Question

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8.2.19 E Question Help According to a study conducted by an organization, the proportion of Americans who were afraid to fly in 2006 was 0.10. A random sample of 1,200 Americans results in 108 indicating that they are afraid to fly. Explain why this is not necessarily evidence that the proportion of Americans who are afraid to fly has decreased. Select the correct choice below and, if necessary, fill in the answer box to complete your choice OA. This is not necessarily evidence that the proportion of Americans who are afraid to fly has decreased below 0.10 because the probability of obtaining a value equal to or more extreme than the sample proportion is 1200, which is not unusual Round to four decimal places as needed.) B. Th s is not necessa y ev dence that the proportion o Ame cans who are afraid to y has decreased below 0 10 because the sample propo tion ,s close to 0.10 (??? Type an integer or a decimal.) ° C. This is not necessarily evidence that the proportion of Americans who are afraid to fly has decreased below 010 because the sample size n is more than 5% of the population. ? D. This is not necessarily endence that the proportion of Americans who are afraid to fly has decreased below 010 because the value of np(1-pis less than 10

Answer 1

Here p = population proportion = 0.10

x = Number of Americans afraid to fly = 108

n = Total number of Americans = 1200

$\hat p = \frac{x}{n} = \frac{108}{1200} = 0.09$

Here n *p * (1 - p) = 1200 * 0.1 * 0.9 = 108 which is greater than 10 and sample size is also sufficiently large.

So the distribution of phat is normal with mean $\mu _{\hat p} = p = 0.10$

and the standard deviation is $\sigma _{\hat p} = \sqrt{\frac{p*(1-p)}{n}} = 0.00866$

If 108 out of 1200 is "not unusual" certainly any things larger would also be not unusual.

So have to find $P(\hat p \geq 0.09)$

First find z score for phat = 0.09

$Z = \frac{\hat p - \mu _{\hat p}}{\sigma _{\hat p}} = \frac{0.09 - 0.10}{0.00866} = -1.15$

That is $P(\hat p \geq 0.09) \ becomes \ P(Z \geq -1.15)$

To find the probability for z score -.15 use z table.

The probability using z table for z = -1.15 is 0.1251

But the z table always provides the less than probability and here we need greater than,

So just subtract less than probability from 1 to get greater than the probability that is

1 - 0.1251 = 0.8749

This probability is more than 0.05 that is not unusual.

Therefore option A is correct.

A. This is not necessarily evidence that the proportion of Americans who are afraid to fly has decreased below 0.10 because the probability of obtaining a value equal to or more extreme than the sample proportion is 0.8749, which is not unusual.