Suppose we want a 99% confidence interval for the average amount spent on books by freshmen in their first year at college. The amount spent has a normal distribution with standard deviation $34. (a) How large should the sample be to guarantee the margin of error will be no more than $3?(b) If we wanted a smaller margin of error, we would need a sample. (Enter: ''LARGER'', ''SMALLER'' or ''SAME SIZE'', without the quotes.)

a) 855

b) We would need a Larger sample.

Step-by-step explanation:

(a) How large should the sample be to guarantee the margin of error will be no more than $3?

The formula for Margin of Error = z × Standard deviation/√sample size

z = z score of 99% confidence interval = 2.58

Standard deviation = $34

Sample size = n = unknown

Margin of Error = $3

√n = z × Standard deviation/Margin of Error

Square both sides

n = (z × standard deviation/Margin of Error)²

n =( 2.58 ×34/ 3)²

n = (29.24)²

n = 854.9776

Approximately, the size of the sample be to guarantee the margin of error will be no more than $3 is 855

(b) If we wanted a smaller margin of error, we would need a sample. (Enter: ''LARGER'', ''SMALLER'' or ''SAME SIZE'', without the quotes.)

We would need a Larger sample.

This is because, the larger the sample, the smaller the margin of Emerror and the smaller the sample, the larger the margin of error. Therefore, for the question b, we would be needing a larger sample.

ok

step-by-step explanation:

asdfghkl

step-by-step explanation:

asdfghkl