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 3.1 pounds per year with a standard deviation of 1.06 pounds. A a= 0.01 can you reject the claim?

Identify the null hypothesis and alternative hypothesis:

Identify the standardized test statistic
z=?
Find the P-value
P=?
Decide whether to reject or fail to reject of fail the null hypothesis

Answer 1

Solution:

Claim : mean tuna consumption by a person is 3.2 pounds per year

i.e. = 3.2

So the null and alternative hypothesis are

H0 : = 3.2 (null hypo.)

H1 :     3.2

The test statistic z is given by

z =

= (3.1 - 3.2) / (1.06/50)

= -0.67

The test statistic z = -0.67

Now , observe that ,there is   sign in H1. So , the test is two tailed.

p value = P(Z < -z) + P(Z > +z)

= P(Z < -0.67) + P(Z > +0.67)

= 0.2514 + 0.2514

= 0.5028

p value = 0.5028

Since p value is greater than significance level alpha = 0.01 , so

fail to reject the null hypothesis.