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

`\hat \mu = \bar X = 25000`

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((\sum_(i=1)^n X_i)/(n))`

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

`E(\bar X) =(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) =(1)/(n) \sum_(i=1)^n \mu =(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`