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%

The margin of error is given below:

Where:

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

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v