Question

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.

Answer 1

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
$x\pm z\times (\frac{s}{\sqrt{n}}) \\ \\10 \pm 1.96(\frac{0.262}{\sqrt{7}})=10\pm 0.1941$