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I have a data set of 10 sampling points for ethyl-benzene concentration in soils (mg/kg). The data is

vector <- c(14.77, 59.05, 94.91, 157.55, 33.89, 0.00, 0.00, 11.36, 4.35, 0.00)

Can a one sample t test be used to test statistical significance difference of the concentrations from the ethyl-benzene Dutch intervention value of 110 mg/kg? How should the zero concentrations be treated? The 0.00 is for concentrations that were Below Detection Limit (BDL)

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  • $\begingroup$ I would say no - you have 0-inflated values and your sample size of 10 is too small. You can take (number of values smaller than 110, number of values bigger than 110) and make a sign test - if the null hypothesis that the mean is equal to 110 is true, you should have approx equal number of values that exceed the mean and that are below (I guess it is called not sign test but Binomial). $\endgroup$
    – German Demidov
    Commented Apr 10, 2019 at 13:09
  • $\begingroup$ You should not treat the BDL values as zeros. This introduces two biases: first, it (obviously) biases the mean downwards. More subtly, it biases the standard deviation upwards. The amount of bias depends on the reporting limits. See Dennis Helsel's book Nondetections and data analysis. Nevertheless, unless your reporting limits are all substantially greater than 110 mg/Kg, there's little question that these data are from a population with mean less than 110. $\endgroup$
    – whuber ♦
    Commented Apr 10, 2019 at 14:26
  • $\begingroup$ @GermanDemidov I disagree with your points here. One can easily calibrate the T-test when the distribution is small and non-normal, even multimodal (use simulation). It's easy to show the type-1 error rate is conserved and that the test has some power. At any rate, a jackknife could be recommended as an alternative. The important point is coding LLD values as 0 when they are positively valued. $\endgroup$
    – AdamO
    Commented Apr 10, 2019 at 14:44

3 Answers 3

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How the zero concentration should be treated depends on what zero values mean. Is a concentration of zero theoretically possible or does it indicate that something went wrong at that measurement? If the latter is the case I would suggest replacing those values with missings since they show some mistake in the measurement and not the actual concentration.

Assuming zero values make theoretically sense you could use a one sample t test. If you doubt the assumptions of the t test to be met you can alternatively use the one sample sign test. Since your data is really small it is difficult to acually check the assumptions of the t test because tests for normality depend on sample size and I further I would argue that in small sample sizes like yours values easily appear as outliers. Thus I would go with the one sample sign test that uses the median rather than the mean which is a more robust parameter. For this you would need the median concentration of the dutch sample.

If zero is a valid value the R code would be

t.test(vector, mu = 110)
# t(9)= -4.3708, p=.002

library(BSDA)
# assuming that he dutch sample is normally distributed and thus mean= median (or you provide median of dutch data)
SIGN.test(vector, md = 110)
# s= 1, p=.021

EDIT

There has been some discussion and this is why I want to make an update.

It was suggested in the comments that in this case where human and environmental safety is important the mean is to be used and not the median. I would argue that the opposite can be true and this is why. Assume country A has 100 soil-samples with all of them being just right under some cutoff for some risky chemicals in the soil. On the other hand, country B has 99 chemical-free samples while only one soil-sample contains a massive concentration. So in average country B could have a significantly higher mean concentration although only a few people are at risk in this country. But in country A are all people at risk and may all suffer in the long term. Thus to me it is not obvious why the median would be inappropriate here and the mean would be the better measure.

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  • $\begingroup$ Thanks, the zero is for concentrations that were Below Detection Limit (BDL) $\endgroup$
    – Bert Nyarenchi
    Commented Apr 10, 2019 at 14:11
  • $\begingroup$ So in fact the values may be not zero but 2 or 3, for example? At what value is the BDL? $\endgroup$
    – user213325
    Commented Apr 10, 2019 at 14:19
  • $\begingroup$ The BDL is <0.01 $\endgroup$
    – Bert Nyarenchi
    Commented Apr 10, 2019 at 14:24
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    $\begingroup$ The problem with the sign test is that it evaluates the wrong hypothesis. This is more than just a statistical subtlety: typically the concern is about potential risk to human health or the environment and that risk is proportional to the mean concentration, not its median or some other statistic. Thus, the challenge here is to test the null hypothesis that the population mean--not something else--exceeds 110 mg/Kg and to treat the BDL values appropriately. $\endgroup$
    – whuber ♦
    Commented Apr 10, 2019 at 14:37
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    $\begingroup$ Although actually the question doesn't explicitly say that it is about the mean. It only asks whether a t test can be used with no restriction to the mean being the hypothesis. $\endgroup$
    – user213325
    Commented Apr 11, 2019 at 18:07
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I think that the intervention value is a regulatory limit, and it is not obvious what summary statistic should be used unless it is specified in the regulation. For example, in the U.S., for bacteria in water, often the geometric mean is indicated in the regulation. For example in New Jersey, the limit for E. coli for some types of waterbodies is

"E. Coli levels shall not exceed a geometric mean of 126 / 100ml or a single sample maximum of 235 / 100ml."

Without this kind of guidance, I don't think you can assume that the relevant statistic is the mean or the median or some other statistic. In the case of the ethyl-benzene values in the original question, we get a different answer if we use the mean or use the maximum observation.

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There are many proposed methods for handling truncated data. But we need context.

Ethyl-benzene is toxic, flammable, possibly carcinogenic, etc. The idea of using null hypothesis significance testing is questionable here because... what's the burden of proof? You mention a one sided test, but as a null are concentrations above or below a toxic threshold? The costs on a type 1 and type 2 error are different than in many conventional scientific fields.

In either case, imputing 0 is bad. Better to impute LLD with its lowest detectable value as a conservative measure. Don't know it? Read up on the technology. Then you could apply a log transform to the sample, which is an important convention for concentration data. Inspecting a simple density smoother shows that the distributions are much better behaved on a log scale, so the T-test will better operating characteristics.

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  • $\begingroup$ Of course in soil or groundwater, there are traces of petroleum hydrocarbons and the DIV is the recommended acceptable limit for the concentration of these contaminants, beyond which a red flag is raised. The LLD is 0.01. $\endgroup$
    – Bert Nyarenchi
    Commented Apr 10, 2019 at 16:11
  • $\begingroup$ @BertNyarenchi Good to know LLD, and we all understand toxicology and the issues of precise assays. If I were the toxicologist, I wouldn't be concerned with the mean, and consequently sod the T-test. I would report that least one sample that exceeds the DIV. That's significant. As I said: what's the burden of proof? $\endgroup$
    – AdamO
    Commented Apr 10, 2019 at 18:11
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    $\begingroup$ I'm afraid this post gives bad advice. These are censored data, not truncated data. Also, "better to impute LLD with its lowest detectable value" is specifically known to be one of the worst possible procedures. It's not necessarily "conservative," as I pointed out in a comment to the question--especially when applying a log transformation. $\endgroup$
    – whuber ♦
    Commented Apr 11, 2019 at 14:36
  • $\begingroup$ @whuber I think you are talking about estimation and I am talking about inference. Imputing LLD gives a biased estimate of the mean, but as a significance test for the hypothesis "Mean concentrations are below DIV", LLD imputation increases false positive error rate but it increases power. It's justified as a pragmatic solution because the cost of a type 2 error is greater than in most scenarios. $\endgroup$
    – AdamO
    Commented Apr 12, 2019 at 14:29
  • $\begingroup$ Thank you for the explanation. I am concerned about testing, too, because that is what the question is about. $\endgroup$
    – whuber ♦
    Commented Apr 12, 2019 at 16:08

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