$
then some textbooks point out that you can use
a cut-off value from z tables instead of t tables as an approximation.
For example, if $n = 35,$ the exact 95% CI is of the form
$bar X pm 2.032 S/sqrt35,$ where 2.032 cuts area 2.5% from the upper
tail of $mathsfT(df = 34).$ But for some authors it is good enough
to use the CI $bar X pm 1.96 S/sqrt35,$ where 1.96 cuts area 2.5% from
the upper tail of the standard normal distribution.
Properly used, the approximation usually causes no harm. (In the example, 2.032 and 1.96 both round to 2.0.)
The only real problem with this is that some people have trouble remembering
all three conditions (i), (ii), and (iii). Then they get to thinking that t methods
are only good for small samples. Or they use z methods for anything where $n ge 30.$
answered Jul 29 at 3:54
BruceET
33.1k61440
add a comment |Â
up vote
0
down vote
Data must be normal (or nearly normal) in order to use either z or t test (or confidence interval).
Then the choice between z and t is easy:
- If the population standard deviation $sigma$ is known, then use z methods (test or CI).
- If the population standard deviation $sigma$ is unknown and estimated by
the sample standard deviation $S,$ then use t methods (test or CI).
The sample size $n$ has nothing to do with this choice between z and t.
Note: If (i) the significance level of a test is $alpha = 0.05$ or a confidence interval has confidence level 95%,
(ii) the sample standard deviation is unknown, and (iii) the sample size is $n ge 30,$
then some textbooks point out that you can use
a cut-off value from z tables instead of t tables as an approximation.
For example, if $n = 35,$ the exact 95% CI is of the form
$bar X pm 2.032 S/sqrt35,$ where 2.032 cuts area 2.5% from the upper
tail of $mathsfT(df = 34).$ But for some authors it is good enough
to use the CI $bar X pm 1.96 S/sqrt35,$ where 1.96 cuts area 2.5% from
the upper tail of the standard normal distribution.
Properly used, the approximation usually causes no harm. (In the example, 2.032 and 1.96 both round to 2.0.)
The only real problem with this is that some people have trouble remembering
all three conditions (i), (ii), and (iii). Then they get to thinking that t methods
are only good for small samples. Or they use z methods for anything where $n ge 30.$
answered Jul 29 at 3:54
BruceET
33.1k61440
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Data must be normal (or nearly normal) in order to use either z or t test (or confidence interval).
Then the choice between z and t is easy:
- If the population standard deviation $sigma$ is known, then use z methods (test or CI).
- If the population standard deviation $sigma$ is unknown and estimated by
the sample standard deviation $S,$ then use t methods (test or CI).
The sample size $n$ has nothing to do with this choice between z and t.
Note: If (i) the significance level of a test is $alpha = 0.05$ or a confidence interval has confidence level 95%,
(ii) the sample standard deviation is unknown, and (iii) the sample size is $n ge 30,$
then some textbooks point out that you can use
a cut-off value from z tables instead of t tables as an approximation.
For example, if $n = 35,$ the exact 95% CI is of the form
$bar X pm 2.032 S/sqrt35,$ where 2.032 cuts area 2.5% from the upper
tail of $mathsfT(df = 34).$ But for some authors it is good enough
to use the CI $bar X pm 1.96 S/sqrt35,$ where 1.96 cuts area 2.5% from
the upper tail of the standard normal distribution.
Properly used, the approximation usually causes no harm. (In the example, 2.032 and 1.96 both round to 2.0.)
The only real problem with this is that some people have trouble remembering
all three conditions (i), (ii), and (iii). Then they get to thinking that t methods
are only good for small samples. Or they use z methods for anything where $n ge 30.$
answered Jul 29 at 3:54
BruceET
33.1k61440
Data must be normal (or nearly normal) in order to use either z or t test (or confidence interval).
Then the choice between z and t is easy:
- If the population standard deviation $sigma$ is known, then use z methods (test or CI).
- If the population standard deviation $sigma$ is unknown and estimated by
the sample standard deviation $S,$ then use t methods (test or CI).
The sample size $n$ has nothing to do with this choice between z and t.
Note: If (i) the significance level of a test is $alpha = 0.05$ or a confidence interval has confidence level 95%,
(ii) the sample standard deviation is unknown, and (iii) the sample size is $n ge 30,$
then some textbooks point out that you can use
a cut-off value from z tables instead of t tables as an approximation.
For example, if $n = 35,$ the exact 95% CI is of the form
$bar X pm 2.032 S/sqrt35,$ where 2.032 cuts area 2.5% from the upper
tail of $mathsfT(df = 34).$ But for some authors it is good enough
to use the CI $bar X pm 1.96 S/sqrt35,$ where 1.96 cuts area 2.5% from
the upper tail of the standard normal distribution.
Properly used, the approximation usually causes no harm. (In the example, 2.032 and 1.96 both round to 2.0.)
The only real problem with this is that some people have trouble remembering
all three conditions (i), (ii), and (iii). Then they get to thinking that t methods
are only good for small samples. Or they use z methods for anything where $n ge 30.$
answered Jul 29 at 3:54
BruceET
33.1k61440
edited Jul 29 at 4:20
answered Jul 29 at 3:54
BruceET
33.1k61440
answered Jul 29 at 3:54
BruceET
33.1k61440
answered Jul 29 at 3:54
BruceET
33.1k61440
33.1k61440
add a comment |Â
add a comment |Â
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