SOLUTION: Each bottle of fruit juice from a small manufacturing plant is supposed to contain exactly 12 fluid ounces of juice. Susan is in charge of quality control and decided to test this

Question 1010549: Each bottle of fruit juice from a small manufacturing plant is supposed to contain exactly 12 fluid ounces of juice. Susan is in charge of quality control and decided to test this claim by gathering a SRS of 30 bottles. She will recalibrate the machinery if the average amount of juice per bottle differs from 12 fluid ounces at the 1% level of significance. Her sample of 30 bottles has an average of 11.92 fluid ounces and a sample standard deviation of 0.26 fluid ounces. Conduct a hypothesis test to determine if the machinery needs recalibrated.
Calculate test statistic and/or p-value. (Round to 2 decimal places.)
Conclude your test. Justify your answer.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
you might be interest in this reference.

http://blog.minitab.com/blog/adventures-in-statistics/understanding-hypothesis-tests%3A-significance-levels-alpha-and-p-values-in-statistics

your significant level is .01.

since your sample statistics can be more then or less than the desired level of 12, this would be a two tailed distribution.

therefore the alpha is equal to .01/2 = .005 on each end.

since you are using the sample standard deviation rather than the population standard deviation, then you will be using a t-test rather than a z-test.

the t-test uses the same size to calculate the t-score.

it uses what is called degrees of freedom.

with a sample size of 30, the degrees of freedom is equal to 30 - 1 = 29.

your critical t-score will be calculated using 29 degrees of freedom.

with an alpha of .005 on each end, your critical t-score will be plus or minus 2.756.

if the t-score of your sample is beyond these limits, you would need to re-calibrate your equipment.

if it is within these limits, there would be no need for re-calibration.

the mean of your sample is 11.92 with a standard deviation of .26

the standard error is equal to the standard deviation divided by the square root of the sample size.

this becomes .26 / sqrt(30) which is equal to .0475.

your t-score will be (x-m)/s.

x is the mean of your sample.
m is the mean of the population which is the desired measurement.
s is the standard error.

your t-score will be (12-11.92)/.0475 = 1.684.

this is well within the limits of t = plus or minus 2.756.

therefore, no calibration is required.

it makes no difference to this study if you had said x was the mean of the sample and m was the desired score.

your t-score would then have been (11.92-12)/.0475 = -1.684.

it will still have bee well within the limits of t = plus or minus 2.756.

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