Question

A nutritionist claims that the mean tuna consumption by a person is 35 pounds per year A sample of 90 people shows that the mean tuna consumption by a person is 3.2 pounds per year. Assume the population standard deviation is 103 pounds. At a = 0.03, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. O A. Horus 3.5 Ha> 3.5 OD. Hou732 Hau=32 OB. Ho 33.5 H, HS35 O E. Ho u$32 Ha H> 32 OC. H. 4-3.5 Ha #3.5 OF. H, > 32 H, 53.2 (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. O A Reject Ho. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.5 pounds O C. Fail to reject H, There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.5 pounds O B. Reject Hg There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.5 pounds. OD. Fail to reject He There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.5 pounds

Answer 1

Solution :

$\mu$= 3.5

= 3.2

T

$\sigma$ = 1.03

n = 90

This is the two tailed test .

a) The null and alternative hypothesis is

c) H₀ :  $\mu$ = 3.5

H_a : $\mu$ $\neq$ 3.5

b) Test statistic = z

= (
T
- $\mu$ ) / $\sigma$ / $\sqrt$ n

= (3.2-3.5) / 1.03 / $\sqrt$ 90

= -2.76

P (Z < -2.76 ) = 0.0058

c) P-value = 0.006

$\alpha$ = 0.03

0.006 < 0.03

d)A. Reject H_O. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.5pounds.