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 0.01 significance level.

(a) What type of test is this? This is a two-tailed test. This is a left-tailed test.     This is a right-tailed test. (b) What is the test statistic? Round your answer to 2 decimal places. tx=   (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)
To test the hypothesis is that the mean amount of real fruit juice in all 32 ounce cans
is different from 6.40 ounces at 10% significance level. The hypothesis is two-tailed,
The null and alternative hypothesis is.

H0:μ = 6.40

Ha: μ≠6.40

b)
The t-test statistics is,
By using MINITAB. find t-test statistics with the help of following steps is:
~ Import the data.
~ Select the Stat and choose the Basic Statistics option.
~Select the 2 sample ¢ and choose variable option and put Sample in column
~Give the Hypothesized mean
~Click option button choose level of significance and alternative hypothesis.
~Click OK.

From the MINITAB output, the t-test statistics is -1.53 .

c)

The p-value for this tes tis,

From the MINITAB output in part (b), the p value for this test is 0.137

4)

Decision

The conclusion is that the p-value in this context is higher than 0.10 which is 0.137, so
the null hypothesis is not rejected at 10% level of significance.

e)

There is insufficient evidence to indicate that the mean amount of real fruit juice in all 32
ounce cans is different from 6.40 ounces. The result is not statistically significant.