. A clinic offers a weight-loss program. A review of is records found the following amounts of weight loss, in pounds, for a random sample of 24 of its clients at the conclusion of a 4-month program.182516111520161928252631454 036192825361635201619a) Assume the customer base of this program is very large (>1000). Find a 99% confidence interval for the population mean. b) Without doing the calculations, explain whether a 90% confidence interval for the population mean would be wider than, narrower than, or the same as that found in part a). c) Now assume the total number of customers who have participated in this weight loss program is 400. What is the 99% confidence interval now

A) = (  19.2501, 29.4999 )

B) 90% confidence interval would be narrower than 99% confidence interval

C)  ( mean - margin of error ,  mean + margin of error )

Step-by-step explanation:

customer base > 10000

sample = 24 clients

period ( time ) = 4 months

A) Determine 99% confidence interval for population mean

mean = 24.3750

std = 8.9434

sample size = 24

initial margin of error = 1 - 0.99 = 0.01

99% confidence interval for population mean = (  19.2501, 29.4999 )

B) 90% confidence interval would be narrower than 99% confidence interval

c) The 99% confidence interval  when sample size = 400

we cannot determine the mean value for 400 customers because the values given are for just 24 customers

 ( mean - margin of error ,  mean + margin of error )

attached below is the detailed solution

45.3592

step-by-step explanation:

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18