Question

A sample of 8 golf balls is randomly selected and the following weights are measured in grams. Give a point estimate for the population variance. Round your answer to three decimal places. 45.15,45.12,45.19,45.08,45.21,45.17,45.14,45.24

Answer 1

Answer:

$\bar X = 45.1625$

And the sample variance is given by:

$s^2 = \frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}$

And replacing we got:

$s^2= 0.00262\approx 0.003$

And this one is the best estimator for the population variance $\sigma^2$

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:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X = 45.1625$

And the sample variance is given by:

$s^2 = \frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}$

And replacing we got:

$s^2= 0.00262\approx 0.003$

And this one is the best estimator for the population variance $\sigma^2$