Question

A new flame atomic absorption spectroscopic method of determining antimony in the atmosphere was compared with the recommended calorimetric method. For samples from an urban atmosphere, the following results were obtained.

Sample No.	Antimony found (mg m-3)
	New Method	Old Method
1 2 3 4 5 6	22.2 19.2 15.7 20.4 19.6 15.7	25.0 19.5 16.6 21.3 20.7 16.8

Do the results obtained by the two methods differ significantly?

Clear answer required please, showing which tests used

Answer 1

In order to determine if the two methods are significantly differing wrt antimony's presence, we construct our null and alternative hypotheses as H0: mu1 = mu2 vs Ha: mu1 not equal to mu2 where mu1 and mu2 are the unknown means of the presence of antimony in the population of two methods.

The test statistic used for this test is T= (x1bar-x2bar)/sqrt((s1*s1/n1)+(s2*s2/n2)) where x1bar, x2bar are the sample means, s1,s2 are the sample standard deviations and n1,n2 are the sample sizes. sqrt refers to the square root function.

We reject H0 if |T(observed)| > t(alpha/2,v) where t(alpha/2,v) is the upper alpha/2 point of a Student's t distribution with "v" degrees of freedom. Alpha is the level of significance.

v = (((s1*s1/n1)+(s2*s2/n2))^2) / (((s1*s1/n1)^2)/(n1-1))+(((s2*s2/n2)^2)/(n2-1))

We see that x1bar = 18.8

x2bar = 19.98333

s1 = 2.613044

s2 = 3.137781

n1 = n2 = 6

So, here T(observed) = -0.7098509 and t(alpha/2,v) = 2.238071. (Obtained from the probability tables of Student's t distribution).

Thus we see that |T(observed)| < t(alpha/2,v). Hence we fail to reject H0 and conclude on the basis of the given sample at a 5% level of significance that there is not enough evidence to say that the two methods significantly differ in the terms of presence of antimony