I am absolutely stumped for part c). I was able to use the software program statcrunch to complete parts a abs b but I'm unsure how to find C either by working it out or statcrunch, please help

A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.54 and the standard 61.47 47.01 55.390 deviation is $21.83. 21.78 67.64 75.53 a) Construct a 98% confidence interval for the mean purchases of all customers, assuming that the assumptions 23.13 9.66 -36.00 and conditions for the confidence interval have been met. 23.96 49.26 88.63 b) How large is the margin of error? 82.28 51.65 37.42 c) How would the confidence interval change if you had assumed that the population standard deviation was 46.86 32.93 79.86 known to be $22? 46.98 53.26 a) What is the confidence interval? ( 37.14 , 61.93 ) (Round to two decimal places as needed.) b) What is the margin of error? The margin of error is 12.39 . (Round to two decimal places as needed.) c) What is the confidence interval using the given population standard deviation? Select the correct choice below and fill in the answer boxes within your choice. (Round to two decimal places as needed.) O A. The new confidence interval ( ) is narrower than the interval from part a. O B. The new confidence interval ( ) is wider than the interval from part a