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Sagot :
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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
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 [tex]n = 400, \pi = \frac{280}{400} = 0.7[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
Give the margin of error for the 90% confidence interval.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.645\sqrt{\frac{0.7*0.3}{400}}[/tex]
[tex]M = 0.038[/tex]
The margin of error for the 90% confidence interval is of 0.038.
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