Question

The yearbook committee polled 80 randomly selected students from a class of 320 ninth graders to see if they would be willing to pay more for a yearbook if their names were printed on the front. Of the students who were surveyed, 26 of them said they would be willing to pay extra. With a desired confidence interval of 90%, which has a z*-score of 1.645, what is the margin of error of this survey? E = z* 4% 6% 8% 9%

Answer 1

Answer:

The margin of error is given below:

$E=z \times \sqrt{\frac{\hat{p}(1-\hat{p}}{n} }$

Where:

$\hat{p}=\frac{26}{80}=0.325$

$\therefore E = z \times \sqrt{\frac{0.325(1-0.325}{80} }$

                           $=z \times 0.0523659$

Therefore, the option E = z*6% is correct as it closest to the given answer

Answer 2

Answer:

9%

Find the sample size and sample proportion then multiply the sample proportion by 1-p divide the result by n take the square root of the calculated value multiply the result by the appropriate z value.

Sample size = 80 sample proportion is the number in the sample with the characteristic of interest, divided by n, so it is computed by = 26/80 = .325 or .33 The margin of error for this polling question is calculated in this matter: Z* square root of ((p(1-p)) divided by sample size=1.645* square root of (.33)(.67) divided by 80= 1.645 * 0.0526= 0.0865 or 8.7% According to this data, you conclude with 90% confidence that 33% of the students who were surveyed are willing to pay extra have a margin of error of 8.7% which rounds out to 9%