Q:

Help with my two math problems please:)

Accepted Solution

A:
The confidence interval is given by[tex](\text{sample proportion}-\text{margin of error},\text{sample proportion}+\text{margin of error})[/tex]For the 114-person sample, the confidence interval is[tex]\left((1-0.04)\dfrac{87}{114},(1+0.04)\dfrac{87}{114}\right)\approx(0.733,0.794)[/tex]which is to say, if [tex]\alpha\%[/tex] is the confidence level for this interval, then out of 100 samples taken, we expect [tex]\alpha[/tex] 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[tex]\left((1-0.04)\dfrac{87}{114},(1+0.04)\dfrac{87}{114}\right)9918\approx(7266,7874)[/tex]making A the closest answer.###Same basic problem as before. We construct the confidence interval:[tex]\left((1-0.02)\dfrac{2004}{3421},(1+0.02)\dfrac{2004}{3421}\right)\approx(0.574,0.598)[/tex]which corresponds to a range of[tex]\left((1-0.02)\dfrac{2004}{3421},(1+0.02)\dfrac{2004}{3421}\right)20000\approx(11481,11950)[/tex]so D is the most likely answer.