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):
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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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