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10 bootstrap samples of the amount of weight change of college students in freshman year are considered. There are four possible sample values: 11, 3, 0 and -2. Here, -2 denotes a loss of 2 kgs, and all other positive values represent weight gained.

The 10 samples are:

{11, 11, 11, 0}, {11, -2, 0, 11}, {11, -2, 3, 0}, {3, -2, 0, 11}, {0, 0, 0, 3}, {3, -2, 3, -2}, {11, 3, -2, 0}, { -2, 3, -2, 3}, { -2, 0, -2, 3} and {3, 11, 11, 11}.

The 80% confidence interval estimate of the population proportion has the following expression:

$P_{10}<p<P_{90}$

Here, $P_{10}$denotes the 10^th percentile of the sorted sample proportions and $P_{90}$denotes the 90^th percentile of the sorted sample proportions.

The 80% confidence interval estimate of the population standard deviation has the following expression:

$P_{10}<蟽<P_{90}$

Here, $P_{10}$denotes the 10^th percentile of the sorted sample standard deviations and $P_{90}$denotes the 90^th percentile of the sorted sample standard deviations.

The following formula of the sample mean is used to compute the sample mean of each of the 10 bootstrap samples:

$\stackrel{炉}{x}=\frac{{鈭�}_{i=1}^n{x}_i}{n}$

Here, $x_i$denotes the i-th sample observation, and n is the sample size.

The sample mean for the first bootstrap sample is computed below.

$$\begin{gathered} {\stackrel{炉}{x}}_1=\frac{11+11+11+0}{4} \\ =8.25\text{kgs} \end{gathered}$$

Similarly, the sample means of the remaining 9 bootstrap samples are computed.

The following table shows the sample mean of each of the 10 bootstrap samples:

The following table shows the sorted values of the sample means in ascending order:

For finding the percentile, first, compute the value of the locator (L) as follows.

$L=\frac{k}{100}脳n$

Here, k is the percentile value, and n is the total number of observations.

a.

For finding the 10^th percentile of the sorted values, first, find the value of L.

Here, n is equal to 10, and k is equal to 5.

Thus,

$$\begin{gathered} L=\frac{k}{100}脳n \\ =\frac{10}{100}脳10 \\ =1 \end{gathered}$$

As L is a whole number, the value of $P_{10}$is the sum of the ${\text{L}}^{\text{th}}$,and the ${\left(\text{L}+1\right)}^{\text{th}}$sample means divided by 2.

$$\begin{gathered} P_{10}=\frac{1^{st}\text{sample}\text{mean}+2^{nd}\text{sample}\text{mean}}{2} \\ =\frac{\left(-0.25\right)+0.5}{2} \\ =0.125\text{kg} \end{gathered}$$

Thus, $P_{10}$is equal to 0.125 kg.

For finding the 90^th percentile, first, find the value of L.

$$\begin{gathered} L=\frac{k}{100}脳n \\ =\frac{90}{100}脳10 \\ =9 \end{gathered}$$

As L is a whole number, the value of $P_{90}$is the sum of the ${\text{L}}^{\text{th}}$and the ${\left(\text{L}+1\right)}^{\text{th}}$sample means divided by 2.

$$\begin{gathered} P_{90}=\frac{9^{th}\text{sample}\text{mean}+{10}^{th}\text{sample}\text{mean}}{2} \\ =\frac{8.25+9}{2} \\ =8.625\text{kg} \end{gathered}$$

Thus, $P_{90}$is equal to 8.625 kg.

a.

The 80% confidence intervalestimate of the population meanusing the bootstrap samples is as follows.

$$\begin{gathered} CI=P_{10}<渭<P_{90} \\ =0.125\text{kg}<渭<8.625\text{kg} \end{gathered}$$

Thus, the 80%confidence intervalestimate of the population mean is equal to (0.125 kg, 8.625kg).

The following formula of the sample standard deviation is used to compute the sample standard deviations of each of the 10 bootstrap samples.

$s=\sqrt{\frac{{鈭�}_{i=1}^n{(x_i-\stackrel{炉}{x})}^2}{n-1}}$

Here,$x_i$denotes the ith sample observation, is the sample mean, and n is the sample size.

The sample standard deviation for the first bootstrap sample is computed below.

$$\begin{gathered} s_1=\sqrt{\frac{{鈭�}_{i=1}^n{(x_i-{\stackrel{炉}{x}}_1)}^2}{n-1}} \\ =\sqrt{\frac{{\left(11-8.25\right)}^2+{\left(11-8.25\right)}^2+{\left(11-8.25\right)}^2+{\left(0-8.25\right)}^2}{4-1}} \\ =5.50\text{kg} \end{gathered}$$

Similarly, the sample standard deviations of the remaining 9 bootstrap samples are computed.

The following table shows the sample standard deviations of each of the 10 bootstrap samples:

The sorted sample standard deviations in ascending order are shown below.

The value of L will remain the same for the 10^thand 90^th percentiles.

The value of the 10^th percentile of the sorted sample standard deviations is equal to

$$\begin{gathered} P_{10}=\frac{1^{st}\text{sample}\text{standard}\text{deviation}+2^{nd}\text{sample}\text{standard}\text{deviation}}{2} \\ =\frac{1.50+2.36}{2} \\ =1.93\text{kg} \end{gathered}$$

Thus, $P_{10}$is equal to 1.93 kg.

The value of the 90t^h percentile of the sorted sample standard deviations is equal to

$$\begin{gathered} P_{90}=\frac{9^{th}\text{sample}\text{standard}\text{deviation}+{10}^{th}\text{sample}\text{standard}\text{deviation}}{2} \\ =\frac{5.72+6.98}{2} \\ =6.35\text{kg} \end{gathered}$$

Thus, $P_{90}$is equal to 6.35 kg.

b.

The 80% confidence intervalestimate of the population standard deviationusing the bootstrap samples is as follows.

$P_{10}<蟽<P_{90}=1.93\text{kg}<蟽<6.35\text{kg}$

Therefore, the 80% confidence intervalestimate of the population standard deviation of weight gained is equal to (1.93 kg, 6.35 kg).