Question

NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6.Suppose a sample of 518 people is drawn. Of these people, 150 passed out. Using the data, construct the 85% confidence interval for the population proportion of people who black out at G forces greater than 6. Round your answers to three decimal places.

Answer 1

Answer:

The 85% confidence interval would be given by (0.261;0.319)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Data given

$X=150$ number of people who pass

$n=518$ random sample taken

$\hat p=\frac{150}{518}=0.290$ estimated proportion of people who black put at G forces greater then 6

$p$ true population proportion of people who black put at G forces greater then 6

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 85% of confidence, our significance level would be given by $\alpha=1-0.85=0.15$ and $\alpha/2 =0.075$. And the critical value would be given by:

$z_{\alpha/2}=-1.440, z_{1-\alpha/2}=1.440$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.290 - 1.440\sqrt{\frac{0.290(1-0.290)}{518}}=0.261$

$0.290 + 1.440\sqrt{\frac{0.290(1-0.290)}{518}}=0.319$

The 85% confidence interval would be given by (0.261;0.319)