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STA 210-7521 Spring 17 Homework: Section 7.2 Lecture Video & Homework Score: 2 o

ID: 3180741 • Letter: S

Question


STA 210-7521 Spring 17 Homework: Section 7.2 Lecture Video & Homework Score: 2 of 4 pts. 13 of 14 (13 complete) v HW Score: 84.44%, 38 of 4 7.2.33-T 2 Assigned Media E Question Help A nutritionist claims that the mean tuna consumption by a person is 31 pounds per year A sample of 50 people shows that the mean tuna c by a person is 29Pounds per year Assume the population standard deviation is 1 19 pounds. At aao.06. can you reject the claim? Ha P 31 Ha 73 1 Ha 2.9 D Ho, H 29 Ho u 29 3,1 Ha H-29 3.1 Ha HS 29 (b) identify the standardized test statistic 19 (Round two decimal places as needed Find the P-value [gound to three decma places as needed Enter your answer in the answer box and then cick Check Answer

Explanation / Answer

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 3.1 pounds per year.” Expressed mathematically is µ = 3.1

The order hypothesis is the complement of the claim. The complement of µ = 3.1 is µ 3.1.

The null hypothesis H0 is a statement that contain a statement of equality, such as , =, or .

The alternative hypothesis Ha 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, H0 and Hacan be stated as follows.

H0: µ = 3.1

Ha : µ 3.1

b)z-statistic=(x bar-mu)/(sigma/sqrt(n))

so (2.9-3.1)/(1.19/sqrt(50))=-1.19.

c)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 = -1.19. 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 = -1.19.

The area to the left of z = -1.19 is 0.117

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

P-value = 0.2340

Since P-value>significance value so we fail to reject the null hypothesis.