Question

Two water samples were analyzed for their copper content by atomic absorption spectroscopy. After 5 replicate analyses, the first water sample was found to have 12.6± 0.9 mg/L copper. After 6 replicate analyses, the second water sample was found to have 13.1± 0.6 mg/L copper. Is the copper content in these water samples significantly different? Report your answer in a full sentence, including all necessary information to explain how confident you are about your answer.

Answer: Since the calculated t value is less than the tabulated value (95%, 9 degrees of freedom, tcalc = 0.6, ttable = 2.3) , the two copper concentrations are not statistically different to 95% confidence.

Can you show me how he got to that answer?

Answer 1

t-value calculations :

for Method comparison :

Cu in water sample by AAS:

wo water samples were analyzed for their copper content by atomic absorption spectroscopy. After 5 replicate analyses, the first water sample was found to have 12.6± 0.9 mg/L copper. After 6 replicate analyses, the second water sample was found to have 13.1± 0.6 mg/L copper .

here results, are reported as : mean ± standard deviation

s1 = 0.9 mg/L ; s2 = 0.6 mg/L

first calculate the pooled standard deviation, s :

where s₁ and s₂ are the standard deviations of the two methods with sample sizes n₁ = 5 and n₂ = 6.

The standard error of the difference between the two means is calculated as:

The significance level, is calculated using the t-test, with the value t calculated as:

Cu in water : df = n₁ + n₂ − 2 = 9

s = 0.748; S.E. = 0.453

difference of means = 0.500

t _calc (value) city water : 1.103

since calculated t-value is less than value at 95 % confidence level : 2.26

If the absolute value of the t-value is greater than the critical value, you reject the null hypothesis. If the absolute value of the t-value is less than the critical value, you fail to reject the null hypothesis.

the conclusion is that the two means are not significantly different.