A machine is used to fill plastic bottles with bleach. A sample of 18 bottles had a mean fill volume of 2.007 L and a standard deviation of 0.010 L. The machine is then moved to another location. A sample of 10 bottles filled at the new location had a mean fill volume of 2.001 L and a standard deviation of 0.012 L. It is believed that moving the machine may have changed the mean fill volume, but it is unlikely to have changed the standard deviation. Assume that both samples come from approximately normal populations and have equal variance. The 99% confidence interval for the difference between the mean fill volumes at the two locations is approximately

Required:
Find a 99% confidence interval for the difference between the mean fill volumes at the two locations.

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, [tex]\bar{x}_{1}[/tex] = 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, [tex]\bar{x}_{2}[/tex] = 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

[tex]\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}}}[/tex]

(1 - C)/2 = (1 - 0.99)/2 = 0.005, the degrees of freedom, f = n₁ - 1 = 10 - 1 = 9

∴ [tex]t_{\alpha /2}[/tex] = 3.25

Therefore, we have;

[tex]\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}}[/tex]

Therefore, we have the difference of the two means given as follows;

-0.1175665 L < μ₁ - μ₂ < 0.1295665 L

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