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 6.32 with standard
deviation of 0.23. 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 0.05significance level.

(a) What type of test is this?

This is a right-tailed test.This is a two-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 *H*_{0}fail to reject
*H*_{0}

(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) it is a two tailed hypothesis

because we have to test whether the mean weight is 6.4 ounces or not.

(B) Using TI 84 calculator

press stat then tests then TTest

enter the data

press calculate

test statistic = -2.31

(c) p value = 0.0258

(D) Reject Ho because p value is less than significance level

(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.

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
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A 32 ounce can of a popular fruit drink claims to contain 20%
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Real Fruit Juice (Raw Data, Software
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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 6.4ounces 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...

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...

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...

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