The Ohio Department of Agriculture tested 203 fuel samples across the state in 1999 for accuracy of the reported octane level.For premium grade,14 out of 105 samples failed.(They didn't meet ASTM specification and the FTC octane posting rule. )How many samples would be needed to create a 99 percent confidence interval that is within 0.02 of the true proportion of premium grade fuel-quality failures?

Question

Quizplus · Accepted Answer

We can use the formula for the sample size for a proportion:

n = [(z_crit)^2 * p * (1-p)] / E^2

Where:
- z_crit is the critical value of the standard normal distribution for the desired confidence level (99%)
- p is an estimate of the proportion of failures, which we don't know yet
- E is the margin of error, which is 0.02 in this case

To estimate the proportion of failures, we can use the sample proportion from the 1999 data:

p̂ = 14/105 ≈ 0.1333

Using a calculator or a table, we find the critical value for a 99% confidence interval:

z_crit ≈ 2.58

Now we can plug in the numbers and solve for n:

n = [(2.58)^2 * 0.1333 * (1-0.1333)] / 0.02^2 ≈ 1912.85

We need a sample size of at least 1913 to create a 99% confidence interval with a margin of error of 0.02. Therefore, the best choice is C) 1913.

The Ohio Department of Agriculture Tested 203 Fuel Samples Across