Question

Real Fruit Juice: 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 42 such cans of this fruit drink. Of these, the mean volume of fruit juice is 6.34 with standard deviation of 0.21. 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) What are the critical values of t? Use the answer found in the t-table or round to 3 decimal places.
t_α/2 = ±  

(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.

Answer 1

a) this is a two-tailed test

null hypothesis:

alternate hypothesis:

b)

c)The degree of freedom = n-1= 42-1=41

Based on the information provided, the significance level is and the critical value for a two-tailed test is

The rejection region for this two-tailed test is

d) Since it is observed that it is then concluded that the null hypothesis is not rejected.

Fail to reject

e)There is not enough data to justify the rejection of the claim that the mean amount of real fruit juice in all 32-ounce cans is 6.4 ounces.