Question

(1) As a consequence of fertilizer runoff from heavy farming, nitrogen levels are increasing in local bodies of water. Two towns have taken competing approaches to deal with the problem, and you’ve been hired as a consultant to test whether either town has outperformed the other. (The goal is low N levels.) Here are the data, in the form of x1000mg/m3 .

What can you say?

Town A

average N conc 10.0

sample size 14

standard dev 2.075

Town B

average N conc 15

sample size 11

standard dev 2.449

(2)When lightning strikes, it increases CO2 in the atmosphere. (This can have serious effects on ozone levels.) But what if lightning strikes water? The amount of dissolved CO2 in the water might go down if it is released into the air… or maybe it goes up, due to the formation of nitrogen oxide. You traverse the rural Hudson Valley during thunderstorms and amass a collection of water samples from 25 ponds, collected just before and just after lightning strikes. You find that the average difference in dissolved CO2 is that there is 32ppm more after a strike, with a standard deviation of 70ppm. What do you conclude?

Answer 1

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μ1​ =μ2​

Ha:μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are FL​=0.308 and FU​=3.583, and since F=0.718, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=23. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc​=2.069, for α=0.05 and df=23.

The rejection region for this two-tailed test is R={t:∣t∣>2.069}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=2.462>tc​=2.069, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0218, and since p=0.0218<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1 for CITY A​ is different than μ2 for CITY B​, at the 0.05 significance level.

SOLUTION 2:

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:

Ha:

This corresponds to a left-tailed test, for which a z-test for one population proportion needs to be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is

The rejection region for this left-tailed test is R={t:}

(3) Test Statistics

The t-statistic is computed as follows:

t= dbar/sd/sqrt(n)

t= 32/70/sqrt(25)

t= 32/14

t= 2.29

t statistic greater than t critical therefore REJECT NULL HYPOTHESIS H0.

Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that there is difference , the amount of dissolved CO2 in the water at the α=0.05 significance level.