Question

A soil scientist collected random soil samples from a field and measured the dissolved organic carbon (mg/L) level of each sample. The measurements are recorded in DissolvedOrganicCarbon.xlsx (Links to an external site.)Links to an external site.. The researcher would like to test if the mean dissolved organic carbon level in the field is less than 14 mg/L. Use a significance level of 0.05.

Part 1: Which hypothesis test is appropriate for this situation?  

The difference of means, using independent samples.   

The mean of differences, using dependent samples.   

One mean sigma known

What is the test statistic for the test mentioned above? (Show step by step)

What is the p-value for your test above? (Show step by step)

What is your conclusion based on the p-value above?

What is the 20th percentile of carbon level?

Dissolved Organic Carbon (mg/L)
14.367
11.958
12.038
12.875
14.690
10.297
9.901
11.325
12.105
16.519
6.185
24.275
20.792
15.197
9.879
18.820
13.296
9.845
10.906
8.587

Answer 1

1. It will be one sample mean test with sigma unknown(because sigma is not given). Hence it's a t-test.

2. The test statistic for the test will be   ,

where      sample size,      the claimed poputation mean in null hypothesis, i.e. 14 mg/L,      sample mean,      sample standard deviation.

Hence the calculated value of the test statistic is  

3. Here we have to test against   .

Under null hypothesis    and the test s left-tailed test.

So the p-value of the test is  

4. Here the p-value of the test is greater than the significance level 0.05. Hence My conclusion based on the sample information is to reject null hypothesis, i.e. dissolved organic carbon level in the field is less than 14 mg/L.

5. The 20th percentile of carbon level is

-0.8609506.