Question

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 6.4 ounces 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 0.05 significance level.

(a) What type of test is this? This is a two-tailed test.This is a right-tailed test.    This is a left-tailed test. (b) What is the test statistic? Round your answer to 2 decimal places. t x = (c) Use software to get the P-value of the test statistic. Round to 4 decimal places. P-value = (d) What is the conclusion regarding the null hypothesis? reject H0fail to reject H0     (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 ounces    6.31 6.54 6.55 6.38 6.26 6.01 6.28 6.05 6.10 6.04 6.50 6.01 6.50 6.00 6.64 6.33 6.53 6.45 5.96 6.11 6.22 6.26 5.79 6.58 5.71 6.70 6.45 6.75 6.31 6.25

Answer 1

(a) What type of test is this?

This is a two-tailed test

(b) What is the test statistic? Round your answer to 2 decimal places.
t= -2.32

(c) Use software to get the P-value of the test statistic. Round to 4 decimal places.
P-value = 0.0276

(d) What is the conclusion regarding the null hypothesis?

reject H₀

(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.

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Explanation:

Hypothesis Ho:mean amount of real fruit juice = 6.4

H_1:  mean amount of real fruit juice   6.4

sample mean = 6.29

sample SD (s)= 0.27

Test Statistic t=(-)/ (s/) under Ho has t distribution with df (30-1)=29 as number of sample is 30 and population SD unknown

= (6.29-6.4)/ (0.27/)

= -2.32

P value = 2*P[T>-2.32] = 0.0276

AS P Value is less than 0.05 we have to reject Ho at 5% level alpha.