Question

NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6.

Step 2 of 2 :  

Suppose a sample of 543543 people is drawn. Of these people, 217 passed out. Using the data, construct the 98% confidence interval for the population proportion of people who black out at G forces greater than 6. Round your answers to three decimal places.

phat = 0.400

Answer 1

Confidence Intervals: Confidence Intervals are one way to draw the conclusion about the population based on a random sample data.

• The Population Proportion is denoted by 'p'.
• The Sample Proportion is denoted by '$\hat{p}$ '. (Natural Estimate)
• The Sample Size is denoted by 'n'.

Constructing an Interval around $\hat{p}$ which has higher probability including p, is known as Confidence Interval (CI).

To arrive at the confidence interval, we have to know Estimate $\pm$ Margin of Error which is formulated as

  $\hat{p}$$\pm$ z* SE($\hat{p}$)

In an Interval, It is expressed as ($\hat{p}$ - z* SE($\hat{p}$) , $\hat{p}$ + z* SE($\hat{p}$))

The factor z* is based on the normal distribution.

• $\hat{p}$ = 0.400
• p = 217/ 543543 = 0.000399
• SE = sqrt [ p(1 - p) / n ] = sqrt [0.000399(1-0.000399)/ 543543] = 7.315
• z*=2.33 from t-distribution calculation

Therefore, The confidence interval range is 0.4 $\pm$2.33(7.315) = 0.4 $\pm$17.044

The confidence interval is 16.644 to 17.444