Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations sigma(1) = 0.010 liter and sigma(2) = 0.015 liter, respectively. A random sample of n(1) = 25 bottles from machine 1 and n(2) = 20 bottles from machine 2 results in average net contents of x-bar = 2.04 liters and x-bar = 2.07 liters.

(a) Test the hypothesis that both machines fill to the same net contents, using alpha = 0.05. What are your conclusions?
(b) Find the P-value for this test.
(c) Construct a 95% confiedence interval on the difference in mean fill volume.

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.  

Then we have , , and , , . The pooled estimate is given by  

a. We want to test vs (two-tailed alternative).  

The test statistic is and the observed value is . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

, i.e.,

where is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

, i.e.,

(-0.0225, -0.0375)

the answer is 5636.7