91Ó°ÊÓ

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

Therefore,$\raisebox{1ex}{$\mathrm{Î\pm }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{0.05}{2}$

=0.025

. Now the value of $z_{\raisebox{1ex}{$\mathrm{Î\pm }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$can be found from the z table and so the values is 1.96.

The margin of error is calculated below:

The calculation of the confidence intervals of the boundaries are calculated below:

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

Therefore,$\raisebox{1ex}{$\mathrm{Î\pm }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{0.05}{2}$

=0.025

. Now the value of can be found from the z table and so the values is 1.96.

The margin of error is calculated below:

The calculation of the confidence intervals of the boundaries are calculated below:

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

Therefore,$\raisebox{1ex}{$\mathrm{Î\pm }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{0.05}{2}$

=0.025

. Now the value of can be found from the z table and so the values is 1.96.

The margin of error is calculated below:

The calculation of the confidence intervals of the boundaries are calculated below:

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

Therefore,$\raisebox{1ex}{$\mathrm{Î\pm }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{0.05}{2}$

=0.025

.

Now the value of$Z_{\raisebox{1ex}{$\mathrm{Î\pm }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$can be found from the z table and so the values is 1.96.

The margin of error is calculated below:

The calculation of the confidence intervals of the boundaries are calculated below:

e.

The answer is no in this case. According to the theory of central limit theorem, the sample mean remains exactly normal. This applies specifically for the large samples where the size remains above 30.