Connect with knowledgeable individuals and get your questions answered on IDNLearn.com. Get accurate and timely answers to your queries from our extensive network of experienced professionals.
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.
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com provides the answers you need. Thank you for visiting, and see you next time for more valuable insights.
