Answer:
The 99% confidence interval for the difference between the mean fill volumes at the two locations is;
-0.1175665 L < μ₁ - μ₂ < 0.1295665 L
Step-by-step explanation:
The number of bottles in the sample at the first location, n₁ = 18 bottles
The mean fill volume,
= 2.007 L
The standard deviation, σ₁ = 0.010 L
The number of bottles in the sample at the second location, n₂ = 10 bottles
The mean fill volume,
= 2.001 L
The standard deviation, σ₂ = 0.012 L
The nature of the variance of the two samples = Equal variance
The confidence interval of the statistics, C = 99%
The difference between the mean
![\mu_1 - \mu_2 = \left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2} \times \sqrt{\dfrac{\sigma _{1}^{2}}{n_{1}}+\dfrac{\sigma _{2}^{2}}{n_{2}}}](https://tex.z-dn.net/?f=%5Cmu_1%20-%20%5Cmu_2%20%3D%20%5Cleft%20%28%5Cbar%7Bx%7D_%7B1%7D-%20%5Cbar%7Bx%7D_%7B2%7D%20%20%5Cright%20%29%5Cpm%20t_%7B%5Calpha%20%2F2%7D%20%5Ctimes%20%5Csqrt%7B%5Cdfrac%7B%5Csigma%20_%7B1%7D%5E%7B2%7D%7D%7Bn_%7B1%7D%7D%2B%5Cdfrac%7B%5Csigma%20_%7B2%7D%5E%7B2%7D%7D%7Bn_%7B2%7D%7D%7D)
(1 - C)/2 = (1 - 0.99)/2 = 0.005, the degrees of freedom, f = n₁ - 1 = 10 - 1 = 9
∴
= 3.25
Therefore, we have;
![\mu_1 - \mu_2 = \left (2.007- 2.001 \right )\pm 3.25 \times \sqrt{\dfrac{0.01^{2}}{18}+\dfrac{0.12^{2}}{10}}](https://tex.z-dn.net/?f=%5Cmu_1%20-%20%5Cmu_2%20%3D%20%5Cleft%20%282.007-%202.001%20%20%5Cright%20%29%5Cpm%203.25%20%5Ctimes%20%5Csqrt%7B%5Cdfrac%7B0.01%5E%7B2%7D%7D%7B18%7D%2B%5Cdfrac%7B0.12%5E%7B2%7D%7D%7B10%7D%7D)
Therefore, we have the difference of the two means given as follows;
-0.1175665 L < μ₁ - μ₂ < 0.1295665 L