Chapter 8.1, Problem 1E

To determine

To find:

The best point estimate for the mean amount of money spent per week on lunch for all teachers, if 42 randomly selected teachers spend a mean of $18 per week on lunch

Answer to Problem 1E

Solution:

Best point estimate for the mean amount of money spent per week on lunch for all 42 teachers is $18.

Explanation of Solution

Given:

Sample size is 42.

Sample mean is 18.

Description:

Given a population with standard deviation $\sigma$, while estimating a population parameter, it is better to consider a range of values using samples of the same size (say $n$), also known as an interval estimate.

The true population parameters probability lies in this interval range is known as the confidence level $\left(c=1-\alpha \right)$.

At a certain level of confidence, the interval in which the maximum error can be observed is known as the confidence interval.

The maximum distance from point estimate that the confidence interval covers is margin of error and is given by:

$E=\frac{z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sigma }{\sqrt{n}}$,

Where, $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is the critical value for the level of confidence $c=1-\alpha$, in case of t distribution $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is replaced with $t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Given the sample mean $\left(\overline{x}\right)$, the population mean confidence interval is calculated as:

$\left(\overline{x}-E,\overline{x}+E\right)$.

In this question, an unbiased estimator that is the best point estimate of population mean.

Conclusion:

Best point estimate for the mean amount of money spent per week on lunch for all 42 teachers is $18.