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opudodennis

Answer:

ME=0.014

Sample Proportion=0.725

Step-by-step explanation:

-The margin of error is defined as the percentage by which obtained results differ from the real population value.

-It is calculated as half the difference between the confidence interval levels:

[tex]ME=\frac{1}{2}[CI_u-CI_l]\\\\=0.5[0.739-0.711]\\\\=0.014[/tex]

Hence, the margin of error is 0.014

-The sample proportion is mathematically half the sum of the confidence intervals:

[tex]ME=\frac{1}{2}[CI+u-CI_l]\\\\=0.5[0.739+0.711]\\\\=0.725[/tex]

Hence, the sample proportion is 0.725

Using confidence interval concepts, it is found that:

  • The sample proportion is of 0.725.
  • The margin of error is of 0.014.

What are the point estimate and the margin of error of a confidence interval:

Considering the interval having bounds (a,b), we have that:

  • The point estimate is (a + b)/2.
  • The margin of error is (b - a)/2.

In this problem, the interval is (0.711​,0.739​), hence a = 0.711, b = 0.739, and:

(0.711 + 0.739)/2 = 0.725

(0.739 - 0.711)/2 = 0.014.

Then:

  • The sample proportion is of 0.725.
  • The margin of error is of 0.014.

More can be learned about confidence interval concepts at https://brainly.com/question/25779324