A local brewery whishes to ensure that an average of 14 ounces of beer is used t

Question

A local brewery whishes to ensure that an average of 14 ounces of beer is used to fill a bottle. In order to analyze the accuracy of the bottling process, the bottler takes a random sample of 49 bottles. The sample mean weight and the sample standard deviation of the bottles is 12 ounces and 0.5 ounces, respectively. (20 points) (a) State the null and alternative hypotheses for the test. (b) Do you need to make any assumptions regarding the population for testing? (c) At the = 0.05, specify the critical value(s). What is the decision rule? (d) Make a recommendation to the bottler.
Need all work to be shown A local brewery whishes to ensure that an average of 14 ounces of beer is used to fill a bottle. In order to analyze the accuracy of the bottling process, the bottler takes a random sample of 49 bottles. The sample mean weight and the sample standard deviation of the bottles is 12 ounces and 0.5 ounces, respectively. (20 points) (a) State the null and alternative hypotheses for the test. (b) Do you need to make any assumptions regarding the population for testing? (c) At the = 0.05, specify the critical value(s). What is the decision rule? (d) Make a recommendation to the bottler.
Need all work to be shown A local brewery whishes to ensure that an average of 14 ounces of beer is used to fill a bottle. In order to analyze the accuracy of the bottling process, the bottler takes a random sample of 49 bottles. The sample mean weight and the sample standard deviation of the bottles is 12 ounces and 0.5 ounces, respectively. (20 points) (a) State the null and alternative hypotheses for the test. (b) Do you need to make any assumptions regarding the population for testing? (c) At the = 0.05, specify the critical value(s). What is the decision rule? (d) Make a recommendation to the bottler.
Need all work to be shown

Explanation / Answer

(a) Ho: The mean fill is 14 ounces ( = 14) versus Ha: The mean fill is different from 14 ounces ( 14)

(b) The fill amounts are assumed to be normally distributed.

(c)

= 0.05

Degrees of freedom = 49 - 1 = 48

Lower Critical t- score = -2.010634722

Upper Critical t- score = 2.010634722

(d)

SE = s/Ön = 0.5/49 = 0.071428571

t = (x-bar - )/SE = (12 - 14)/0.0714285714285714 = -28

p- value = 2.17902E-31   

Decision (in terms of the hypotheses):

Since 28 > 2.010635 we reject Ho and accept Ha

Conclusion (in terms of the problem):

It appears that the mean fill amount is different from 14 ounces. The dispenser must be reset.

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