Question

In Exercises 5–8, use the relatively small number of given bootstrap samples to construct the confidence interval.Seating Choice In a 3M Privacy Filters poll, respondents were asked to identify their favourite seat when they fly, and the results include these responses: window, window, other, other. Letting “window” = 1 and letting “other” = 0, here are ten bootstrap samples for those responses: {0, 0, 0, 0}, {0, 1, 0, 0}, {0, 1, 0, 1}, {0, 0, 1, 0}, {1, 1, 1, 0}, {0, 1, 1, 0}, {1, 0, 0, 1}, {0, 1, 1, 1}, {1, 0, 1, 0}, {1, 0, 0, 1}. Using only the ten given bootstrap samples, construct an 80% confidence interval estimate of the proportion of respondents who indicated their favourite seat is “window.”

Short answer

The 80% confidence interval estimate of the population proportion of respondents who prefer the window seat is equal to (0.125,0.75).

Step-by-step solution

Given information

10 bootstrap samples of responses on the favourite seat of the respondents when they fly are considered. “Window” is denoted by 1, and “Other” is denoted by 0.

The ten samples are:

{0, 0, 0, 0}, {0, 1, 0, 0}, {0, 1, 0, 1}, {0, 0, 1, 0}, {1, 1, 1, 0}, {0, 1, 1, 0}, {1, 0, 0, 1}, {0, 1, 1, 1}, {1, 0, 1, 0} and {1, 0, 0, 1}.

Expression of the confidence intervals

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.

Find the sample proportion from each bootstrap sample

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

$\hat{p}=\frac{x}{n}$

Here, denotes the number of respondents who prefer window seats, and n is the sample size.

The value of n is the same for all samples and is equal to 4.

The sample proportion for the first bootstrap sample is computed below

The number of 1s in the first sample is the value of $x_1$.

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{0}{4} \\ =0 \end{gathered}$$

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

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

Sample number	Bootstrap samples				Sample proportion
1	0	0	0	0	0
2	0	1	0	0	0.25
3	0	1	0	1	0.5
4	0	0	1	0	0.25
5	1	1	1	0	0.75
6	0	1	1	0	0.5
7	1	0	0	1	0.5
8	0	1	1	1	0.75
9	1	0	1	0	0.5
10	1	0	0	1	0.5

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

0

0.25

0.25

0.5

0.5

0.5

0.5

0.5

0.75

0.75

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 proportions divided by 2.

$$\begin{gathered} P_{10}=\frac{1^{st}\text{sample}\text{proportion}+2^{nd}\text{sample}\text{proportion}}{2} \\ =\frac{0+0.25}{2} \\ =0.125 \end{gathered}$$

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

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 ${\left(\text{L}+1\right)}^{\text{th}}$the sample proportions divided by 2.

$$\begin{gathered} P_{90}=\frac{9^{th}\text{sample}\text{proportion}+{10}^{th}\text{sample}\text{proportion}}{2} \\ =\frac{0.75+0.75}{2} \\ =0.75 \end{gathered}$$

Thus, $P_{90}$is equal to 0.75.

Confidence interval estimate of the population proportion

The 80% confidence interval estimate of the population proportion of respondents who prefer the window seat using the bootstrap samples is as follows.

$P_{10}<p<P_{90}=0.125<p<0.75$

Thus, the 80%confidence intervalestimate of the population proportion of respondents who prefer the window seat is equal to (0.125,0.75).