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

Answer:

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

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

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 $n = 400, \pi = \frac{280}{400} = 0.7$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

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

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 1.645\sqrt{\frac{0.7*0.3}{400}}$

$M = 0.038$

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