Question

A nutritionist claims that the mean tuna consumption by a person is 3.3 pounds per year. A sample of 50 people shows that the mean tuna consumption by a person is 2.9 pounds per year. Assume the population standard deviation is 1.08 pounds. At a = 0.07, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. B. Ho:u> 3.3 Ha: u 53.3 A. Ho: u = 3.3 Ha:u#3.3 O D. Ho:u#2.9 Hai u = 2.9 O C. Ho: u> 2.9 Ha:us 2.9 OF. Horus 3.3 H:> 3.3 E. Ho:us 2.9 Ha:u> 2.9 (b) Identify the standardized test statistic. Z= (Round to two decimal places as needed.)

Answer 1

The null and alternative hypothesis :

$\\H_o:\mu = 3.3 \\ H_a:\mu \neq 3.3$

[This is a two tailed test]

Values are given as, mean= 3.3 , n= 50 , x= 2.9 , sd= 1.08

Calculating test -statistic:

$\\z=(x-\mu)/(\sigma/\sqrt{n}) \\\\z=(2.9-3.3)/(1.08/\sqrt{50})$

z= -2.6189

z= -2.62

The test -statistic is : -2.62

P-value = 0.0088 (using z table with z= -2.62 for a two tailed test)

Decision rule: If [p <0.07 ] then reject the null hypothesis .

Conclusion : Yes, we can reject the claim that the mean tuna consumptions by a person is 3.3 pounds per year at $\alpha$ =0.07.