Answer:
[tex]\bar X= \frac{1.83+1.85+1.79+1.73+1.69+1.74+1.76+1.70}{8}= 1.76125[/tex]
Now we can estimate the population variance with the sample variance 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 = 0.0033839[/tex]
And the estimator for the population deviation [tex]\sigma[/tex] is given by :
[tex]\hat \sigma = \sqrt{s^2}= \sqrt{0.0033839}= 0.058172[/tex]
Step-by-step explanation:
For this case we have the following data given:
1.83,1.85,1.79,1.73,1.69,1.74,1.76,1.70
First 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{1.83+1.85+1.79+1.73+1.69+1.74+1.76+1.70}{8}= 1.76125[/tex]
Now we can estimate the population variance with the sample variance 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 = 0.0033839[/tex]
And the estimator for the population deviation [tex]\sigma[/tex] is given by :
[tex]\hat \sigma = \sqrt{s^2}= \sqrt{0.0033839}= 0.058172[/tex]
