The problems on this page concern the data set
S = \{ 0.28, 0.2, 0.01, 0.80, 0.1 \} .
Let x_ i denote the i'th element of the data set S.
What is the sample mean \hat{\mu } of the data set S? 0.278
What is the sample variance \widehat{\sigma ^2} of the data set S?
(You may use either the unbiased sample variance or the MLE of the variance.)
\widehat{\sigma ^2}=
The unbiased sample variance is calculated using the formula:
\widehat{\sigma ^2} = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \hat{\mu})^2
where n is the number of elements in the data set and \hat{\mu} is the sample mean.
Using the given data set S, the sample mean is \hat{\mu} = 0.278.
Substituting the values into the formula:
\widehat{\sigma ^2} = \frac{1}{5-1} [(0.28 - 0.278)^2 + (0.2 - 0.278)^2 + (0.01 - 0.278)^2 + (0.80 - 0.278)^2 + (0.1 - 0.278)^2]
Calculating the values inside the parentheses:
\widehat{\sigma ^2} = \frac{1}{4} [(0.002)^2 + (-0.078)^2 + (-0.268)^2 + (0.522)^2 + (-0.178)^2]
Simplifying the terms and calculating the sum:
\widehat{\sigma ^2} = \frac{1}{4} [(0.000004) + (0.006084) + (0.071824) + (0.272484) + (0.031684)]
\widehat{\sigma ^2} = \frac{1}{4} (0.38208)
\widehat{\sigma ^2} = 0.09552
Therefore, the sample variance \widehat{\sigma ^2} of the data set S is 0.09552.
To calculate the sample variance, you can use the formula:
\(\widehat{\sigma ^2} = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \hat{\mu})^2\)
where \(n\) is the number of elements in the data set and \(\hat{\mu}\) is the sample mean.
Given that the data set S = {0.28, 0.2, 0.01, 0.80, 0.1} and \(\hat{\mu} = 0.278\), we can substitute these values into the formula:
\(\widehat{\sigma ^2} = \frac{1}{5-1} \sum_{i=1}^{5} (x_i - 0.278)^2\)
Calculate the squared differences for each element:
\(0.28 - 0.278 = 0.002\)
\(0.2 - 0.278 = -0.078\)
\(0.01 - 0.278 = -0.268\)
\(0.80 - 0.278 = 0.522\)
\(0.1 - 0.278 = -0.178\)
Now, plug in these values and calculate the sum:
\(\widehat{\sigma ^2} = \frac{1}{4} (0.002^2 + (-0.078)^2 + (-0.268)^2 + 0.522^2 + (-0.178)^2)\)
Simplify the expression:
\(\widehat{\sigma ^2} = \frac{1}{4} (0.000004 + 0.006084 + 0.071824 + 0.272484 + 0.031684)\)
Add the values together:
\(\widehat{\sigma ^2} = \frac{1}{4} (0.382080)\)
Simplify further:
\(\widehat{\sigma ^2} = 0.09552\)
Therefore, the sample variance \(\widehat{\sigma ^2}\) of the data set S is approximately 0.09552.