What is the correct method for determining confidence intervals for the mean?

[tzx9633]

Homework Statement

Both images are 2 consecutive pages of my notes . In this theory , i was told to use z-distribution when the sample size is large ( more than 30) and the standard deviation of the population , σ is known) . However , in the 2nd image , i was told to replace σ with s ( sample standard deviation) . Which is correct ?

Homework Equations

The Attempt at a Solution

I think the first theory of using z-distribution when the sample size is large ( more than 30) and the standard deviation of the population , σ is known) is correct , and the[/B] replace σ with s ( sample standard deviation) . is wrong . Am i right ?

 

[FactChecker]

If you know the value of σ from some theory or some larger external set of data, then use it. Otherwise, you have to use your sample data to estimate σ. That tends to give a slightly smaller estimate than it should for the general population, so a better estimate of σ is obtained by dividing by n-1 instead of n.

 

[Ray Vickson]

Homework Statement

Both images are 2 consecutive pages of my notes . In this theory , i was told to use z-distribution when the sample size is large ( more than 30) and the standard deviation of the population , σ is known) . However , in the 2nd image , i was told to replace σ with s ( sample standard deviation) . Which is correct ?

Homework Equations

The Attempt at a Solution

I think the first theory of using z-distribution when the sample size is large ( more than 30) and the standard deviation of the population , σ is known) is correct , and the[/B] replace σ with s ( sample standard deviation) . is wrong . Am i right ?

I would say the second attachment is wrong: the correct distribution to use is the t-distribution with $n-1$ degrees of freedom. However, for large $n$ there is not much difference between the t-distribution and the normal distribution, so it is a reasonable approximation to replace the t by the normal. For example, when $n=31$ (30 degrees of freedom) we have
$$
\begin{array}{cc}
P(N \leq -1) = 0.15866, & P(T \leq -1) = 0.16265\\
P(N \leq 1) = 0.84134, & P(T \leq 1) = 0.83735
\end{array}
$$

 

[tzx9633]

If you know the value of σ from some theory or some larger external set of data, then use it. Otherwise, you have to use your sample data to estimate σ. That tends to give a slightly smaller estimate than it should for the general population, so a better estimate of σ is obtained by dividing by n-1 instead of n.

Do you mean that for large sample size n > 30 , it's okay to use the assumption that the standard deviation of the population equal to the standard deviation of the sample ?

 

[tzx9633]

If you know the value of σ from some theory or some larger external set of data, then use it. Otherwise, you have to use your sample data to estimate σ. That tends to give a slightly smaller estimate than it should for the general population, so a better estimate of σ is obtained by dividing by n-1 instead of n.

I would say the second attachment is wrong: the correct distribution to use is the t-distribution with $n-1$ degrees of freedom. However, for large $n$ there is not much difference between the t-distribution and the normal distribution, so it is a reasonable approximation to replace the t by the normal. For example, when $n=31$ (30 degrees of freedom) we have
$$
\begin{array}{cc}
P(N \leq -1) = 0.15866, & P(T \leq -1) = 0.16265\\
P(N \leq 1) = 0.84134, & P(T \leq 1) = 0.83735
\end{array}
$$

Do you mean that for large sample size n > 30 , it's okay to use the assumption that the standard deviation of the population equal to the standard deviation of the sample ?

 

[FactChecker]

For any size sample, if you don't have some other way of knowing the value of σ, then you have no alternative to using the sample standard deviation.