The contents of seven similar containers of sulfuric acid are 9.8, 10.2, 10.4, 9.8, 10.0, 10.2, and 9.6 liters. find a 95% confidence interval for the mean contents of all such containers, assuming an approximately normal distribution.

The confidence interval is from 9.81 to 10.19.

We first find the mean of the data:

(9.8+10.2+10.4+9.8+10.0+10.2+9.6)/7 = 10

Next we find the standard deviation:
σ=√([(9.8-10)^2+(10.2-10)^2+(10.4-10)^2+(9.8-10)^2+(10-10)^2+(10.2-10)^2+(9.6-10)^2]/7) = 0.262

The z-score for 95% confidence is found by
1-0.95 = 0.05; 0.05/2 = 0.025; from the z-table, it is 1.96.

The confidence interval is calculated using

If you divide 0.1 and 0.6 you wii get 0.166666 which rounds to .167 which is answer b