Normal distribution and t-distribution, confidence intervals

Question

In many cases, the $σ$ of a normal distribution is not known. So looking for an interval x_n +- (z_α/2)$\dfrac{σ}{\sqrt{n} }$ without knowing the σ leads to a similar expression with σ=S_n . Since S_n depends on the sample size, we no longer have a normal distribution. It is said that we have a $t$-distribution. What exactly this means, also how could we interpret the degrees of freedom, and eventually why it converges to a normal distribution ? ( as the degrees of freedom increase )

Answer 1

Comment continued with Example.

In R statistical software, the procedure t.test provides a 95% confidence interval for $\mu$ as part of the output.

Suppose we have a sample of size $n = 20$ from a normal population:

x = rnorm(20, 100, 15)
summary(x); length(x);  sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  79.91   86.73  100.52  101.92  109.29  140.36 
[1] 20           # sample size
[1] 17.97414     # sample standard deviation

t.test(x)$conf.int
[1]  93.50513 110.32944 
attr(,"conf.level")
[1] 0.95

You can use the formula in my Comment along with the value of $t^*$ given there and $\bar X = 101.92$ and $S = 17.97$ to verify this confidence interval $(93.505, 110.329).$

Notes: (a) Here is the sample used in the example--rounded to 3 places and sorted from smallest to largest:

sort(round(x,3))    
 [1]  79.910  80.391  82.185  82.339  86.338  86.867  92.912  93.487  94.925  99.789
[11] 101.253 104.700 107.627 107.857 108.890 110.477 111.960 129.211 136.867 140.360

(b) Here is the same confidence interval from Minitab statistical software:

One-Sample T 

 N    Mean  StDev  SE Mean       95% CI
20  101.92  17.97     4.02  (93.51, 110.33)