Respuesta :
Answer:
The correct option is A) This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the probability of obtaining a value equal to or more extreme than the sample proportion is . which is not unusual.
Step-by-step explanation:
Consider the provided information.
The formula for testing a proportion is based on the z statistic.
[tex]z=\frac{\hat p-p_0}{\sqrt{p_0\frac{1-p_0}{n}}}[/tex]
Were [tex]\hat p[/tex] is sample proportion.
[tex]p_0[/tex] hypothesized proportion and n is the sample space,
The proportion of Americans who were afraid to fly in 2006 was 0.10. A random sample of 1,300 Americans results in 143 indicating that they are afraid to fly.
Therefore, n = 100 [tex]\hat p = \frac{143}{1300}=0.11[/tex] , [tex]p_0 = 0.10[/tex] and [tex]1 - p_0 = 1 - 0.10 = 0.90[/tex]
Substitute the respective values as shown:
[tex]z=\frac{0.11-0.10}{\sqrt{\frac{0.10\times 0.90}{1300}}}[/tex]
[tex]z=1.201850[/tex]
P(x>sample proportion)=p(z>1.20)=0.11507
Hence, the correct option is A) This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the probability of obtaining a value equal to or more extreme than the sample proportion is . which is not unusual.
Using the normal distribution and the central limit theorem, it is found that the probability is of 0.1151, which is not unusual.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex].
In this problem, we have that the proportion is of p = 0.1, while the sample size is of n = 1300, hence the mean and the standard error are given by:
[tex]\mu = p = 0.1[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.1(0.9)}{1300}} = 0.0083[/tex]
The sample proportion was of:
[tex]\overline{p} = \frac{143}{1300} = 0.11[/tex]
The probability of obtaining a value equal to or more extreme than the sample proportion is one subtracted by the p-value of Z when X = 0.11, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.11 - 0.1}{0.0083}[/tex]
[tex]Z = 1.2[/tex]
[tex]Z = 1.2[/tex] has a p-value of 0.8849.
1 - 0.8849 = 0.1151 probability, which is not unusual, as it is greater than 0.05.
To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213
