Question

Through careful genetic engineering, you have produced what you call the Super Tree – a tree that can absorb extraordinary amounts of carbon dioxide (CO₂), thereby helping to slow the process of global warming. Past research has shown that a regular tree absorbs 48 pounds of (CO₂) per year but you find that in a random sample of 36 of your own trees, the average amount of CO₂ absorbed is 54.3 pounds with a standard deviation of 18. Using an alpha level of .05, test the hypothesis that the average CO₂ absorption in your population of Super Trees is better than the typical tree.

a.What is the null hypothesis for this test?

b.What is the alternative hypothesis for this test?

c.Is this a one-sided or two-sided test? How do you know?

d.Will you use a z-test or a t-test in this situation? Why?

e.What is the critical value of the test statistic?

f.What is the value of the test statistic calculated from the sample results?

g.What decision about the null hypothesis does your test lead to?

h.What does this indicate about the CO₂ absorption of your Super Trees?

i.Would your decision be different if you used an alpha level of .01? Explain

j.Would your decision be different if you used an alpha level of .001? Explain.

Answer 1

A) The null hypothesis, ho: regular tree absorbs 48 pounds of (CO₂) per year. U=48

B) an alternative hypothesis, h1: regular tree doesn't absorb 48 pounds of (CO₂) per year. U=/=48

C) since the direction of the claim is not mentioned, I can see that it is a two-tailed hypothesis

D) The sample size is large (>30), it is sufficient to use a z test

E) critical value, z(a/2)
z(0.05/2)
1.960

F) test statistic, z = (mean-u)/(sd/sqrt(n))
= (54.3-48)/(18/sqrt(36))
2.1000

G) With |z| > z(a/2), I reject the null hypothesis at 5% level of significance.

H) there is NO sufficient evidence to conclude that regulatory absorb 48 pounds of co2 per year. In other words, I can say that regular tree doesn't absorb 48 pounds of (CO₂) per year. U=/=48.

I)

For Alpha is equal to 1%, the critical value is given by:
z(a/2)
z(0.01/2)
2.576

Since |z| < z(a/2), I fail to reject the null hypothesis at 1% level of significance and conclude that regular tree absorbs 48 pounds of (CO₂) per year. U=48.

The results are different at the 1% level of significance as compared to the 5% level of significance.

J)

Alpha level of 0.1%, the critical value is given by
z(a/2)
z(0.001/2)
3.291

In this case also, |Z| < Z(a/2), hence I fail to reject the null hypothesis and 0.1% level of significance and conclude that regulatory absorbs 48 pounds of CO2 per year. U=48

Hence the results are different and 0.1% level of significance as compared to the 5% level of significance.