Ask Question
2 July, 06:30

A nutritionist claims that the mean tuna consumption by a person is 3.5 pounds per year. A sample of 70 people shows that the mean tuna consumption by a person is 3.3 pounds per year. Assume the population standard deviation is 1.22 pounds. At alphaequals0.02 , can you reject the claim?

+2
Answers (1)
  1. 2 July, 06:46
    0
    Let μ = mean tuna consumption (in pounds per year)

    H0: μ = 3.1

    HA: μ ≠ 3.1

    Sample mean = 2.9

    Standard deviation = 0.94

    Standard error of mean = s / √ n

    Standard error of mean = 0.94 / √ 60

    SE = 0.94/7.746

    Standard error of mean 0.1214

    z = (xbar - μ) / SE

    z = (2.9-3.1) / 0.1214

    z = - 1.6481

    p-value = P (z 1.6481) = 2 (0.0495) = 0.099

    Fail to reject null hypothesis since 0.099 >0.08

    The nutritionist's claim that the mean tuna consumption by a person in the U. S is 3.1 pounds per year is not rejected.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A nutritionist claims that the mean tuna consumption by a person is 3.5 pounds per year. A sample of 70 people shows that the mean tuna ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers