Why we use t-distribution when sample size is small? (n < 30)What is the reason that Student-t Distribution is used when the number of samples is smallHow is it that the required sample size for a specified error and confidence is not dependent on population size?How do I use student's-t distribution without the sample size?Sampling distribution with large sample sizeWhen the population variance is unknown, we should use t-distribution.Should I use the t distribution instead of normal for large samples?What is the standard deviation of the sample distribution for the sample standard deviation?Why prefer the t-score when the sample size is low?Is a t distribution for a certain degree of freedom equivalent to the sample mean distribution for the corresponding sample size?Why do we use the t-distribution for this problem? Isn't the sample size too large?Sample Size Proof
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Why we use t-distribution when sample size is small? (n
What is the reason that Student-t Distribution is used when the number of samples is smallHow is it that the required sample size for a specified error and confidence is not dependent on population size?How do I use student's-t distribution without the sample size?Sampling distribution with large sample sizeWhen the population variance is unknown, we should use t-distribution.Should I use the t distribution instead of normal for large samples?What is the standard deviation of the sample distribution for the sample standard deviation?Why prefer the t-score when the sample size is low?Is a t distribution for a certain degree of freedom equivalent to the sample mean distribution for the corresponding sample size?Why do we use the t-distribution for this problem? Isn't the sample size too large?Sample Size Proof
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When population is normal, distribution of sample mean is also normal regardless of sample size n. Then, why we use t-distribution instead of normal distribution when sample size is small? (n < 30)
statistics normal-distribution
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add a comment |
$begingroup$
When population is normal, distribution of sample mean is also normal regardless of sample size n. Then, why we use t-distribution instead of normal distribution when sample size is small? (n < 30)
statistics normal-distribution
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1
$begingroup$
But you have an unknown standard deviation I'm guessing. When this is the case and you replace the standard deviation with the sample standard deviation, you end up with a $t$-distribution. (Remember if $X_i$ are iid $Nleft(mu,sigma^2right)$, $overlineX_n =frac1nsumlimits _i =1^nX_i$ and $S = sqrtfrac1n-1sumlimits_i=1^nleft(X_i- overlineX_nright)^2$, then $fracoverlineX_n - muS/sqrtnsim t_n-1$.)
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– Minus One-Twelfth
Mar 29 at 11:10
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When population variance $sigma^2$ is unknown and estimated by sample variance $S^2,$ you should always use a t test for most accurate results. For $n > 30$ results from an incorrect z test may be a serviceable approximation. The approx 'rule of 30' works only for 95% CIs or 5% significance levels because 1.96 cuts 2.5% from upper tail of std normal and for $n > 30$ numbers around 2.0 cut 2.5% from upper tail of Student's t dist'n with $n - 1 ge 30$ degrees of freedom.
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– BruceET
Mar 29 at 16:14
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See math.stackexchange.com/questions/2992333
$endgroup$
– cgiovanardi
Mar 29 at 22:09
add a comment |
$begingroup$
When population is normal, distribution of sample mean is also normal regardless of sample size n. Then, why we use t-distribution instead of normal distribution when sample size is small? (n < 30)
statistics normal-distribution
$endgroup$
When population is normal, distribution of sample mean is also normal regardless of sample size n. Then, why we use t-distribution instead of normal distribution when sample size is small? (n < 30)
statistics normal-distribution
statistics normal-distribution
asked Mar 29 at 11:06
firia2000firia2000
1011
1011
1
$begingroup$
But you have an unknown standard deviation I'm guessing. When this is the case and you replace the standard deviation with the sample standard deviation, you end up with a $t$-distribution. (Remember if $X_i$ are iid $Nleft(mu,sigma^2right)$, $overlineX_n =frac1nsumlimits _i =1^nX_i$ and $S = sqrtfrac1n-1sumlimits_i=1^nleft(X_i- overlineX_nright)^2$, then $fracoverlineX_n - muS/sqrtnsim t_n-1$.)
$endgroup$
– Minus One-Twelfth
Mar 29 at 11:10
$begingroup$
When population variance $sigma^2$ is unknown and estimated by sample variance $S^2,$ you should always use a t test for most accurate results. For $n > 30$ results from an incorrect z test may be a serviceable approximation. The approx 'rule of 30' works only for 95% CIs or 5% significance levels because 1.96 cuts 2.5% from upper tail of std normal and for $n > 30$ numbers around 2.0 cut 2.5% from upper tail of Student's t dist'n with $n - 1 ge 30$ degrees of freedom.
$endgroup$
– BruceET
Mar 29 at 16:14
$begingroup$
See math.stackexchange.com/questions/2992333
$endgroup$
– cgiovanardi
Mar 29 at 22:09
add a comment |
1
$begingroup$
But you have an unknown standard deviation I'm guessing. When this is the case and you replace the standard deviation with the sample standard deviation, you end up with a $t$-distribution. (Remember if $X_i$ are iid $Nleft(mu,sigma^2right)$, $overlineX_n =frac1nsumlimits _i =1^nX_i$ and $S = sqrtfrac1n-1sumlimits_i=1^nleft(X_i- overlineX_nright)^2$, then $fracoverlineX_n - muS/sqrtnsim t_n-1$.)
$endgroup$
– Minus One-Twelfth
Mar 29 at 11:10
$begingroup$
When population variance $sigma^2$ is unknown and estimated by sample variance $S^2,$ you should always use a t test for most accurate results. For $n > 30$ results from an incorrect z test may be a serviceable approximation. The approx 'rule of 30' works only for 95% CIs or 5% significance levels because 1.96 cuts 2.5% from upper tail of std normal and for $n > 30$ numbers around 2.0 cut 2.5% from upper tail of Student's t dist'n with $n - 1 ge 30$ degrees of freedom.
$endgroup$
– BruceET
Mar 29 at 16:14
$begingroup$
See math.stackexchange.com/questions/2992333
$endgroup$
– cgiovanardi
Mar 29 at 22:09
1
1
$begingroup$
But you have an unknown standard deviation I'm guessing. When this is the case and you replace the standard deviation with the sample standard deviation, you end up with a $t$-distribution. (Remember if $X_i$ are iid $Nleft(mu,sigma^2right)$, $overlineX_n =frac1nsumlimits _i =1^nX_i$ and $S = sqrtfrac1n-1sumlimits_i=1^nleft(X_i- overlineX_nright)^2$, then $fracoverlineX_n - muS/sqrtnsim t_n-1$.)
$endgroup$
– Minus One-Twelfth
Mar 29 at 11:10
$begingroup$
But you have an unknown standard deviation I'm guessing. When this is the case and you replace the standard deviation with the sample standard deviation, you end up with a $t$-distribution. (Remember if $X_i$ are iid $Nleft(mu,sigma^2right)$, $overlineX_n =frac1nsumlimits _i =1^nX_i$ and $S = sqrtfrac1n-1sumlimits_i=1^nleft(X_i- overlineX_nright)^2$, then $fracoverlineX_n - muS/sqrtnsim t_n-1$.)
$endgroup$
– Minus One-Twelfth
Mar 29 at 11:10
$begingroup$
When population variance $sigma^2$ is unknown and estimated by sample variance $S^2,$ you should always use a t test for most accurate results. For $n > 30$ results from an incorrect z test may be a serviceable approximation. The approx 'rule of 30' works only for 95% CIs or 5% significance levels because 1.96 cuts 2.5% from upper tail of std normal and for $n > 30$ numbers around 2.0 cut 2.5% from upper tail of Student's t dist'n with $n - 1 ge 30$ degrees of freedom.
$endgroup$
– BruceET
Mar 29 at 16:14
$begingroup$
When population variance $sigma^2$ is unknown and estimated by sample variance $S^2,$ you should always use a t test for most accurate results. For $n > 30$ results from an incorrect z test may be a serviceable approximation. The approx 'rule of 30' works only for 95% CIs or 5% significance levels because 1.96 cuts 2.5% from upper tail of std normal and for $n > 30$ numbers around 2.0 cut 2.5% from upper tail of Student's t dist'n with $n - 1 ge 30$ degrees of freedom.
$endgroup$
– BruceET
Mar 29 at 16:14
$begingroup$
See math.stackexchange.com/questions/2992333
$endgroup$
– cgiovanardi
Mar 29 at 22:09
$begingroup$
See math.stackexchange.com/questions/2992333
$endgroup$
– cgiovanardi
Mar 29 at 22:09
add a comment |
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$begingroup$
But you have an unknown standard deviation I'm guessing. When this is the case and you replace the standard deviation with the sample standard deviation, you end up with a $t$-distribution. (Remember if $X_i$ are iid $Nleft(mu,sigma^2right)$, $overlineX_n =frac1nsumlimits _i =1^nX_i$ and $S = sqrtfrac1n-1sumlimits_i=1^nleft(X_i- overlineX_nright)^2$, then $fracoverlineX_n - muS/sqrtnsim t_n-1$.)
$endgroup$
– Minus One-Twelfth
Mar 29 at 11:10
$begingroup$
When population variance $sigma^2$ is unknown and estimated by sample variance $S^2,$ you should always use a t test for most accurate results. For $n > 30$ results from an incorrect z test may be a serviceable approximation. The approx 'rule of 30' works only for 95% CIs or 5% significance levels because 1.96 cuts 2.5% from upper tail of std normal and for $n > 30$ numbers around 2.0 cut 2.5% from upper tail of Student's t dist'n with $n - 1 ge 30$ degrees of freedom.
$endgroup$
– BruceET
Mar 29 at 16:14
$begingroup$
See math.stackexchange.com/questions/2992333
$endgroup$
– cgiovanardi
Mar 29 at 22:09