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.
a. \\( h _ { 0 } : \mu = 3.9 \\)
\\( h _ { a } : \mu \
eq 3.9 \\)
b. \\( h _ { 0 } : \mu \leq 3.9 \\)
\\( h _ { a } : \mu > 3.9 \\)
c. \\( h _ { 0 } : \mu > 3.7 \\)
\\( h _ { a } : \mu \leq 3.7 \\)
d. \\( h _ { 0 } : \mu \leq 3.7 \\)
\\( h _ { a } : \mu > 3.7 \\)
e. \\( h _ { 0 } : \mu > 3.9 \\)
\\( h _ { a } : \mu \leq 3.9 \\)
f. \\( h _ { 0 } : \mu \
eq 3.7 \\)
\\( h _ { a } : \mu = 3.7 \\)
(b) identify the standardized test statistic.
\\( z = \square \\) (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for z - statistic

The formula for the z - statistic in a one - sample z - test is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

Step2: Substitute the given values into the formula

We are given that \(\bar{x} = 3.7\), \(\mu=3.9\), \(\sigma = 1.19\), and \(n = 60\).

$$
LATEXBLOCK0
$$

Answer:

\(z=-1.29\)