Students at a certain school were surveyed, and it was estimated that 28% of college students abstain from drinking alcohol. To estimate this proportion in your school, how large a random sample would you need to estimate it to within 0.02 with probability 0.95, if before conducting the study

(a) you are unwilling to predict the proportion value at your school and
(b) you use the results from the surveyed school as a guideline.

a) We need a sample size of at least 2401.

b) We need a sample size of at least 1936.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

95% confidence level

So , z is the value of Z that has a pvalue of , so .

(a) you are unwilling to predict the proportion value at your school and

We use , which is when we are going to need the largest sample size.

So we need to find n when . So

We need a sample size of at least 2401.

(b) you use the results from the surveyed school as a guideline.

Now the same calculation, just with . So

We need a sample size of at least 1936.

total amount of money that he should pay if he has not used his duty free exemption is $98

step-by-step explanation:

given a united states citizen returning to the states declares the following items at the customs office: 3 shirts at $8.50 each 2 dresses at $27.50 each 1 pair of gold cuff links at $17.50 per pair. we have to find how much he pay

cost of 3 shirts if one cost at $8.50 is

cost of 2 dresses if one cost at $27.50 is  

cost of 1 pair of gold cuff at $17.50

so, total amount of money that he should pay if he has not used his duty free exemption is   $25.5+$55+$17.50=$98