# I have to write a abstract for my paper I was wondering if it was better to write it at the beginning or the end of the lab report.

If I was to eliminate 1 value as an outlier would I go and include that in calculations of the RSD? I would assume not since It would affect the RSD

And how would I decide on a confidence level for the Q test? How would I know if it was 90/95/or 99% conficence? It's not explained how would you choose

this.

I would write the abstract, introduction, introductory sections, summary of results etc. at the end. You should start to write the main body of the paper first.

I would remove the outlier and then compute the standard deviation. The choice of the confidence interval is subjective. You want to try to minimize the effects of getting false positives and retaining outliers. But to do that you must know the probability of getting outliers.

The best way to test the methodology is to do a Monte Carlo simulation. You just generate artificial data using a random generator. Then you use the method to analyze the data, including the procedures you use to remove outliers. This does mean that you have to generate the outliers the correct probability...

You can then see how well your method is performing and you can optimize the procedure for removing outliers.

An abstract isn't the body of the paper..

And I don't have a random generator just lying around...unfortunately

Abstracts are usually put at the beginning. Then comes the introduction, introductory sections, the main body of the paper and you conclude with the conclusion, appendices and references.

But this is not the order at which you should write...

From practical experience, this is what I would do.

Most chemistry scientific papers I see (those journals that are refereed) have the abstract first followed by the full text of the article. So for your set up you would write the abstract and follow it with the lab report.

If you intend to eliminate an outlier, you do that before calculating the RSD as you have suggested.

As for the Q test, many make the mistake of performing the test, determining the value of Q_{exp}, then looking in the table of confidence limits to see if a value can be rejected at the 90, 95, or 99% confidence level. Then they choose the level that fits. But that isn't the way to do it. Choose the confidence level FIRST, then go through the steps described in your text. My recommendation is to use the 95% confidence level. Most chemistry data I have seen use the 95% level as a "routine" level with 99% or 99.9% being used for VERY PRECISE calculations. Thus you will determine, at the 95% confidence interval, if the suspect outlier is or is not an outlier, and you will be 95% certain that it is (or it isn't). Finally, please remember that the Q test is not a good test for removing suspected outliers unless you have 40 or 50 values in the set of data. I think ALL of the statistical tests for removing suspected outliers fall into this category. I understand the t test is somewhat better for small numbers of values in the set of data but I am not a statistician. From my own personal experience, I always used the t test. But here is a quote from the 6th edition of your text; "The blind application of statistical tests to retain or reject a suspect measurement in a samll set of data is not likely to be much more fruitful than an arbitrary dedcision. The apliation of good judgement based on broad experience with an analytical method is usually a sounder approach. In the end, the only valid reason for rejecting a result from a small set of data is the sure inowledge that a mistake was made in the mesurment process. Without this knowledge, a *cautions approach to rejection of an outlier is wise.* The italics WERE NOT added; that's part of the quote.

By the way, I may have been a little hasty in my last response to you concerning precision. It is MUCH more difficult to get great precision for samples containing small amounts of a material. For example, suppose we determine the amount of copper as 1.0 ng, 1.1 ng, and 0.90 ng respectively in three separate determinations. The mean is 1.0 ng and the precision from the mean is (0.1/1)*100 = +/- 10%. That is pretty good work in my book. But if we had 50.0%, 50.1%, and 49.9%, the mean is 50.0% and the precision from the mean is (0.1/50.0)*100 = +/- 0.2% or 2 ppt. The point here is that the same 0.1 absolute difference makes a huge difference when compared with the value. Small values make the difference look bad. But looking again at the 1 ng, if we have determined one as 0.000000001 g and the other as 0.0000000011 g, and the third as 0.00000000090 g, is the agreement really all that bad (10%)? I don't think so. So your values were about 0.6% or so; that gives you a little more leeway in estimating how your numbers agreed with one another. I hope this helps.

It would make more sense if it was at the END right?

No. Abstracts go at the beginning. BUT you write the entire paper, THEN you write the abstract. The purpost of an abstract is to give the reader a quick take of what the article (lab report) is all about. In a journal, we read the abstract to determine if we want to read the entire article to get the details.

For the test to eliminate an outlier the thing is IF I use common sense judgement that wouldn't suffice for a lab report where you need an actual thing to base your elimination of a value on, but if I'm just handing in a value like for the unknown that would be a different thing.

Qexp is the same as Q right?

In my text they say Q> Qcrit=reject

I just didn't know which test to use b/c it wasn't specified or I guess we're expected to know which test to use off the top of our head but...since you said the t test

I would appreciate it if you looked at the calculations and see if I got it correct since I'm just reading about this t test right now (we skipped over this in class)

wine % = 0.6402, 0.5814,& 0.6008

Ok...this is sad but I don't get the text...getting a brain-ache

Uo=? how do they get that value ??

Then they say U=Uo null hypothesis

but if not then you determine rejection region...and I'm officially lost.

according to the text

t= (mean-Uo)/(s/[radical N])

I know that:

s= 0.02998

N= 3

mean= 0.6074

Uo=?

And about the precision I get what you said until the "precision from the mean" how did you get the top value for the (X/mean)? what eqzn is this?

By the way I was wondering what would happen if someone didn't heat the KMnO4 solution for the specified amount of time and basically only heated off about 50ml from a 1300ml sol..I would think that since the ammount of g of KMnO4 was the same 3.200g of solid wouldn't the concentration end up as less when standardizing? And you wouldn't end up with less ppt of MnO4? Just an observation from what another classmate did, and was curious what would happen.

But I did heat & boil my solution for the appropriate time and got 0.02002M for the KMnO4 solution. And what you said about having a idea of the concentration of the solution before I actually standardized...well I saw after looking at the lab I'm doing now of the Determination of iron in an iron ore sample tat it said to titrate with..0.02M KMnO4. I also used my common sense when I eliminated a outlier of 0.01763 since the other values were 0.01996 & 0.02007.

Lastly for the RSD I was wondering IF I was to eliminate a value I wouldn't include that in the calculation of the RSD right so for the mean if I had 3 values but only used 2 I would just average the 2 and act like I only had 2 trial runs right?

Thanks Dr.Bob

Oh..I get it so I do the whole paper then try to squeeze the abstract in at the begining?

If that's correct then I get it

For the test to eliminate an outlier the thing is IF I use common sense judgement that wouldn't suffice for a lab report where you need an actual thing to base your elimination of a value on, but if I'm just handing in a value like for the unknown that would be a different thing.

**I think the passage I quoted from the book is just trying to emphasize that common sense is just as good (or better in some cases) as trying to use statistical formulas WHEN THE NUMBER OF VALUES IS SMALL. And he goes on to say that common sense really doesn't work UNLESS you are sure something went wrong in the measurement. Common sense doesn't include a hunch that such and such a number shouldn't be included. And of course, you are right, hunches don't count when turning in the results.**

Qexp is the same as Q right?

In my text they say Q> Qcrit=reject

I just didn't know which test to use b/c it wasn't specified or I guess we're expected to know which test to use off the top of our head but...since you said the t test

I would appreciate it if you looked at the calculations and see if I got it correct since I'm just reading about this t test right now (we skipped over this in class)

wine % = 0.6402, 0.5814,& 0.6008

Ok...this is sad but I don't get the text...getting a brain-ache

Uo=? how do they get that value ??

Then they say U=Uo null hypothesis

but if not then you determine rejection region...and I'm officially lost.

according to the text

t= (mean-Uo)/(s/[radical N])

I know that:

s= 0.02998

N= 3

mean= 0.6074

Uo=?

**The t test in my sixth edition is defined as T _{n} = |X - mean|/s where T_{n} is the calcualted value for t, X is the suspect value, the mean in this case is 0.6074, and s is 0.02998. The text emphasizes that s includes the suspect value. I have forgotton which value you wanted to throw away, so I tested both 0.6402 AND 0.5814. You may want to go through the calculation; if I didn't make an error I found T_{n} = 1.09 for 0.6402 and 0.867 for 0.5814. Then I compared these with values of T_{critical} in the table and the table value is 1.15 for 95% confidence level. The calculated value of T_{n} must be larger than T_{critical} to be rejected and it isn't for either value; therefore, my recommendation is to keep all three values. In your text, I'm sure the Uo is the suspect value. Since the denominator is s/sqrt N, your text is using a different table so you won't be able to compare my calculation to your table. That's why I included the value from my table. **

And about the precision I get what you said until the "precision from the mean" how did you get the top value for the (X/mean)? what eqzn is this?

**It's just taking the three values of 1.0, 1.1, and 0.9, the mean is 1.0, subtact the high value from the mean (1.1-1.0 = 0.1) and divide by the mean of 1.0 so 0.1/1.0 = 0.1, then multiply by 100 to get 10%. That's on the positive side. For the negative side, use the (0.9 - 1.0)/1 * 100 = -10%. This is just a simple way to determine how much the highest value in the set and how much the lowest value in the set differ from the mean. The closer the high and low values are to the mean the better I can feel about the numbers; i.e., if I get good precision then I have a good feeling that my answers for an "unknown" are good. Of course that doesn't take into account that chemistry errors may play a part in the analytical procedure. And it's a lot faster than calculating s (althouth calculators do a fast job now).**

By the way I was wondering what would happen if someone didn't heat the KMnO4 solution for the specified amount of time and basically only heated off about 50ml from a 1300ml sol..I would think that since the ammount of g of KMnO4 was the same 3.200g of solid wouldn't the concentration end up as less when standardizing? And you wouldn't end up with less ppt of MnO4? Just an observation from what another classmate did, and was curious what would happen.

**More volume means a more dilute solution. If the solution were heated for at least 30 minutes, I think most of the MnO2 would have pptd, and the only consequency, then, would be that a more dilute solution MIGHT make the end point a little less distinct although KMnO4 is a very good self indicator.**

But I did heat & boil my solution for the appropriate time and got 0.02002M for the KMnO4 solution. And what you said about having a idea of the concentration of the solution before I actually standardized...well I saw after looking at the lab I'm doing now of the Determination of iron in an iron ore sample tat it said to titrate with..0.02M KMnO4. I also used my common sense when I eliminated a outlier of 0.01763 since the other values were 0.01996 & 0.02007.

**You can use the t test here, too, (the table value is 1.15) to see if your common sense holds up to the t test. :-0)**

Lastly for the RSD I was wondering IF I was to eliminate a value I wouldn't include that in the calculation of the RSD right so for the mean if I had 3 values but only used 2 I would just average the 2 and act like I only had 2 trial runs right?

**That's right.**

Yes, you do the whole paper, then write the abstract and place it at the beginning. I don't get the "squeeze the abstract in at the beginning" since there is no squeezing here that I see. The abstract takes up the volume it needs to take to say what you want it to say. I would center, at the very beginning, the

Title of report or article

Double space

Next line is centered the word "ABSTRACT" without the quotation marks.

Double space

Write the abstract

Double space

Do the lab report in whatever style your instructor has requested.

The t test in my sixth edition is defined as Tn = |X - mean|/s where Tn is the calcualted value for t, X is the suspect value, the mean in this case is 0.6074, and s is 0.02998. The text emphasizes that s includes the suspect value. I have forgotton which value you wanted to throw away, so I tested both 0.6402 AND 0.5814. You may want to go through the calculation; if I didn't make an error I found Tn = 1.09 for 0.6402 and 0.867 for 0.5814. Then I compared these with values of Tcritical in the table and the table value is 1.15 for 95% confidence level. The calculated value of Tn must be larger than Tcritical to be rejected and it isn't for either value; therefore, my recommendation is to keep all three values. In your text, I'm sure the Uo is the suspect value. Since the denominator is s/sqrt N, your text is using a different table so you won't be able to compare my calculation to your table. That's why I included the value from my table.

**actually my text has that eqzn and the one with the sqrt N right next to each other but in a different part than I was looking at since I used the index to look for the t test. In my bk it isn't described as the t test specifically, only the one with the sqrt is and when I went hunting for the table for t values it isn't called Tcritical but "Values of t for various levels of probability" and yet they still have that eqzn you used for calculating the t value but say it's only for single measurements while the sqrt N is for the mean of N measurements. How can my text be using the same eqzn + 1 other and use a table different from yours? The value you got for t isn't even on the table at all in my bk.**

I don't think that the Uo is the suspect value since in an example calculation they had 3 values (procedure for the rapid determination of the % of sulfur in kerosenes was tested on a sample known from it's method of preparation to contain 0.123%(Uo=0.123%) The results were % S= 0.112, 0.118, 0.115, and 0.119. Do the data indicate there is a bias in the method at the 95% confidence level?) They used the

t= (mean-Uo)/(s/sqrtN) and compared it to my table 7-3 (t for various levels of probability) the value they got for t= -4.375 while the value in the table for 3 degrees of freedom and 95% confidence level was 3.18 and they said since t was less than or = -3.18 there was bias....I really am confused now about Uo

I don't think that the Uo is the suspect value since in an example calculation they had 3 values (procedure for the rapid determination of the % of sulfur in kerosenes was tested on a sample known from it's method of preparation to contain 0.123%(Uo=0.123%) The results were % S= 0.112, 0.118, 0.115, and 0.119. Do the data indicate there is a bias in the method at the 95% confidence level?) They used the

t= (mean-Uo)/(s/sqrtN) and compared it to my table 7-3 (t for various levels of probability) the value they got for t= -4.375 while the value in the table for 3 degrees of freedom and 95% confidence level was 3.18 and they said since t was less than or = -3.18 there was bias....I really am confused now about Uo

It's just taking the three values of 1.0, 1.1, and 0.9, the mean is 1.0, subtact the high value from the mean (1.1-1.0 = 0.1) and divide by the mean of 1.0 so 0.1/1.0 = 0.1, then multiply by 100 to get 10%. That's on the positive side. For the negative side, use the (0.9 - 1.0)/1 * 100 = -10%. This is just a simple way to determine how much the highest value in the set and how much the lowest value in the set differ from the mean. The closer the high and low values are to the mean the better I can feel about the numbers; i.e., if I get good precision then I have a good feeling that my answers for an "unknown" are good. Of course that doesn't take into account that chemistry errors may play a part in the analytical procedure. And it's a lot faster than calculating s (althouth calculators do a fast job now).

**it is alot faster than calculating s and I'll do that in the future when looking at my values that I get**

More volume means a more dilute solution. If the solution were heated for at least 30 minutes, I think most of the MnO2 would have pptd, and the only consequency, then, would be that a more dilute solution MIGHT make the end point a little less distinct although KMnO4 is a very good self indicator.

**interesting b/c this person while filtering was told by the lab tech that they wouldn't get alot of precip and since they left the filter in the sink after filtering I saw that the filter was basically still white while I had alot of stuff on mine and it got all dirty. I guess they didn't heat it for 30 min. I guess they would have to use alot more solution to titrate. shrug***

But I did heat & boil my solution for the appropriate time and got 0.02002M for the KMnO4 solution. And what you said about having a idea of the concentration of the solution before I actually standardized...well I saw after looking at the lab I'm doing now of the Determination of iron in an iron ore sample tat it said to titrate with..0.02M KMnO4. I also used my common sense when I eliminated a outlier of 0.01763 since the other values were 0.01996 & 0.02007.

You can use the t test here, too, (the table value is 1.15) to see if your common sense holds up to the t test. :-0)

**first how do you put in the subscript?**

For my calculations

mean= 0.01922

s= 1.376X10^-3

Tn=|0.01763-0.01922|/1.376x10^-3 =

1.156

Hm... my value compared to the table value you gave is 1.156> 1.15 so I guess I should eliminate right?

For my calculations

mean= 0.01922

s= 1.376X10^-3

Tn=|0.01763-0.01922|/1.376x10^-3 =

1.156

Hm... my value compared to the table value you gave is 1.156> 1.15 so I guess I should eliminate right?

What you found in your text I have in my edition, also, but it is a slightly different thing. For example, I have the table called Values of t for Various levels of probability, also. That table is to be used in connection with the formula t=(x-mu)/s and it has been rearranged in confidence level form, when sigma; i.e, when large numbers of samples are run) is not known, as mu = mean +/- (ts)/sqrt N. Using t and s and the table provided, tells you how close the value of x is to the TRUE VALUE mu. The t test, used for discarding suspected outliers, is in my edition just a few pages later, and is discussed right after the Q test. I calculated t for your KMnO4 solution using

Tn = (x - mean)/s and found

mean = 0.01922 (that includes the suspect value).

s = 0.00138 including the suspect value, so Tn = 1.152. The tabular value is 1.15 for 3 observations at 95% CL. I would reject the value of 0.01763 although purists will say that the 2 on 1.152 is not significant. Probably they are right but I would reject it any way (since the numbers are the same), THEN I would recalculate the mean for the two values I kept as 0.02002 and use that value in all of my titrations. As for a different table, look closely at the two tables and you will see that the tables are in fairly good agreement at larger numbers of replicates. That is, for 10 samples in the T(critical) table, we see 2.18 for the value at 95% CL. The Values for t for Various levels of Probability for 9 degrees of freedom is 2.26 which is close. The T(critical) table doesn't go to large numbers of observations but if it did, the values for the two tables would agree at infinity when s = sigma and the mean = mu. Furthermore, a page earlier has the quote, "It is essential to keep in mind at all times that confidence intervals based on equation 3-8 apply only in the absence of determinate errors AND ONLY IF WE CAN ASSUME THAT S = SIGMA." Equation 3-8 is

mu = mean +/- (z*sigma)/sqrt N. In the t values you are discussing, t has been substituted for z, s has been substituted for sigma, N has been adjusted in T(critical) table to account for the N-1 as well as for sqrt N. And of course, three samples doesn't come close to being sigma. That is why I like the t test a little better because it gives some indication of what can be done with suspect values even with small numbers of replicates. By the way, when I took quant as a student (undergraduate) I usually ran six samples, usually messed up on the first one, and took the other five as good. Never failed me.

I think you will find the T(critical) table in your edition, also. It is within a page or two of the Q table and the Various.....table.

When I said squeeze the abstract in all I meant was to type everything else in and literally just type in the abstract in at the beginning.

thank you Dr Bob =)

Well I'm looking at the text and after the Values of t for various levels of probability I see the z test and the t test and no chart after that. Again, they refer back to the values of t for various levels of probability. After that is the t test for differences in means, F critical values chart, analysis of variance, Q test, Q crit chart, and END OF CHAPTER Questions.

I looked at the chapter, previous chapter, and the next chapter and no t critical values chart at all so I would really like to look at that other table but.. I can't.

And your chapter 3 is my bk's chapter 7, but I did see that statement.

And about you running 6 samples...I would never be in possession of that much equipment and would never have that much time to have that many trial runs.

In addition to no Tcritical table I also have no Various chart. As stated before all that's after the Q crit chart is the 1pg description of Other statistical tests and Recomendation of treating outliers and...End of chapter Questions and Problems

Do you think I could find this Tcrit table on the web?

And from before...how do you type in subscripts???

Thanks Dr Bob =)

From Fundamentals of Analytical Chemistry, Skoog, West, and Holler, 6th edition, page 47.

#observations followed by value @ 95% CL

3   1.15

4   1.46

5   1.67

6   1.82

7   1.94

8   2.03

9   2.11

10  2.18

There are also columns for 97.5% CL and 99% CL but all you are likely to use is the 95 so I'll not type the others.

As for subscripts and superscripts.

< followed by sub or sup followed by >. To turn sub or sup off, < followed by /sub or /sup followed by >.

_{19}K

^{39}

You might also go to your library or ask your instructor for another quant book and look in there. The tables will be the same if the same subject is discussed. Undoubtedly that table is somewhere on the web but I wouldn't know how to find it. I'll look at Google and see and if so I'll post it.

Thanks alot =)

I found many tables but not the one I wanted.

Well Thank You for spending your time to look

When I have time I'll go on a hunt for the T

_{critical value table }