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 44 such cans of this fruit drink. Of these, the mean volume of fruit juice is 6.33 with standard deviation of 0.19. 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 left-tailed test.

This is a right-tailed test.

This is a two-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 H0

fail 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.

b) The test statistic t = ()/(s/)

                                 = (6.33 - 6.4)/(0.19/)

                                 = -2.44

c) P-value = 2 * P(T < -2.44)

                 = 2 * 0.0094

                 = 0.0188

d) Since the P-value is less than the significance level(0.0188 < 0.05), so we should reject the null hypothesis.

Reject H₀

e) 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.