Question

Why can we not use z values to estimate the mean for a normal distribution with unknown standard deviation?

As sample size n increases, does the variance of the corresponding t distribution increase or decrease? Why?

Answer 1

When σ Is Unknown
If the standard deviation, σ, is unknown, we cannot transformto standard normal. However, we can estimate σ using the sample standard deviation, s, and transformto a variable with a similar distribution, the t distribution. There are actually many t distributions, indexed by degrees of freedom (df). As the degrees of freedom increase, the t distribution approaches the standard normal distribution.
If X is approximately normally distributed, then has a t distribution with (n-1) degrees of freedom (df) The z table gives detailed correspondences of P(Z>z) for values of z from 0 to 3, by .01 (0.00, 0.01, 0.02, 0.03,…2.99. 3.00). The (one-tailed) probabilities are inside the table, and the critical values of z are in the first column and top row.
The t-table is presented differently, with separate rows for each df, with columns representing the two-tailed probability, and with the critical value in the inside of the table
if we look at the last row for z=1.96 and follow up to the top row, we find that
P(|Z| > 1.96) = 0.05

Exercise: What is the critical value associated with a two-tailed probability of 0.01?
Now, suppose that we want to know the probability that Z is more extreme than 2.00. The t-table gives us
P(|Z| > 1.96) = 0.05
And
P(|Z| > 2.326) = 0.02
So, all we can say is that P(|Z| > 2.00) is between 2% and 5%, probably closer to 5%! Using the z-table, we found that it was exactly 4.56%