QuestionAugust 25, 2025

What is the standard error of this data set? 0,1,2,2,3,3,3,4,4,5,6

What is the standard error of this data set? 0,1,2,2,3,3,3,4,4,5,6
What is the standard error of this data set? 0,1,2,2,3,3,3,4,4,5,6

Solution
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Answer

0.4563 Explanation 1. Calculate the Mean The mean \bar{x} is calculated as \bar{x} = \frac{\sum x_i}{n} = \frac{0+1+2+2+3+3+3+4+4+5+6}{11} = 3.0909. 2. Calculate the Variance Variance s^2 is calculated using s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}. Compute each squared deviation, sum them, and divide by 10. (0-3.0909)^2 + (1-3.0909)^2 + (2-3.0909)^2 + (2-3.0909)^2 + (3-3.0909)^2 + (3-3.0909)^2 + (3-3.0909)^2 + (4-3.0909)^2 + (4-3.0909)^2 + (5-3.0909)^2 + (6-3.0909)^2 = 22.9091 s^2 = \frac{22.9091}{10} = 2.29091 3. Calculate the Standard Deviation Standard deviation s is \sqrt{s^2} = \sqrt{2.29091} = 1.5136. 4. Calculate the Standard Error **Standard error** SE = \frac{s}{\sqrt{n}} = \frac{1.5136}{\sqrt{11}} = 0.4563.

Explanation

1. Calculate the Mean<br /> The mean $\bar{x}$ is calculated as $\bar{x} = \frac{\sum x_i}{n} = \frac{0+1+2+2+3+3+3+4+4+5+6}{11} = 3.0909$.<br />2. Calculate the Variance<br /> Variance $s^2$ is calculated using $s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$. Compute each squared deviation, sum them, and divide by $10$.<br /> $(0-3.0909)^2 + (1-3.0909)^2 + (2-3.0909)^2 + (2-3.0909)^2 + (3-3.0909)^2 + (3-3.0909)^2 + (3-3.0909)^2 + (4-3.0909)^2 + (4-3.0909)^2 + (5-3.0909)^2 + (6-3.0909)^2 = 22.9091$<br /> $s^2 = \frac{22.9091}{10} = 2.29091$<br />3. Calculate the Standard Deviation<br /> Standard deviation $s$ is $\sqrt{s^2} = \sqrt{2.29091} = 1.5136$.<br />4. Calculate the Standard Error<br /> **Standard error** $SE = \frac{s}{\sqrt{n}} = \frac{1.5136}{\sqrt{11}} = 0.4563$.
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