91影视

A sample of measured radiation emissions for cell phones is as given as follows.

38, 55, 86, 145

The ten bootstrap samples are as follows.

{38, 145, 55, 86}, {86, 38, 145, 145}, {145, 86, 55, 55}, {55, 55, 55, 145}, {86, 86, 55, 55}, {38, 38, 86, 86}, {145, 38, 86, 55}, {55, 86, 86, 86}, {145, 86, 55, 86}, {38, 145, 86, 556}.

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

$P_{10}<渭<P_{90}$

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

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 means of each of the 10 bootstrap samples:

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

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

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

$$\begin{gathered} {\stackrel{炉}{x}}_1=\frac{38+145+55+86}{4} \\ =\frac{324}{4} \\ =81.0\text{cW}/\text{kg} \end{gathered}$$

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

The following table shows the sample means 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.

$$\begin{gathered} P_{10}=\frac{1^{st}\text{sample}\text{mean}+2^{nd}\text{sample}\text{mean}}{2} \\ =\frac{62+70.5}{2} \\ =66.25\text{cW}/\text{kg} \end{gathered}$$

Thus, $P_{10}$is equal to 66.25 cW/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.

$$\begin{gathered} P_{90}=\frac{9^{th}\text{sample}\text{mean}+{10}^{th}\text{sample}\text{mean}}{2} \\ =\frac{93+103.5}{2} \\ =98.25\text{cW}/\text{kg} \end{gathered}$$

Thus, $P_{90}$is equal to 98.25 cW/kg.

a.

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

$$\begin{gathered} CI=P_{10}<渭<P_{90} \\ =66.25\text{cW}/\text{kg}<渭<98.25\text{cW}/\text{kg} \end{gathered}$$

Thus, the 80%confidence intervalestimate of the population mean is equal to (66.25 cW/kg, 98.25 cW/kg).

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, $\stackrel{炉}{x}$is the sample mean, and n is the sample size.

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

$$\begin{gathered} s_1=\sqrt{\frac{{鈭�}_{i=1}^n{(x_i-{\stackrel{炉}{x}}_1)}^2}{n-1}} \\ =\sqrt{\frac{{\left(38-81\right)}^2+{\left(145-81\right)}^2+{\left(55-81\right)}^2+{\left(86-81\right)}^2}{4-1}} \\ =47.1\text{cW}/\text{kg} \end{gathered}$$

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^th and 90^th percentiles.

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

$$\begin{gathered} P_{10}=\frac{1^{st}\text{sample}\text{standard}\text{deviation}+2^{nd}\text{standard}\text{deviation}}{2} \\ =\frac{15.5+17.9}{2} \\ =16.70 \end{gathered}$$

Thus, $P_{10}$is equal to 16.70 cW/kg.

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

$$\begin{gathered} P_{90}=\frac{9^{th}\text{sample}\text{standard}\text{deviation}+{10}^{th}\text{standard}\text{deviation}}{2} \\ =\frac{47.1+51.8}{2} \\ =49.42 \end{gathered}$$

Thus,$P_{90}$ is equal to 49.42 cW/kg.

b.

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

$$\begin{gathered} CI=P_{10}<蟽<P_{90} \\ =16.70\text{cW}/\text{kg}<渭<49.42\text{cW}/\text{kg} \end{gathered}$$

Therefore, the 80% confidence intervalestimate of the population standard deviation is equal to (16.70 cW/kg, 49.42 cW/kg).