Homework SourseA research firm conducted a survey to determine the mean amo
| A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. They found the distribution of amounts spent per week followed the normal distribution with a population standard deviation of $5. A sample of 64 steady smokers revealed that |
Solution
a)
Solution :
Number of smokers n =64
Mean Mu= $20
Standard Deviation SD = $5
Z FOR 96% CONFIDNCE INTERVAL = 1.96
CI : (Mu- z* SD/sqrt(n)) , Mu+ z* SD/sqrt(n) )
=(20- 1.96* 5/sqrt(64)) , 20+ 1.96* 5/sqrt(64) )
= (18.775,21.225) Answer
