Respuesta :
Answer:
[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
But we need to calculate the mean with the following formula:
[tex]\bar X = \frac{\sum_{i=1}^n X_I}{n}[/tex]
And replacing we got:
[tex]\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072[/tex]
And for the sample variance we have:
[tex] s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003[/tex]
And thi is the best estimator for the population variance since is an unbiased estimator od the population variance [tex]\sigma^2[/tex]
[tex] E(s^2) = \sigma^2 [/tex]
Step-by-step explanation:
For this case we have the following data:
1.04,1.00,1.13,1.08,1.11
And in order to estimate the population variance we can use the sample variance formula:
[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
But we need to calculate the mean with the following formula:
[tex]\bar X = \frac{\sum_{i=1}^n X_I}{n}[/tex]
And replacing we got:
[tex]\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072[/tex]
And for the sample variance we have:
[tex] s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003[/tex]
And thi is the best estimator for the population variance since is an unbiased estimator od the population variance [tex]\sigma^2[/tex]
[tex] E(s^2) = \sigma^2 [/tex]
