A nutritionist claims that the mean tuna consumption by a person is 3.9 pounds p

Question

A nutritionist claims that the mean tuna consumption by a person is 3.9 pounds per year. A sample of 60 people shows that the mean tuna consumption by a person is 3.6 pounds per year. Assume the population standard deviation is 1.06 pounds. At =0.07, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. How-3.9 D. Ho :us 3.9 OB. Ho : -3.6 Ha Hs3.6 OE. Ho : s3.6 ° C. Ho : > 3.9 Ha: Hs3.9 Ha : * 3.9 Ha : >3.9 Ha : = 3.6 (b) Identify the standardized test statistic. z-L (Round to two decimal places as needed.)

Explanation / Answer

We have the answer to the question below:

Ho: Mu = 3.6
Ha: Mu !=3.6

n = 60
SampleMean = 3.9
sigma = 1.06

a. A is correct

b. z = (3.9-3.6)/(1.06/sqrt(60)) = 2.192

c. The P-Value is 0.028524.
The result is significant at p < 0.05

d. B is correct. So, we reject Ho. There is sufficient evidence to reject the claim that mean tuna
consumption is equal to 3.9

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.