SOLUTION: A local juice manufacturer distributes juice in bottles labeled 12 ounces. A government agency thinks that the company is cheating its customers. The agency selects 50 of these bot

Question 1204611: A local juice manufacturer distributes juice in bottles labeled 12 ounces. A government agency thinks that the company is cheating its customers. The agency selects 50 of these bottles, measures their contents, and obtains a sample mean of 11.6 ounces with a standard deviation of 0.70 ounce. Use a 0.01 significance level to test the agency's claim that the
company is cheating its customers.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
ckauned iybces are 12.
sample mean of 50 of these bottles is 11.6 with a standard deviation of .7
test is to see if the sample is less than 12.
.01 significance is one tailed on the low end of the normal distribution curve.
standard error is equal to standard deviation / sqrt(sample size) = .7/sqrt(50) = .0990 rounded to 4 decimal places.
t-score = (11.6 - 12) / .0990 = -4.0404.
11.6 is the sample mean
12 is the claimed mean
.0990 is the standad error.
t-score of -4.0404 with 49 degrees of freedom has .00009 proportion of the are aunder the normal distribuion curve to the left of it.
this is significantly less than the critical area of .01, so the results are considered significant and the conclusion is that the actual lmean is less than the claimed mean, indicating that the company is cheating on its customer.
here's swhat the results look like on the t-score test calculator i found online at https://www.statskingdom.com/130MeanT1.html

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