Question

A university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduates returned the survey. The average salary was $25,000, with a standard deviation of $2,500. What is the best point estimate of the population mean?

Answer 1

Answer:

$\hat \mu = \bar X = 25000$

Step-by-step explanation:

Data given

$\bar X = 25000$ represent the sample mean obtained

$s= 2500$ represent the sample deviation obtained

$\mu =?$ represent the true mean (variable of interest)

n=400 represent the sample size selected

Solution to the problem

The best estimator for the true mean $\mu$ is given by the sample mean $\bar X$, and the reason is: If we assume that in tha random sample $X_1,X_2,....,X_{400}$ each observation follows a normal distribution like this:

$X_i \sim N(\mu, \sigma)$

Then when we find the expected value for the sample mean we have this:

$E(\bar X)= E(\frac{\sum_{i=1}^n X_i}{n})$

If we use the properties of the expected value we have:

$E(\bar X) =\frac{1}{n} \sum_{i=1}^n E(X_i)$

And since the expected value of each observation is the same we have:

$E(\bar X) =\frac{1}{n} \sum_{i=1}^n \mu =\frac{1}{n} n\mu = \mu$

So then we have that the best estimator for the population mean is given by:

$\hat \mu = \bar X = 25000$