Students took n=35 samples of water from the east basin of Lake Macatawa and mea
ID: 3204797 • Letter: S
Question
Students took n=35 samples of water from the east basin of Lake Macatawa and measured the amount of sodium in parts per million. For their data, they calculated x = 24.11 and s^2 = 24.44.
A) Find and interpret 90% confidence interval for , the mean of the amount of sodium in parts per million in the east basin of Lake Macatawa.
B) True mean amount of sodium in west basin is 20.5 in parts per million. Can you claim a difference that of from the east basin of the lake?
C) By estimating the mean amount of sodium, they measure the effect from the lake in salting the city street in the winter. Ecologists suggest that any difference of 3.45 in the mean sodium amount is not practically significant. If there is any difference in the mean amount of sodium concentration in the east and west basins of the lake, is this a practically important difference?
For part A) I got the CI to be between 22.74 and 25.48.
B) I would assume that since 20.5 is not included in the CI, there is a claim of difference. Is this correct?
C) I am highly confused about this one. I really don't know the answer to this.
Thank you for your help :)
Explanation / Answer
a) mean =24.11
std deviation =(24.44)1/2 =4.9437
std error =std deviation/(n)1/2 =0.8356
for 90% CI and (n-1=34) degrees of freedom t =1.6909
hence confidence interval =mean +/- t*std error =22.697 ; 25.523
B) you are correct.
C) as our mean difference =24.11-20.5 =3.61 which is greater then 3.45, hence it is practically significant.