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kow [346]
3 years ago
11

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 places.
Mathematics
1 answer:
Eva8 [605]3 years ago
8 0

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 of:

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.

Margin of error:

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.

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