Suppose you have selected a random sample of n = 5 measurements from a normal distribution. Compare the standard normal z-values with the corresponding t-values if you were forming the following confidence intervals.

a. 80% confidence interval

b. 90% confidence interval

c. 95% confidence interval

d. 98% confidence interval

e. 99% confidence interval

f. Use the table values you obtained in parts a–e to sketch the z- and t-distributions. What are the similarities and differences?

As the given confidence interval is 80%, the significance level will automatically be 20% which means 0.20.

Therefore, it can be said that$\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{0.20}{2}$

=0.10

. Now the value of$Z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ can be found from the z table, and so the z value is 1.28.

Now the degrees of freedom 4 and so$t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$from the table can be found out to be t=1.533.

As the given confidence interval is 90%, the significance level will automatically be 10% which means 0.10.

Therefore, it can be said that $\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{0.10}{2}$

=050

Now the degrees of freedom 4 and so$t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ from the table can be found to be t=2.132.

As the given confidence interval is 95%, the significance level will automatically be 5% which means 0.05.

Therefore, it can be said that$\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{0.05}{2}$

=0.25

Now the degrees of freedom 4 and so from the table can be found to be t=2.776.

As the given confidence interval is 98%, the significance level will automatically be 2% which means 0.02.

Therefore, it can be said that$\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{0.02}{2}$

=0.01

Now the degrees of freedom 4 and so from the table can be found to be t=4.604.

As the given confidence interval is 99%, the significance level will automatically be 1% which means 0.01.

Therefore, it can be said that $\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{0.01}{2}$

=0.005

Now the degrees of freedom 4 and so from the table can be found to be t=3.707.

f.

The table results in the following sketch.

The similarities deduced from the diagram are: that both z-distribution(pink) and t-distribution(blue) are symmetric, and the centre is at 0. On the other hand, the dissimilarity Is that the z-distribution's top is greater than t-distribution’s top.