Question

A nutritionist claims that the mean tuna consumption by a person is 3.2 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 1.01 pounds. At alphaequals0.08​, can you reject the​ claim?

​(a) Identify the null hypothesis and alternative hypothesis.

A. Upper H 0​: mugreater than2.9 Upper H Subscript a​: muless than or equals2.9

B. Upper H 0​: mugreater than3.2 Upper H Subscript a​: muless than or equals3.2

C. Upper H 0​: muequals3.2 Upper H Subscript a​: munot equals3.2

D. Upper H 0​: muless than or equals2.9 Upper H Subscript a​: mugreater than2.9

E. Upper H 0​: munot equals2.9 Upper H Subscript a​: muequals2.9

F. Upper H 0​: muless than or equals3.2 Upper H Subscript a​: mugreater than3.2 ​

(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.

A. Reject Upper H 0. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.2 pounds.

B. Fail to reject Upper H 0. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.2 pounds.

C. Fail to reject Upper H 0. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.2 pounds.

D. Reject Upper H 0. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.2 pounds.

Answer 1

a)

Since, the null hypothesis is of no difference between the population parameter and some hypotheised value, therefore, the null hypothesis is:

And the alternative hypothesis is:

Therefore, option C) is correct.

b)

The sample size is n=50.

The sample mean is .

Assume the population standard deviation is .

Under null hypothesis, the formula of the standardized test statistic is:

Therefore, it is obtained as:

Therefore, the standardized test statistic is -2.10.

c)

The P-value can be obtained as:

Therefore, the P-value is 0.018.

d)

The level of significance is given as .

Since, the P-value is less than the level of significance, therefore, Reject Upper H0. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.2 pounds.

Therefore, option A) is correct.