# 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.

• There is no need to delete your post. Our goal here is to build a permanent repository of high quality statistical information in the form of questions & answers. You get your question answered along the way, but that is almost a side effect. The fact that this has been resolved very much does not mean it should be deleted. On the other hand, if you have figured this out, you can answer your own question & mark that as accepted. – gung Sep 21 '15 at 22:57

## 2 Answers

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.

• Sounds right. I'm also wondering about the $s->\sigma$ terminology. – Antoni Parellada Sep 22 '15 at 1:58

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.

• But with either t or z test, I still can do neither, since I need the means from both samples right? – Ro Siv Sep 20 '15 at 14:26
• This is a test of proportions. Here's a link which may help you. stattrek.com/m/hypothesis-test/difference-in-proportions.aspx – user89929 Sep 20 '15 at 15:58
• I added more work, if you could check it i would appreciate it. Thank you. – Ro Siv Sep 20 '15 at 18:48
• This advice about sample size, although common, is poor. Sample size is not dispositive and should not be the sole determinant of choice of test. Moreover, the information in the question is not sufficient to perform a t-test anyway. The whole point is to conduct a z-test using an assumed standard deviation $\sigma$. Finally, this is not a "test of proportions" as claimed in a comment. The values of $0.12\%$ and $0.16\%$ are means of measurements of chemical concentrations. – whuber Sep 21 '15 at 14:42