Question

In: Math

In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of...

In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.

Is fishing better from a boat or from the shore? Pyramid Lake is located on the Paiute Indian Reservation in Nevada. Presidents, movie stars, and people who just want to catch fish go to Pyramid Lake for really large cutthroat trout. Let row B represent hours per fish caught fishing from the shore, and let row A represent hours per fish caught using a boat. The following data are paired by month from October through April.

Oct Nov Dec Jan Feb March April
B: Shore 1.7 1.9 2.0 3.2 3.9 3.6 3.3
A: Boat 1.4 1.5 1.7 2.2 3.3 3.0 3.8

Use a 1% level of significance to test if there is a difference in the population mean hours per fish caught using a boat compared with fishing from the shore. (Let d = BA.)

(a) What is the level of significance?

What is the value of the sample test statistic? (Round your answer to three decimal places.)

____________________________________

In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.

The western United States has a number of four-lane interstate highways that cut through long tracts of wilderness. To prevent car accidents with wild animals, the highways are bordered on both sides with 12-foot-high woven wire fences. Although the fences prevent accidents, they also disturb the winter migration pattern of many animals. To compensate for this disturbance, the highways have frequent wilderness underpasses designed for exclusive use by deer, elk, and other animals. In Colorado, there is a large group of deer that spend their summer months in a region on one side of a highway and survive the winter months in a lower region on the other side. To determine if the highway has disturbed deer migration to the winter feeding area, the following data were gathered on a random sample of 10 wilderness districts in the winter feeding area. Row B represents the average January deer count for a 5-year period before the highway was built, and row A represents the average January deer count for a 5-year period after the highway was built. The highway department claims that the January population has not changed. Test this claim against the claim that the January population has dropped. Use a 5% level of significance. Units used in the table are hundreds of deer. (Let d = BA.)

Wilderness District 1 2 3 4 5 6 7 8 9 10
B: Before highway 10.1 7.4 12.7 5.6 17.4 9.9 20.5 16.2 18.9 11.6
A: After highway 9.1 8.2 10.0 4.1 4.0 7.1 15.2 8.3 12.2 7.3

(a) What is the level of significance?

What is the value of the sample test statistic? (Round your answer to three decimal places.)

Solutions

Expert Solution

Question 1

Here, we have to use paired t test.

Level of significance = α = 0.01

H0: µd = 0 versus Ha: µd > 0

Test statistic for paired t test is given as below:

t = (Dbar - µd)/[Sd/sqrt(n)]

From given data, we have

Dbar = 0.3857

Sd = 0.4598

n = 7

df = n – 1 = 6

t = (0.3857 – 0)/[ 0.4598/sqrt(7)]

t = 2.2194

Test statistic = 2.219

P-value = 0.0341

(by using t-table)

P-value > α = 0.01

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that fishing is better from shore than boat.

Question 2

Here, we have to use paired t test.

Level of significance = α = 0.05

H0: µd = 0 versus Ha: µd > 0

Test statistic for paired t test is given as below:

t = (Dbar - µd)/[Sd/sqrt(n)]

From given data, we have

Dbar = 4.4800

Sd = 4.1063

n = 10

df = n – 1 = 9

t = (4.4800 – 0)/[ 4.1063/sqrt(10)]

t = 3.4501

Test statistic = 3.450

P-value = 0.0036

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the January population has dropped.


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