The Retail Advertising and Marketing Association would like to estimate the average amount of money that a person spends for Mother's Day with a 99% confidence interval and a margin of error within plus or minus $6. Assuming the standard deviation for spending on Mother's Day is $36, the required sample size is . A) 313 B) 284 C) 210 D) 239

D) 239

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

Required sample size:

Is n

n is found when

We have that

So

Simplifying by 6

We round up, and the answer is 239.

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation: