Question

a nutritionist claims that the mean tuna consumption by a person is 3.7 pounds per year. a sample of 80 people shows that the mean tuna consumption by a person is 3.4 pounds per year. assume the population standard deviation is 1.04 pounds. at $alpha = 0.08$, can you reject the claim?
(a) identify the null hypothesis and alternative hypothesis.
a. $h_0:muleq3.4$ $h_a:mu > 3.4$ b. $h_0:mu = 3.7$ $h_a:mu
eq3.7$ c. $h_0:mu > 3.4$ $h_a:muleq3.4$ d. $h_0:muleq3.7$ $h_a:mu > 3.7$ e. $h_0:mu
eq3.4$ $h_a:mu = 3.4$ f. $h_0:mu > 3.7$ $h_a:muleq3.7$
(b) identify the standardized test statistic $z=square$ (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the z - test statistic

The formula for the z - test statistic when the population standard deviation \(\sigma\) is known 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: Identify the values of \(\bar{x}\), \(\mu\), \(\sigma\), and \(n\)

From the problem, \(\bar{x} = 3.4\) (sample mean), \(\mu=3.7\) (population mean under the null hypothesis), \(\sigma = 1.04\) (population standard deviation), and \(n = 80\) (sample size).

Step3: Substitute the values into the formula

$$
LATEXBLOCK0
$$

Answer:

\(z\approx - 2.59\)