# Problem 407

A nutritionist claim that  the mean tuna consumption by a person is 3.7 pounds per year. A sample of 75 people shown that the mean tuna consumption by a person is 3.4 pounds per year with a standard deviation of 1.14 pounds. At α = 0.07, can you reject the claim?

(a)    Identify the null hypothesis and alternative hypothesis.

(b)   Identify the standardized test statistic.

(c)    Find the P-value.

(d)   Decide whether to reject or fail to reject  the null hypothesis.

Solution

(a)    Identify the null hypothesis and alternative hypothesis.

To find the null and alternative hypothesis, start by writing the claim mathematically. The claim, “the mean tuna consumption by a  person is 3.7 pounds per year.” Expressed mathematically is µ = 3.7.

The order hypothesis is the complement of the claim. The complement of µ = 3.7 is µ ≠3.7.

The null hypothesis , , is a statement that contain a statement of equality, such as ≤, =, or ≥. The alternative hypothesis. , is the complement of the null hypothesis. It is a statement that must be true if is false, and it contain a statement of inequality, such as < ,  ≠ , or >. Thus, and can be stated as follows.

: µ = 3.7

: µ ≠ 3.7

(b)   Identify the standardized test statistic.

The standardized test statistic z is given by the formula

Solve for z, rounding to two decimal places.

z = \frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}

=\frac{3.4 -3.7 }{\frac{1.14 }{\sqrt{75}}}

= -2.28

(c)    Find the P-value.

In order to find the P-value, first determine if the hypothesis test is left-tailed, right-tailed, or two-tailed. The test is two-tailed.

You can either use a standard normal distribution table or technology to find the p-value. For this explanation, a standard normal distribution table is used.

Since the test is two-tailed, the P-value is equal to twice the are in the tail of the standardized test statistic z = -2.28. Because z is negative, the tail is on the left. Use a standard normal distribution table to find the area to the left of z = -2.28.

The area to the left of z = -2.28 is 0.0113.

Now double the area in the left tail to find the P-value, rounding to three decimal places.

P-value = 0.023

(d)   Decide whether to reject or fail to reject  the null hypothesis.

To determine whether to reject or fail to reject the null hypothesis, compare the P-value to the stated α value of 0.07. The P-value is less than α.

If the P-value is less then or equal to the level of significance, α, reject the null hypothesis, which is the claim that mean tuna consumption is equal to 3.7 pounds. If the P-value is greater than the level of significance, α, fail to reject the null hypothesis mean that tuna consumption is equal to 3.7 pounds.