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 α = 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.
0.194 (round to three decimal places as needed.)
(d) decide whether to reject or fail to reject the null hypothesis.
a. fail to reject ( h_0 ). there is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.9 pounds.
b. reject ( h_0 ). there is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.9 pounds.
c. reject ( h_0 ). there is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.9 pounds.
d. fail to reject ( h_0 ). there is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.9 pounds.

Explanation:

Step1: Compare P - value and significance level

We know that the significance level \(\alpha = 0.07\) and the P - value \(P=0.194\).

Step2: Make a decision

Since \(P = 0.194>\alpha=0.07\), we fail to reject the null hypothesis.

Answer:

D. Fail to reject \(H_{0}\). There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.9 pounds.