In 2024, a team of researchers published a study of contaminants in farmed salmon. Random samples of fish from many farms were collected and analyzed for 14 organic contaminants. One of those contaminants is the insecticide mirex which is a carcinogen. One farm in particular had very high levels of mirex. Summary statistics from R for the mirex concentrations in parts per million (ppm) at this farm are listed below. What test should be done to determine if the mirex levels are below the expected value of 0.005 ppm.

𝑋
¯
=
0.019

𝑠
=
0.004

𝑛
=
14

Question

In 2024, a team of researchers published a study of contaminants in farmed salmon. Random samples of fish from many farms were collected and analyzed for 14 organic contaminants. One of those contaminants is the insecticide mirex which is a carcinogen. One farm in particular had very high levels of mirex. Summary statistics from R for the mirex concentrations in parts per million (ppm) at this farm are listed below. What test should be done to determine if the mirex levels are below the expected value of 0.005 ppm. 𝑋 ¯ = 0.019 𝑠 = 0.004 𝑛 = 14单项选择题

A

z test on the proportion

B

binomial condition cannot be met; use some other test not listed here

C

t test on the mean

D

t test on the difference of means

E

nonparametric test because normality cannot be satisfied

𝑋
¯
=
0.019

𝑠
=
0.004

𝑛
=
14

BlackTom AI · Accepted Answer

Because the population standard deviation is unknown and the sample size is small (n = 14), a one‑sample t test for the mean should be used.

1) Relevant knowledge points and concepts
- A one‑sample t test compares the sample mean to a hypothesized population mean when sigma is unknown and n is small (typically n < 30).  
- Normality of the underlying variable is required (or approximately satisfied); a Q–Q plot here shows points close to the line, supporting approximate normality.  
- Other listed tests (z on a proportion, binomial, two‑sample t, nonparametric) apply to different data types or scenarios.

2) Complete reasoning process step by step
- Hypothesis form for checking "below 0.005 ppm": H0: μ = 0.005 vs Ha: μ < 0.005.  
- Use t statistic: t = (x̄ − μ0) / (s / √n).

3) Calculations with intermediate steps
- x̄ = 0.019, μ0 = 0.005, s = 0.004, n = 14, √n = √14 ≈ 3.741657.  
- Standard error = s / √n = 0.004 / 3.741657 ≈ 0.001069.  
- t = (0.019 − 0.005) / 0.001069 = 0.014 / 0.001069 ≈ 13.10.  
- Degrees of freedom = n − 1 = 13. (This huge positive t indicates the sample mean is far above 0.005, so the data provide strong evidence μ > 0.005 rather than μ < 0.005.)

4) Why the correct answer is correct
- The t test on the mean is the appropriate procedure because we compare a single sample mean to a known value with unknown population standard deviation and a small sample, and the normality check is satisfied.

5) Why other options are incorrect
- z test on the proportion: wrong data type (continuous concentration, not a proportion).  
- Binomial condition: irrelevant (binomial applies to counts of successes).  
- t test on the difference of means: requires two independent samples, not applicable.  
- Nonparametric test: unnecessary here because the Q–Q plot supports approximate normality and the t test is valid.