Question

part 1 - Suppose a sample of 518 people is drawn. Of these people, 368 didn't pass out at G forces greater than 6. Using the data, estimate the proportion of people who pass out at more than 6 Gs. Enter your answer as a fraction or a decimal number rounded to three decimal places.

part 2 - Suppose a sample of 518 people is drawn. Of these people, 368 didn't pass 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

Solution:

Part 1

We are given

x = number of items of interest = 518 – 368 = 150

n = sample size = 518

P = x/n = 150/518 = 0.28957529

Estimate for proportion = 0.290

Part 2

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = number of items of interest = 518 – 368 = 150

n = sample size = 518

P = x/n = 150/518 = 0.28957529

Confidence level = 85%

Critical Z value = 1.4395

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.28957529 ± 1.4395* sqrt(0.28957529*(1 – 0.28957529)/518)

Confidence Interval = 0.28957529 ± 1.4395* 0.0199

Confidence Interval = 0.28957529 ± 0.0287

Lower limit = 0.28957529 - 0.0287 = 0.261

Upper limit = 0.28957529 + 0.0287 = 0.318

Confidence interval = (0.261, 0.318)