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 = - 1.30 \\) (round to two decimal places as needed.)
(c) find the p - value.
(round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for P - value in a two - tailed z - test

For a two - tailed z - test, \(P - value = 2\times(1 - \Phi(|z|))\), where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution and \(z=- 1.30\)

Step2: Calculate \(P - value\)

First, find \(\Phi(1.30)\). Looking up in the standard normal table or using a calculator, \(\Phi(1.30)=0.9032\)
Then \(P - value = 2\times(1 - 0.9032)\)

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Answer:

\(0.194\) (rounded to three decimal places)