Answer:
[tex]\bar X = 45.1625[/tex]
And the sample variance is given by:
[tex]s^2 = \frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}[/tex]
And replacing we got:
[tex] s^2= 0.00262\approx 0.003[/tex]
And this one is the best estimator for the population variance [tex]\sigma^2[/tex]
Step-by-step explanation:
For this case we have the following data:
45.15,45.12,45.19,45.08,45.21,45.17,45.14,45.24
The first step would be calculate the sample mean given by:
[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And replacing we got:
[tex]\bar X = 45.1625[/tex]
And the sample variance is given by:
[tex]s^2 = \frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}[/tex]
And replacing we got:
[tex] s^2= 0.00262\approx 0.003[/tex]
And this one is the best estimator for the population variance [tex]\sigma^2[/tex]
