Show me your work Two machines are used for filling glass bottles with a soft-dr
ID: 3266149 • Letter: S
Question
Show me your work Two machines are used for filling glass bottles with a soft-drink beverage. The filling processes have known standard deviations 1 = 0.010 liter and 2 = 0.015 liter, respectively. A random sample of n1 = 25 bottles from machine 1 and n2 = 20 bottles from machine 2 results in average net contents of = 2.04 liters and = 2.07 liters. a. Test the hypothesis that both machines fill to the same net contents, using = 0.05 = 0.05. What are your conclusions? b. Find the P-value for this test. c. Construct a 95% confidence interval on the difference in mean fill volume.Explanation / Answer
a. State the hypotheses: The null hypothesis is hypothesis of no difference. Therefore, it states that there is nodifference in mean filling content of two machines. It is of interest to know whether there is difference in mean filling content of two machines. The hypotheses are therefore as follows:
H0:mu1-mu2=0
Ha: mu1-mu2=/=0
Assumptions: Mean filling content of machine 1 is independent of mean filling content of machine 2, filling content of both machines are normally distributed. Use, Student's t distribution to compute the t test statistic for drawing inference between two population means.
Test statistic
t=(x1bar-x2bar)/sqrt[s1^2/n1+s2^2/n2], where, xbar is sample mean, s is sample standard deviation, n is sample size, and 1, 2 denote machine 1 and 2 respectively.
= (2.04-2.07)/sqrt[0.01^2/25+0.015^2/20]
=-7.68
The p value and rejection rule: The p value at 31 degrees of freedom (obtained using technology) is 0.000, assuming unequal variance. Reject null hypothesis if p value is less than 0.05. Here, p value is less than 0.05, therefore, reject null hypothesis.
Conclusion: There is sufficient sample evidence to conclude that there is significant difference in mean filling content for two machines.
b. Using technology, the p value for t(-7.68) at 31 degrees of freedom is 0.0000.
c. The 95% confidence interval for the difference in population mean filling content for machine 1 and 2 is:
(x1bar-x2bar)+-talpha/2 sqrt[s1^2/n1+s2^2/n2]
=(2.04-2.07)+-2.04 sqrt[0.01^2/25+0.015^2/20]
=(-0.03796,-0.02204)
One can be 95% confident that the true mean filling content for machine 1 and 2 is between -0.03796 litres and -0.02204 litres respectively.
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