Real Fruit Juice (Raw Data, Software
Required):
A 32 ounce can of a popular fruit drink claims to contain 20%
real fruit juice. Since this is a 32 ounce can, they are
actually claiming that the can contains 6.4ounces of real fruit
juice. The consumer protection agency samples 30 such cans of
this fruit drink. The amount of real fruit juice in each can is
given in the table below. Test the claim that the mean amount of
real fruit juice in all 32 ounce cans is 6.4 ounces. Test this
claim at the 0.01 significance level.
(a) What type of test is this? This is a righttailed test.This is a twotailed test. This is a lefttailed test. (b) What is the test statistic? Round your answer to 2 decimal places. tx= (c) Use software to get the Pvalue of the test statistic. Round to 4 decimal places. Pvalue = (d) What is the conclusion regarding the null hypothesis? reject H_{0}fail to reject H_{0} (e) Choose the appropriate concluding statement. There is enough data to justify rejection of the claim that the mean amount of real fruit juice in all 32 ounce cans is 6.4 ounces.There is not enough data to justify rejection of the claim that the mean amount of real fruit juice in all 32 ounce cans is 6.4 ounces. We have proven that the mean amount of real fruit juice in all 32 ounce cans is 6.4 ounces.We have proven that the mean amount of real fruit juice in all 32 ounce cans is not 6.4 ounces. 
DATA ( n = 30 ) Real Juice

The statistical software output for this problem is:
Hence,
a) Two tailed test
b) Test statistic = 2.56
c) Pvalue = 0.0160
d) Fail to reject Ho
e) There is not enough data to justify rejection of the claim that the mean amount of real fruit juice in all 32 ounce cans is 6.4 ounces.
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