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10 bootstrap samples of responses on the favourite seat of the respondents when they fly are considered. 鈥淲indow鈥� is denoted by 1, and 鈥淥ther鈥� 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}.

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 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,\(x\)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\({x_1}\).

\(\begin{array}{c}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}}\\ = \frac{0}{4}\\ = 0\end{array}\)

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:

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

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,

\(\begin{array}{c}L = \frac{k}{{100}} \times n\\ = \frac{{10}}{{100}} \times 10\\ = 1\end{array}\).