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 3.7 pounds per year. assume the population standard deviation is 1.19 pounds. at \\( \alpha = 0.07 \\), can you reject the claim?
(a) identify the null hypothesis and alternative hypothesis.
\\( \bigcirc \\) a. \\( h _ { 0 } : \mu = 3.9 \\)
\\( h _ { a } : \mu \
eq 3.9 \\)
\\( \bigcirc \\) b. \\( h _ { 0 } : \mu \leq 3.9 \\)
\\( h _ { a } : \mu > 3.9 \\)
\\( \bigcirc \\) c. \\( h _ { 0 } : \mu > 3.7 \\)
\\( h _ { a } : \mu \leq 3.7 \\)
\\( \bigcirc \\) d. \\( h _ { 0 } : \mu \leq 3.7 \\)
\\( h _ { a } : \mu > 3.7 \\)
\\( \bigcirc \\) e. \\( h _ { 0 } : \mu > 3.9 \\)
\\( h _ { a } : \mu \leq 3.9 \\)
\\( \bigcirc \\) f. \\( h _ { 0 } : \mu \
eq 3.7 \\)
\\( h _ { a } : \mu = 3.7 \\)

Explanation:

In hypothesis testing, the null hypothesis \(H_0\) represents the claim being tested. Here, the nutritionist's claim is that the mean tuna consumption \(\mu = 3.9\) pounds per year. The alternative hypothesis \(H_a\) is the opposite of the null hypothesis. Since we are checking if we can reject the claim, the alternative hypothesis is \(\mu
eq3.9\) (a two - tailed test).

Answer:

A. \(H_0:\mu = 3.9\), \(H_a:\mu
eq3.9\)