SOLUTION: A producer of a juice drink advertises that it contains 10% real fruit juices. A sample of 75 bottles of the drink is analyzed and the percent of real fruit juices is found to be 6
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Question 495768: A producer of a juice drink advertises that it contains 10% real fruit juices. A sample of 75 bottles of the drink is analyzed and the percent of real fruit juices is found to be 6.9%. If the true proportion is actually 0.10, what is the probability that the sample percent will be 6.9% or less?
You can put this solution on YOUR website! A producer of a juice drink advertises that it contains 10% real fruit juices. A sample of 75 bottles of the drink is analyzed and the percent of real fruit juices is found to be 6.9%. If the true proportion is actually 0.10, what is the probability that the sample percent will be 6.9% or less?
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z(0.069) = (0.069-0.10)/sqrt[0.1*0.9/75] = -0.8949
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P(p-hat < 0.069) = P(z < -0.8949) = 0.1854
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Stan H.
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