Question

Question 12 of 15, Step 2 of 2 Correct NASA is conducting an experiment to find out the fraction of people who black out at forces greater than 6. Step 2 of 2: Suppose a sample of 581 people is drawn. Of these people, 278 passed out. Using the data, construct the 95% confidence interval for the population proportion of people who black out at Gforces greater than 6. Round your answers to three decimal places. Tables Keypad Answer How to enter your answer Previous step answer Lower endpoint Upper endpoint Submit Answer 2020 Hands Learning MacBook Air OBAGO 7 GRTV

Answer 1

Here we are given that the total number of sample=581 people

The frequency of success x=278

278 ::P = 581

..p=0.4785

Since the number of people in the sample n=581 is large and p is close to -.5, we shall use a normal approximation to the binomila distribution.

Hence the 95% confidence interval for the population proportion is therefore

p(1-p) ptvn

where $z_{\frac{\alpha }{2}}$ is the ordinate of the normal distribution such that the area of the normal curve from this point till infinity is 0.025(since it is the confidence interval and the area on the positive axis is 0.025 and the area on the negative axis ie $-\infty$ to $-z_{\frac{\alpha }{2}}$ is 0.025. This value can be obtained from the tables (Fisher & Yates) or from an Excel which is 1.959964 or 1.96.

Substituting all the values in the formula , we shall have the 95% confidence interval as

(0.4785 +0.5215 =0.478541/ 581

¥0.2495 = 0.4785 +1.96/ 581

= 0.4785 + 1.96 * 0.004294

= 0.4785 +1.96 +0.0207

= 0.4785 +0.0406

= (0.4378, 0.5191)

Hence the 95% confidence interval for the population proportion is .

= (0.4378, 0.5191)

Therefore the lower endpoint =0.4378 and the upper end point =0.5191.