Question

a research center claims that 27% of adults in a certain country would travel into space on a commercial flight if they could afford it. in a random sample of 1100 adults in that country, 31% say that they would travel into space on a commercial flight if they could afford it. at \\( \alpha = 0.10 \\), is there enough evidence to reject the research center’s claim? complete parts (a) through (c) below. commercial flight if they could afford it. d. no more than % of adults in the country would travel into space on a commercial flight if they could afford it. let \\( p \\) be the population proportion of successes, where a success is an adult in the country who would travel into space on a commercial flight if they could afford it. state \\( h _ { 0 } \\) and \\( h _ { a } \\). select the correct choice below and fill in the answer boxes to complete your choice (round to two decimal places as needed ) a. \\( h _ { 0 } : p \geq \\) b. \\( h _ { 0 } : p = 0.27 \\) c. \\( h _ { 0 } : p \
eq \\) \\( h _ { a } : p < \\) \\( h _ { a } : p \
eq 0.27 \\) \\( h _ { a } : p = \\) d. \\( h _ { 0 } : p < \\) e. \\( h _ { 0 } : p > \\) f. \\( h _ { 0 } : p \leq \\) \\( h _ { a } : p \geq \\) \\( h _ { a } : p \leq \\) \\( h _ { a } : p > \\) (b) use technology to find the p - value identify the standardized test statistic \\( z = 2.99 \\) (round to two decimal places as needed ) identify the p - value \\( p = \square \\) (round to three decimal places as needed )

Explanation:

Step1: Determine the type of test

Since \(H_{a}:p
eq0.27\), this is a two - tailed test.

Step2: Calculate the P - value for a two - tailed z - test

The formula for the P - value in a two - tailed z - test is \(P = 2\times(1-\Phi(|z|))\), where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution. Given \(z = 2.99\).
We know that \(\Phi(2.99)\) can be found using a standard normal table or technology. Using a standard normal table or a calculator with a normalcdf function (\(normalcdf(-\infty,2.99)\)), \(\Phi(2.99)\approx0.9986\)
Then \(P=2\times(1 - 0.9986)\)

$$P=2\times0.0014=0.0028\approx0.003$$

Answer:

\(P = 0.003\)