Question

Arsenic is a known carcinogen and poison. The standard laboratory procedures for measuring arsenic \(concentration (\mu g/L)\) in water are expensive. Consider the accompanying summary data and Minitab output for comparing a laboratory method to a new relatively quick and inexpensive field method (from the article "Evaluation of a New Field Measurement Method for Arsenic in Drinking Water Samples," J. of Emvir. Engr., 2008: 382-388).

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean

1 3 19.70 1.10 0.65

2 3 10.90 0.60 0.35

Estimate for difference: 8.800

95% CI for difference: 6.498,11.102

T -Test of difference =0 (vs not = ):

T -Value =12.1 P -Value =0.001 DF=3

What conclusion do you draw about the two methods, and why? Interpret the given confidence interval. (Note: One of the article's authors indicated in private communication that they were unsure why the two methods disagreed.)

Short answer

Both methods lead to the same conclusion: There is sufficient evidence to support the claim that the population means are not equal.

Step-by-step solution

To find the  accompanying summary data and Minitab output for comparing a laboratory method to a new relatively quick and inexpensive field method

The difference of the two means was tested here using two-sample t test. The given output suggests that the two methods differ. The hypotheses tested here are \({H_0}:{\mu _1} - {\mu _2} = 0 {\rm{versus}} {H_0}:{\mu _1} - {\mu _2} \ne 0 ,{\rm{where}} {\mu _1},{\mu _2}\) are the true averages of method 1 and 2 , respectively.

The conclusion to reject null hypothesis comes from the fact that

\(P = 0.001 < 0.01 = \alpha \) or any other significance level.

The given confidence interval does not include zero, thus the difference to be zero is highly unlikely.

The methods differ; Confidence interval indicates that it is highly unlikely for difference to be zero.

Explanation B:Step 1: To find the  accompanying summary data and Minitab output for comparing a laboratory method to a new relatively quick and inexpensive field method

TWO-SAMPLE T-TEST

Given in the output:

\(\begin{array}{c}{H_0}:{\mu _1} - {\mu _2} = 0\\{H_0}:{\mu _1} - {\mu _2} \ne 0\\ t = 12.16\\P = 0.001\\ df = 3\end{array}\)

Let us assume:\(\alpha = 0.05.\)

If the P-value is less than the significance level, reject the null hypothesis. \(P < 0.05 \Rightarrow Reject {H_0}\)

There is sufficient evidence to support the claim that the population means are not equal.

To find the accompanying summary data and Minitab output for comparing a laboratory method to a new relatively quick and inexpensive field method

Two sample confidence interval:

Given in the output:\((6.498,11.102)\)

The confidence interval does not contain 0, which indicates that the difference of the population means is not zero and thus the population means appear to be not equal.

We can then conclude: There is sufficient evidence to support the claim that the population means are not equal.

We then note that both methods lead to the same conclusion.