Question

Cell Phone Radiation Here is a sample of measured radiation emissions (cW/kg) for cell phones (based on data from the Environmental Working Group): 38, 55, 86, 145. Here are ten bootstrap samples: {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}.

a. Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the population mean.

b. Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the population standard deviation.

Short answer

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

b. The 80% confidence interval estimate of the population standard deviation is equal to (16.70 cW/kg, 49.42 cW/kg).

Step-by-step solution

Given information

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}.

Expression of the confidence intervals

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

$P_{10}<\mu <P_{90}$

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:

$P_{10}<\sigma <P_{90}$

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.

Find the sample mean from each bootstrap sample

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

$\overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n}$

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.

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

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:

Sample number	Bootstrap samples				Sample mean	Sorted $\overline{X}$
1	38	145	55	86	81	62
2	86	38	145	145	103.5	70.5
3	145	86	55	55	85.25	77.5
4	55	55	55	145	77.5	78.25
5	86	86	55	55	70.5	81
6	38	38	86	86	62	81
7	145	38	86	55	81	81
8	55	86	86	86	78.25	85.25
9	145	86	55	86	93	93
10	38	145	86	55	81	103.5

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

62

70.5

77.5

78.25

81

81

81

85.25

93

103.5

Find the percentiles

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

$L=\frac{k}{100}\times n$

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,

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

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.

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

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.

Confidence interval estimate of the population mean

a.

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

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

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

Find the standard deviation of each bootstrap sample

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

$s=\sqrt{\frac{{\sum }_{i=1}^n{(x_i-\overline{x})}^2}{n-1}}$

Here, $x_i$denotes the ith sample observation, $\overline{x}$is the sample mean, and n is the sample size.

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

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

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:

Sample number	Bootstrap samples				Sample standard deviation
1	38	145	55	86	47.1
2	86	38	145	145	51.8
3	145	86	55	55	42.4
4	55	55	55	145	45.0
5	86	86	55	55	17.9
6	38	38	86	86	27.7
7	145	38	86	55	47.1
8	55	86	86	86	15.5
9	145	86	55	86	37.6
10	38	145	86	55	47.1

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

15.5

17.9

27.7

37.6

42.4

45.0

47.1

47.1

47.1

51.8

Find the percentiles

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.

Confidence interval estimate of the population standard deviation

b.

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

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

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