Question

They main goal is to find either a Z score or T score for the data below

What is the population mean and the sample mean for the elevations (in feet) of the trails below:

Mount Chocorua via Liberty Trail: 2,502 feet of elevation gain

Welch-Dickey Loop: 1,807 feet of elevation gain

Lonesome Lake Trail: 1,040 feet of elevation gain

Mount Willard: 985 feet of elevation gain

Red Hill Fire Tower: 1,350 feet of elevation gain

Pack Monadnock: 840 feet of elevation gain

Mount Cardigan’s Holt Trail: 1,800 feet of elevation gain

Mount Washington via Tuckerman Ravine: 4,238 feet of elevation gain

Presidential Traverse: 4,989 feet of elevation gain

Mount Moosilauke: 2,342 feet of elevation gain

The Carters: 3,305 feet of elevation gain

Mount Carrigain via Signal Ridge: 3,257 feet of elevation gain

Mount Flume + Mount Liberty Loop: 3,099 feet of elevation gain

Mount Isolation via glen boulder trail: 4,931 feet of elevation gain

Mount Monroe Trail: 2,572 feet of elevation gain

Maine

Hunt and Helon Taylor trail: 8,021 feet of elevation gain

Katahdin Loop Trail: 3,894 feet of elevation gain

Abol Trail: 3,950 feet of elevation gain

Hunt Trail: 4,169 feet of elevation gain

Mount Katahdin and Hamlin peak Trail: 4,438 feet of elevation gain

Baxter Peak Via Saddle Trail: 3,832 feet of elevation gain

Knife Edge Trail: 3,987 feet of elevation gain

Dudley Trail: 5,360 feet of elevation gain

Chimney pond Trail: 1,463 feet of elevation

Katahdin North Loop Trail: 4,061 feet of elevation gain

Doubletop Mountain Trail: 4,704 feet of elevation gain

Big Spencer Mountain Trail: 1,820 feet of elevation gain

North traveler Mountain Trail: 3,694 feet of elevation gain

Big Moose Mountain Trail: 1,843 feet of elevation gain

Cranberry Peak Trail: 2,070 feet of elevation gain

I was to choose 30 hiking trails (15 from New Hampshire and 15 from Maine) and record their elevations. My hypothesis for this is I believe that the mean is greater then or equal to 2,500ft. I'm having trouble figuring out my population mean and my sample mean. Also I need to find out my z score or t score and show a graph showing whether its left or right tailed or both.

Answer 1

Let be the true mean elevation of all the hiking trails in New Hamshire and Maine. (Here we are supposing that the poplation is all the trails in New Hamshire and Maine. you could also suppose that your population is all the trails in the USA/Eastern USA/Appalachian Trails etc, if you think that this sample of 30 trails that you have chosen is a representaitve of this population. So the sample of 30 is not the population, but it represents the population, which you need to define)

Your hypothesis is that this true mean elevation of all the hiking trails in New Hamshire and Maine is greater than or equal to 2500 ft, that is we want to test if (equal to is dropped)

The hypotheses are

We have the following sample information

n=30 is the sample size

the sample mean is

Sample standard deviation is

We do not know the population variance/standard deviation. We will estimate the sample standard deviation to estimate the population SD

Next we want the standard error of mean

The hypothesized value of the population mean is

If the sample size is greater than or equal to 30 or the population standard deviation is known, we use z-score for testing the hypotheses.

Since the sample size is 30 and althoguh we do not know the population standard deviation, we will use z test.

The test statistics is

Now the test is a right tailed test (the alternative hypothesis is >2500)

The critical value corresponding to alpha =0.05 is

From the standard normal tables we get for z=1.64, P(Z<1.64) = 0.5+0.4495=0.95

The critical value of z is 1.64

The acceptance region is shown below

We can see that the sample statistics 2.424 is greater than 1.699 and hence falls in the rejection region. We reject the null hypothesis.

We conclude that there is sufficient evidence to support the claim that the true mean elevation of all the hiking trails in New Hamshire and Maine is greater than 2500 ft.