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.DATA ( n
= 30 )
Real Juice
| ounces |
| 6.30 |
| 6.53 |
| 6.54 |
| 6.37 |
| 6.25 |
| 6.00 |
| 6.27 |
| 6.04 |
| 6.09 |
| 6.03 |
| 6.49 |
| 6.00 |
| 6.49 |
| 5.99 |
| 6.63 |
| 6.32 |
| 6.52 |
| 6.44 |
| 5.95 |
| 6.10 |
| 6.21 |
| 6.25 |
| 5.78 |
| 6.57 |
| 5.70 |
| 6.69 |
| 6.44 |
| 6.74 |
| 6.30 |
| 6.24 |
(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 = |
The statistical software output for this problem is :

Test statistics = -2.56
P-value = 0.0160
A 32 ounce can of a popular fruit drink claims to contain 20% real fruit juice....
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 56 such cans of this fruit drink. Of these, the mean volume of fruit juice is 6.32 with standard deviation of 0.23. Test the claim that the mean amount of real fruit juice in all...
A soft-drink manufacturer claims that its 12-ounce cans do not contain, on average, more than 30 calories. A random sample of 68 cans of this soft drink, which were checked for calories, contained a mean of 32 calories with a standard deviation of 3 calories. Does the sample information support the alternative hypothesis that the manufacturer's claim is false? Use a significance level of 5%. Find the range for the p-value for this test. What will your conclusion be using...