Does Chemical Treatment Effectively Reduce Contamination Levels in Waterbodies?

[joemama69]

Homework Statement

Binomial Proportion ...

In a laboratory experiment, water samples from different waterbodies were examined to check the levels of contamination and were subsequently treated with chemicals for reducing the contamination levels.

In one waterbody exposed to industrial pollution, 16 out of 69 sample containers developed High Contamination Level [HCL].
In a second waterbody, 20 out of 72 sample containers were affected by HCL.

All the samples from both the waterbodies were treated with chemicals for reduing the HCL. It was subsequently revealed that 12 out of 69 from the first lot and 15 out of 72 from the second lot still retained HCL.

(i) Test H_01 : First waterbody HCL has been reduced after chemical treatment.

Homework Equations

The Attempt at a Solution

P_1 = 16/69 = 0.2319, P_2 = 12/69 = 0.1739, Test P_1 > P_2

P = (x_1 + x_2)/(n_1+n_2) = (16+12)/(69+69) = 0.2029

Z = (P_1 - P_2)/sqrt(P(1-P)(n_1 + n_2)/(n_1*n_2) = (0.2319 - 0.1739)/sqrt(0.2029(1-0.2029)(69+69)/(69*69) = 0.8471,

looking it up on the z-table gives 0.8023,

p-value = 1-.8023 = 0.1977 = about 20% so we accept the null and declair little to no change has been made...

Is it correct to test if the contamination has gone down (P_1>P_2), then we use the upper tail and have use p-value = 1-zvalue

 

[mighty2000]

or should we use the lower tail and use p-value = zvalue?

Thank you for your post. I would like to address your question and provide my thoughts on your solution attempt.

Firstly, it is important to define the null and alternative hypotheses for your experiment. In this case, the null hypothesis (H_0) would be that there is no difference in the proportion of HCL in the first waterbody before and after treatment. The alternative hypothesis (H_1) would be that the proportion of HCL has decreased after treatment.

Next, we need to determine the appropriate test statistic to use. Since we are comparing two proportions, the appropriate test statistic would be the two-sample z-test. This test compares the difference between two proportions to the expected difference under the null hypothesis.

Using your notation, the test statistic can be calculated as follows:

z = (P_1 - P_2) / √[(P(1-P)) / (1/n_1 + 1/n_2)]

Where P is the pooled proportion, calculated as (x_1 + x_2) / (n_1 + n_2). In this case, P = (16 + 12) / (69 + 69) = 0.2029.

Plugging in the values, we get z = (0.2319 - 0.1739) / √[(0.2029(1-0.2029)) / (1/69 + 1/69)] = 1.027.

Next, we need to calculate the p-value associated with this test statistic. Since we are interested in whether the proportion has decreased (P_1 > P_2), we use the upper tail of the standard normal distribution. This gives us a p-value of 0.152, which is greater than the commonly used significance level of 0.05.

Therefore, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the proportion of HCL in the first waterbody has decreased after treatment.

In summary, your approach was generally correct, but there were a few minor errors. It is important to carefully define the null and alternative hypotheses and use the appropriate test statistic and p-value. Additionally, it is important to interpret the results in the context of the experiment and consider any potential limitations or confounding factors.

I hope this helps clarify your understanding