91Ó°ÊÓ

Find the mean of each given bootstrap sample to later use in constructing the confidence interval. Calculate the mean for each sample: Sample 1: \{38,145,55,86\} Mean = \( \frac{38 + 145 + 55 + 86}{4} = 81\) Sample 2: \{86,38,145,145\} Mean = \( \frac{86 + 38 + 145 + 145}{4} = 103.5\) Sample 3: \{145,86,55,55\} Mean = \( \frac{145 + 86 + 55 + 55}{4} = 85.25\) Sample 4: \{55,55,55,145\} Mean = \( \frac{55 + 55 + 55 + 145}{4} = 77.5\) Sample 5: \{86,86,55,55\} Mean = \( \frac{86 + 86 + 55 + 55}{4} = 70.5\) Sample 6: \{38,38,86,86\} Mean = \( \frac{38 + 38 + 86 + 86}{4} = 62\) Sample 7: \{145,38,86,55\} Mean = \( \frac{145 + 38 + 86 + 55}{4} = 81\) Sample 8: \{55,86,86,86\} Mean = \( \frac{55 + 86 + 86 + 86}{4} = 78.25\) Sample 9: \{145,86,55,86\} Mean = \( \frac{145 + 86 + 55 + 86}{4} = 93\) Sample 10: \{38,145,86,55\} Mean = \( \frac{38 + 145 + 86 + 55}{4} = 81\)

Compute the average of the sample means calculated in Step 1 to find an overall estimate of the population mean. Mean of sample means: \( \frac{81 + 103.5 + 85.25 + 77.5 + 70.5 + 62 + 81 + 78.25 + 93 + 81}{10} = 81.8\)

For a small sample size like 10, the 80% confidence interval often corresponds to the 10th and 90th percentiles of the bootstrap distribution. From sorting the means: [62, 70.5, 77.5, 78.25, 81, 81, 81, 85.25, 93, 103.5], 10th percentile (1st element): 62, 90th percentile (last element): 103.5. Therefore, the 80% confidence interval for the mean is \([62, 103.5]\).

Next, find the standard deviation of each bootstrap sample. Use the formula \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \). Calculate for each sample: Sample 1: \{38,145,55,86\} Standard Deviation = \( \sqrt{\frac{(38-81)^2 + (145-81)^2 + (55-81)^2 + (86-81)^2}{3}} = 45.9\) Sample 2: \{86,38,145,145\} Standard Deviation = \( \sqrt{\frac{(86-103.5)^2 + (38-103.5)^2 + (145-103.5)^2 + (145-103.5)^2}{3}} = 53.37\) Sample 3: \{145,86,55,55\} Standard Deviation = \( \sqrt{\frac{(145-85.25)^2 + (86-85.25)^2 + (55-85.25)^2 + (55-85.25)^2}{3}} = 41.98\) Sample 4: \{55,55,55,145\} Standard Deviation = \( \sqrt{\frac{(55-77.5)^2 + (55-77.5)^2 + (55-77.5)^2 + (145-77.5)^2}{3}} = 43\) Sample 5: \{86,86,55,55\} Standard Deviation = \( \sqrt{\frac{(86-70.5)^2 + (86-70.5)^2 + (55-70.5)^2 + (55-70.5)^2}{3}} = 17.81\) Sample 6: \{38,38,86,86\} Standard Deviation = \( \sqrt{\frac{(38-62)^2 + (38-62)^2 + (86-62)^2 + (86-62)^2}{3}} = 27.63\) Sample 7: \{145,38,86,55\} Standard Deviation = \( \sqrt{\frac{(145-81)^2 + (38-81)^2 + (86-81)^2 + (55-81)^2}{3}} = 45.9\) Sample 8: \{55,86,86,86\} Standard Deviation = \( \sqrt{\frac{(55-78.25)^2 + (86-78.25)^2 + (86-78.25)^2 + (86-78.25)^2}{3}} = 14.38\) Sample 9: \{145,86,55,86\} Standard Deviation = \( \sqrt{\frac{(145-93)^2 + (86-93)^2 + (55-93)^2 + (86-93)^2}{3}} = 37.63\) Sample 10: \{38,145,86,55\} Standard Deviation = \( \sqrt{\frac{(38-81)^2 + (145-81)^2 + (86-81)^2 + (55-81)^2}{3}} = 45.9\)

Compute the average of the standard deviations found in Step 4. Mean of sample standard deviations: \( \frac{45.9 + 53.37 + 41.98 + 43 + 17.81 + 27.63 + 45.9 + 14.38 + 37.63 + 45.9}{10} = 37.75\)

Use the 10th and 90th percentiles of the sorted standard deviations to construct the confidence interval. From sorting the standard deviations: [14.38, 17.81, 27.63, 37.63, 41.98, 43, 45.9, 45.9, 45.9, 53.37], 10th percentile (1st element): 14.38, 90th percentile (last element): 53.37. Therefore, the 80% confidence interval for the standard deviation is \([14.38, 53.37]\).