Question

A nutritionist claims that the mean tuna consumption by a person is 3.4 pounds per year. A sample of 80 people shows that the mean tuna consumption by a person is 3.1 pounds per year. Assume the population standard deviation is 1.23 pounds. At α= 0.08, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. OA, Ho: 3.4 O D. Ho:H> 3.4 ов. Но:us 3.4 Ha H> 3.4 О Е. НО: #3.1 Ос. нон-3.1 Ha HE3.1 HO: $3.1 HH3.1 На: #3.4 OF, На: $3.4 (b) Identify the standardized test statistic. z(Round to two decimal places as needed.) (c) Find the P-value. Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. 0 A. Fail to reject Ho. There is sufficient evidence to reject the O C. Fail to reject Ho- There is not sufficient evidence to reject the O B. Reject HO. There is sufficient evidence to reject the claim that claim that mean tuna consumption is equal to 3.4 pounds mean tuna consumption is equal to 3.4 pounds. D. Reject Ho. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.4 pounds claim that mean tuna consumption is equal to 3.4 pounds

Answer 1

Given that, sample size ( n ) = 80

sample mean = 3.1 pounds per year

population standard deviation = 1.23 pounds

significance level = 0.08

a) The null and the alternative hypotheses are,

Ho : μ = 3.4

b) The standardized test statistic is,

Z = -2.18

c) p-value = 2 * P(Z < -2.18) = 2 * 0.0146 = 0.0292

p-value = 0.0292

d) Here, p-value = 0.0292 <

So, we reject H₀. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.4 pounds.