Question # Here is a sample of amounts of weight change (kg) of college students in their freshman year: 10, 7, 6, -7, where -7 represents a

Confidence intervals
ANSWERED Here is a sample of amounts of weight change (kg) of college students in their freshman year: 10, 7, 6, -7, where -7 represents a loss of 7 kg and positive values represent weight gained. Here are ten bootstrap samples: $$\displaystyle{\left\lbrace{10},{10},{10},{6}\right\rbrace},{\left\lbrace{10},-{7},{6},{10}\right\rbrace},{\left\lbrace{10},-{7},{7},{6}\right\rbrace},{\left\lbrace{7},-{7},{6},{10}\right\rbrace},{\left\lbrace{6},{6},{6},{7}\right\rbrace},{\left\lbrace{7},-{7},{7},-{7}\right\rbrace},{\left\lbrace{10},{7},-{7},{6}\right\rbrace},{\left\lbrace-{7},{7},-{7},{7}\right\rbrace},{\left\lbrace-{7},{6},-{7},{7}\right\rbrace},{\left\lbrace{7},{10},{10},{10}\right\rbrace}$$. a) Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the mean weight change for the population.

$$\Box kg<{\mu}<kg\Box$$ 2020-11-11

Step 1 We have to find the mean of all bootstrap samples and sort them We now obtain from our list of bootstrap sample means a confidence interval. Since we want a 80% confidence interval, we use the 90th and 10th percentiles as the endpoints of the intervals. The reason for this is that we split $$\displaystyle{100}\%-{80}\%={20}\%$$ in half so that we will have the middle 80% of all of the bootstrap sample means.

$$\begin{array}{|c|c|} \hline Bookstrap\ samples & mean & sorted\ means \\ \hline 10,\ 10,\ 10,\ 6 & 9 & -0.25 \\ \hline 10,\ -7,\ 6,\ 10 & 4.75 & 0\\ \hline 10,\ -7,\ 7,\ 6 & 4 & 0 \\ \hline 7,\ -7,\ 6,\ 10 & 4 & 4 \\ \hline 6,\ 6,\ 6,\ 7 & 6.25 & 4\\ \hline 7,\ -7,\ 7,\ -7 & 0& 4\\ \hline 10,\ +7,\ -7,\ 6 & 4 & 4.47\\ \hline -7,\ 7,\ -7,\ 7 & 0 & 6.25\\ \hline -7,\ 6,\ -7,\ 7 & -0.25 & 9\\ \hline 7,\ 10,\ 10,\ 10 & 9.25 & 9.25\\ \hline \end{array}$$

Step 2 From the sorted mean data $$\displaystyle{P}_{{{10}}}={\frac{{{1}^{{{s}{t}}}\text{term}+{2}^{{{n}{d}}}\text{term}}}{{{2}}}}={\frac{{-{0.25}+{0}}}{{{2}}}}={0.125}$$
$$\displaystyle{P}_{{{90}}}={\frac{{{9}^{{{t}{h}}}\text{term}+{10}^{{{t}{h}}}\text{term}}}{{{2}}}}={\frac{{{9}+{9.25}}}{{{2}}}}={9.125}$$

Therefore, $$\displaystyle-{0.125}\text{kg}{<}\mu{<}{9.125}\text{kg}$$