Question

Of the population of all fruit flies we wish to give a 90% confidence interval for the fraction which possess a gene which gives immunity to fungal infections. To this end we have obtained a random sample of 400 fruit flies. We find that 280 of the flies in the sample possess the gene. Give the margin of error for the 90% confidence interval. Round your answer to 3 decimal

Answer 1

The margin of error for the 90% confidence interval is of 0.038.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

To this end we have obtained a random sample of 400 fruit flies. We find that 280 of the flies in the sample possess the gene.

This means that
`n = 400, \pi = (280)/(400) = 0.7`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

Give the margin of error for the 90% confidence interval.

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`M = 1.645\sqrt{(0.7*0.3)/(400)}`

`M = 0.038`

The margin of error for the 90% confidence interval is of 0.038.