Answer:
The margin of error is given below:
[tex]E=z \times \sqrt{\frac{\hat{p}(1-\hat{p}}{n} }[/tex]
Where:
[tex]\hat{p}=\frac{26}{80}=0.325[/tex]
[tex]\therefore E = z \times \sqrt{\frac{0.325(1-0.325}{80} }[/tex]
[tex]=z \times 0.0523659[/tex]
Therefore, the option E = z*6% is correct as it closest to the given answer
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%
