Jackknife standard error with six observations

Question

You jackknife a statistic over n = 6 observations, giving six leave-one-out estimates: 18, \quad 20, \quad 22, \quad 24, \quad 26, \quad 28. Compute the jackknife estimate of the standard error.

Accepted Answer

Use 	ext{SE}{	ext{jack}} = \sqrt{\frac{n-1}{n} \sum{i=1}^{n} \left(\hat{	heta}{(i)} - \bar{	heta}{(\cdot)}ight)^2}. Step 1, mean of the leave-one-out estimates. \bar{	heta}{(\cdot)} = \frac{18 + 20 + 22 + 24 + 26 + 28}{6} = \frac{138}{6} = 23. Step 2, sum of squared deviations from 23. (-5)^2 + (-3)^2 + (-1)^2 + (1)^2 + (3)^2 + (5)^2 = 25 + 9 + 1 + 1 + 9 + 25 = 70. Step 3, apply 	frac{n-1}{n} = 	frac{5}{6} and take the root. 	ext{SE}{	ext{jack}} = \sqrt{\frac{5}{6} 	imes 70} = \sqrt{58.33} \approx 7.64. Takeaway. As n grows the factor 	frac{n-1}{n} approaches 1, so for larger sam…

Jackknife standard error with six observations

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