Null hypothesis testing question? This is a homework question, and I can't see to figure it out. Any hints or suggested formulas would be nice, I am not really stellar at math, and am pretty much a "plug and chug" type of guy. Suggested readings would be nice too, as long as they are "readable".
The question:

An attempt to fix responsibility for an oil spill is to be based upon
  comparison of the sulfur content in the spillage with specimens taken
  from the suspect vessel. Analysis of five samples from each source by
  an established method for which s -> σ = 0.05 has yielded values of
  0.12% S for the collected oil and 0.16% S for oil from this ship. Is there reason to believe that the two specimens had different origins
  at the 95% confidence level?

My attempt:
I was told by my teacher that if  $s$ -> $σ$, then it is reliable, and if s is reliable, then we should use the z test. But I cant use the z test for I do not have any means given for the data samples, and the formula for the z test requires it right? 
$$ z =  X1 - X2 \div{( σ * \sqrt{(N1 + N2 ) \div{ (N1*N2) }})}$$
Where $X1$ = first mean, $X2$ = second mean,  $σ$ = populaton standard deviation?, $N1$ = number of samples from first mean, $N2$ = number of samples from second mean, $z$ = the value you compare to tcrit.
How would I go about solving?
aside: I just had an idea while I was typing this, could I have determined the means by rearranging the formula some how?
In lieu of the comment, I tried to follow the advice on the website:
1) State hypotheses (Ho and Ha)
Ha : P1 !=  P2
Ho : P1 = P2
since these hypothesis correspond to a two tailed test, we will use a two proportioned z test, ( I would of used t test, but I don't have the mean values given to me, so how can I???). Significance level is 0.05.
I calculated pooled sample proportion (p) and the standard error (SE)
pooled sample proportion = $$(p)  =    \dfrac {p_1 * p_1 + p_2 * p_2}{n_1 + n_2}  $$
$$(p) =    \dfrac {0.12 * 5 + 0.16 * 5}{5 + 5}  $$ 
$$ p = 0.14 $$
standard error = $$(SE) =  \sqrt{(p * (1-p) * [\dfrac {1}{N_1}  + \dfrac {1}{N_2}])}  $$
$$(SE) =  \sqrt{(0.14 * (1-0.14) * [\dfrac {1}{5}  + \dfrac {1}{5}])}  $$
$$(SE) =  \sqrt{(0.14 * (1-0.14) * [\dfrac {1}{5}  + \dfrac {1}{5}])}  $$
$(SE) =  0.219453868$ which rounds to $0.22$
Again I can't use the tscore forumla since I lack the mean values, so I guess I use the z score formula?
$$ z =  \dfrac{(p_1 - p_2)} {SE} $$
Where $p_1$ =  sample proportion in sample 1.
And $p_2$ is the sample proportion in sample 2.
$$ z =  \dfrac{(0.12 - 0.16)} {0.219453868} $$
$$ z = -0.182270654 $$
I used the Normal z table to calculate the p value. I provided an image of it below from this pdf.  

At the value closest to my z score, $-0.2$, at a $0.05$ significance level I saw the value $0.40129$ on the z table.
Therefore since $0.40129 > 0.05$ we do not reject the $H_o$? We cannot conclude that the two specimens are of the same origin.
EDIT: I realized my mistake, the values 0.12 and 0.16 were the means of the Sulfur measurement, sulfur standing for S. I flagged my post for deletion. 
 A: You do have means for the data samples. The values given

0.12% S for the collected oil and 0.16% S for oil from this ship

are chemical measurements (as noted by @whuber) rather than percentages of the type you seem to be manipulating in your question. They might, say, have been reported by a machine that determines the number of grams of sulfur contained in a sample of 100 grams of oil. Thus they represent the mean values of the 5 determinations of each of the collected oil and ship oil values. If the value of $\sigma$ is also specified in the same units (% sulfur) then you have a simple problem of comparing 2 means based on 5 determinations each with a known standard deviation, if I interpret s -> $\sigma$ correctly.
A: If the sample size is less that or equal to 30, we go for t-test. So here, sample size is less than 30. There's one more criteria to select the correct test which is if variance is know then we go for normal test (i.e. z test), if not, we go for t-test. I this case, considering both conditions, we'll go for t-test.
