Question

Help with my two math problems please:)

Answer 1

The confidence interval is given by

`(\text{sample proportion}-\text{margin of error},\text{sample proportion}+\text{margin of error})`

For the 114-person sample, the confidence interval is

`\left((1-0.04)(87)/(114),(1+0.04)(87)/(114)\right)\approx(0.733,0.794)`

which is to say, if
`\alpha\%` is the confidence level for this interval, then out of 100 samples taken, we expect
`\alpha` of them to have the sample proportion fall in this interval. Then for the population of 9918 people in the district, this corresponds to a range of

`\left((1-0.04)(87)/(114),(1+0.04)(87)/(114)\right)9918\approx(7266,7874)`

making A the closest answer.

###

Same basic problem as before. We construct the confidence interval:

`\left((1-0.02)(2004)/(3421),(1+0.02)(2004)/(3421)\right)\approx(0.574,0.598)`

which corresponds to a range of

`\left((1-0.02)(2004)/(3421),(1+0.02)(2004)/(3421)\right)20000\approx(11481,11950)`

so D is the most likely answer.