A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2 : Suppose a sample of 322 floppy disks is drawn. Of these disks, 306 were not defective. Using the data, construct the 90% confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places.

The 90% confidence interval would be given by (0.026;0.073)

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

Solution to the problem

The estimated proportion for this case is:

Represent the proportion of defectives for this case

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 90% of confidence, our significance level would be given by and . And the critical value would be given by:

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

If we replace the values obtained we got:

The 90% confidence interval would be given by (0.026;0.073)

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer