Conditions
compute standard error of estimate for data below
x values=3,-2,2,5,10
y values= 4,6,-2,0,-3
Solution
The formula for calculation of standard error:
S E = σ n SE = \frac{\sigma}{\sqrt{n}} SE = n σ
where
σ = 1 N ∑ i = 1 N ( x i − μ ) 2 , where μ = 1 N ∑ i = 1 N x i . \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2}, \quad \text{where} \quad \mu = \frac{1}{N} \sum_{i=1}^{N} x_i. σ = N 1 i = 1 ∑ N ( x i − μ ) 2 , where μ = N 1 i = 1 ∑ N x i . μ x = 1 5 ( 3 − 2 + 2 + 5 + 10 ) = 3 , 6 \mu_x = \frac{1}{5} (3 - 2 + 2 + 5 + 10) = 3,6 μ x = 5 1 ( 3 − 2 + 2 + 5 + 10 ) = 3 , 6 μ y = 1 5 ( 4 + 6 − 2 + 0 − 3 ) = 1 \mu_y = \frac{1}{5} (4 + 6 - 2 + 0 - 3) = 1 μ y = 5 1 ( 4 + 6 − 2 + 0 − 3 ) = 1 σ x = 1 5 ( ( 3 − 3 , 6 ) 2 + ( − 2 − 3 , 6 ) 2 + ( 2 − 3 , 6 ) 2 + ( 5 − 3 , 6 ) 2 + ( 10 − 3 , 6 ) 2 ) = \sigma_x = \sqrt{\frac{1}{5} \left( (3 - 3,6)^2 + (-2 - 3,6)^2 + (2 - 3,6)^2 + (5 - 3,6)^2 + (10 - 3,6)^2 \right)} = σ x = 5 1 ( ( 3 − 3 , 6 ) 2 + ( − 2 − 3 , 6 ) 2 + ( 2 − 3 , 6 ) 2 + ( 5 − 3 , 6 ) 2 + ( 10 − 3 , 6 ) 2 ) = = 1 5 ( 0.36 + 31 , 36 + 2 , 56 + 1 , 96 + 40 , 96 ) = 15 , 44 ≈ 3 , 929 = \sqrt{\frac{1}{5} \left( 0.36 + 31,36 + 2,56 + 1,96 + 40,96 \right)} = \sqrt{15,44} \approx 3,929 = 5 1 ( 0.36 + 31 , 36 + 2 , 56 + 1 , 96 + 40 , 96 ) = 15 , 44 ≈ 3 , 929 σ y = 1 5 ( ( 4 − 1 ) 2 + ( 6 − 1 ) 2 + ( − 2 − 1 ) 2 + ( 0 − 1 ) 2 + ( − 3 − 1 ) 2 ) = \sigma_y = \sqrt{\frac{1}{5} \left( (4 - 1)^2 + (6 - 1)^2 + (-2 - 1)^2 + (0 - 1)^2 + (-3 - 1)^2 \right)} = σ y = 5 1 ( ( 4 − 1 ) 2 + ( 6 − 1 ) 2 + ( − 2 − 1 ) 2 + ( 0 − 1 ) 2 + ( − 3 − 1 ) 2 ) = = 1 5 ( 9 + 25 + 9 + 1 + 16 ) = 12 = 2 3 = \sqrt{\frac{1}{5} (9 + 25 + 9 + 1 + 16)} = \sqrt{12} = 2\sqrt{3} = 5 1 ( 9 + 25 + 9 + 1 + 16 ) = 12 = 2 3 S E x = 3 , 929 5 ≈ 1 , 757 SE_x = \frac{3,929}{\sqrt{5}} \approx 1,757 S E x = 5 3 , 929 ≈ 1 , 757 S E y = 2 3 5 ≈ 1 , 549 SE_y = \frac{2\sqrt{3}}{\sqrt{5}} \approx 1,549 S E y = 5 2 3 ≈ 1 , 549