Answer:
[tex]\hat \mu = \bar X = 25000[/tex]
Step-by-step explanation:
Data given
[tex]\bar X = 25000[/tex] represent the sample mean obtained
[tex] s= 2500[/tex] represent the sample deviation obtained
[tex]\mu =?[/tex] 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 [tex]\mu[/tex] is given by the sample mean [tex]\bar X[/tex], and the reason is: If we assume that in tha random sample [tex]X_1,X_2,....,X_{400}[/tex] each observation follows a normal distribution like this:
[tex]X_i \sim N(\mu, \sigma)[/tex]
Then when we find the expected value for the sample mean we have this:
[tex] E(\bar X)= E(\frac{\sum_{i=1}^n X_i}{n})[/tex]
If we use the properties of the expected value we have:
[tex] E(\bar X) =\frac{1}{n} \sum_{i=1}^n E(X_i)[/tex]
And since the expected value of each observation is the same we have:
[tex] E(\bar X) =\frac{1}{n} \sum_{i=1}^n \mu =\frac{1}{n} n\mu = \mu[/tex]
So then we have that the best estimator for the population mean is given by:
[tex]\hat \mu = \bar X = 25000[/tex]
