Comparing Weight Gain in Rats: An Analysis of Diets A & B

Yes, but it's better to use the degrees of freedom that you get from the F-test.And that's not 10, because you don't have 2 samples.Yes, I have two available tables, one for 10% critical values and one for 5% critical values. Using the 5% one, and 10 degrees of freedom, should my critical value be 3?The critical value is not in the table, it should be calculated.The degrees of freedom are not 10, but depend on the F-test.And the significance level alpha should be 5%, because that's the standard.If you have done the F-test you can look up the critical value from the F-table
  • #1
Maybe_Memorie
353
0

Homework Statement



A study of weight gain in male rats compared two diets, A and B,
which were formulated in terms of level and source of protein.
A total of twenty two rats were randomly assigned to the diets, eleven to each diet.
Summary statistics for the weight gains (grams) by the end of the study are shown.

A B
Mean 99.5 78.7
St. Dev 10.9 16.5
n 11 11


(a) An F-test was carried out to compare the sample standard deviations.
Why might this have been done? Carry out the F-test: specify the null hypothesis,
the significance level and critical value you use in the test and intrepret the
test result.

(b) Carry out a t-test to compare the sample means. State ecplicitly the null hypothesis,
the degrees of freedom for the reference distribution, and the critical values for your test.
Interpret the results of the test in practical terms.

(c) It has been suggested that rather than carry out a test, it would be better to calculate
a confidence interval. Explain why this is either correct or not correct. Calculate and interpret an appropriate confidence interval.

(d) It was asserted that had the raw data been available, a paired t-test would be the appropriate test for analyzing the study results. Discuss.

(e) Before a study such as this one can be carried out, the researchers need to decide on a sample size for the study. Discuss the issues that need to be considered in arriving at a suitable sample size for
this type of study.

Homework Equations





The Attempt at a Solution



(a) I used F = (s1)^2/(s2)^2
This gave (16.5)^2/(10.9)^2 = 2.29

The null hypotheses, Ho: s1=s2

Is this correct?

I don't know how to get the significance level or critical value.


(b) I used t = (x1 - x2)/root[(s1^2 - s2^2)/n)
which gave me 3.49

Ho: u1 = u2

11-1=10 degrees of freedom

Is it okay to do a 95% confidence interval? That's what all the examples do, but I don't know why...


(c) Using (x1 - x2)+-t*(root[(s1^2 - s2^2)/n)

I need to get t* from the previous part.


For (d) and (e) any advice would be greatly appreciated as I'm lost.
 
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  • #2
Hi Maybe_Memorie! :smile:

First I think you need some background information.
This should all be in your textbook, but I'll try to help you along.

To do any statistical test you need to define a null hypothesis Ho, an alternative hypothesis Ha, and a significance level alpha.
An important choice is whether you select a Ha to be 1-sided or 2-sided.
The significance level alpha is usually chosen as 5%, meaning the chance that you draw the wrong conclusion should be less than 5%.

This is a typical experiment where two populations are compared.
To do that we'll do an "Independent two-sample t-test".
For this it matters whether the population variances are the same or not.
If they are the same, they must be "pooled".
So you would either do the "Independent two-sample t-test with equal variances" or the "Independent two-sample t-test with unequal variances".
Do you have the appropriate formulas available?
The formula you used for (b) is not quite the right formula.

To find out if the variances can be assumed equal the F-test is done.
To do an F-test you need to know the degrees of freedom of the populations, and the selected significance level alpha (pick alpha = 5%).
The critical value comes from looking these parameters up in the appropriate table, or use a statistical calculator.
Do you have such a table available?

Does this help you to find the answer to (a)?

For (b) you did not use quite the right formula.
And yes, its okay to do a 95% confidence interval.
That means that you have chosen a significance level alpha of 5%, which is usually chosen.

Btw, a program like SPSS does all these things for you.
Are you supposed to use for instance SPSS?
 
Last edited:
  • #3
Hello again :smile:

I like Serena said:
So you would either do the "Independent two-sample t-test with equal variances" or the "Independent two-sample t-test with unequal variances".
Do you have the appropriate formulas available?
The formula you used for (b) is not quite the right formula.

For unequal variances
[(x1 - x2) - (u1 - u2)]/root[(s1^2 - s2^2)/n)

I'm using n as a common denominator since the number in A and B is the same.

Equal variances, I don't have my book with me at the moment so I can't state the formula off-hand, but I know there is a formula for a pooled t-test.
Btw, I'm using An Introduction to the Practice of Statistics by Moore and McCabe.

So I'm assuming the appropriate formula to use would be determined by the F-test?

I like Serena said:
To find out if the variances can be assumed equal the F-test is done.
To do an F-test you need to know the degrees of freedom of the populations, and the selected significance level alpha (pick alpha = 5%).
The critical value comes from looking these parameters up in the appropriate table, or use a statistical calculator.
Do you have such a table available?

Does this help you to find the answer to (a)?

Each population as 11 members, so 10 degrees of freedom.
Yes, I have two available tables, one for 10% critical values and one for 5% critical values.
Using the 5% one, and 10 degrees of freedom, should my critical value be 3?


I like Serena said:
Btw, a program like SPSS does all these things for you.
Are you supposed to use for instance SPSS?

Unfortunately no. In my exam I have to get all information from the supplied tables.
 
  • #4
Maybe_Memorie said:
Btw, I'm using An Introduction to the Practice of Statistics by Moore and McCabe.

I'm sorry to say that I don't have a good book available, so I'm dependent on Google.
It's best if you pick the right formulas from your own book.


Maybe_Memorie said:
For unequal variances
[(x1 - x2) - (u1 - u2)]/root[(s1^2 - s2^2)/n)

My problem with your formula is that you subtract the sample variances.
Variances are never subtracted, they should be added.



Maybe_Memorie said:
So I'm assuming the appropriate formula to use would be determined by the F-test?

Yes.


Maybe_Memorie said:
Each population as 11 members, so 10 degrees of freedom.
Yes, I have two available tables, one for 10% critical values and one for 5% critical values.
Using the 5% one, and 10 degrees of freedom, should my critical value be 3?

An F-test has 2 parameters that are both degrees of freedom (df).
They are called the df of the nominator and the df of the denominator.
I believe in this case you need 10 and 10, which comes out at alpha=5% as F*=2.97.

So what is your conclusion for the F-test?
 
  • #5
Maybe_Memorie said:

Homework Statement



A study of weight gain in male rats compared two diets, A and B,
which were formulated in terms of level and source of protein.
A total of twenty two rats were randomly assigned to the diets, eleven to each diet.
Summary statistics for the weight gains (grams) by the end of the study are shown.

A B
Mean 99.5 78.7
St. Dev 10.9 16.5
n 11 11


(a) An F-test was carried out to compare the sample standard deviations.
Why might this have been done? Carry out the F-test: specify the null hypothesis,
the significance level and critical value you use in the test and intrepret the
test result.

(b) Carry out a t-test to compare the sample means. State ecplicitly the null hypothesis,
the degrees of freedom for the reference distribution, and the critical values for your test.
Interpret the results of the test in practical terms.

(c) It has been suggested that rather than carry out a test, it would be better to calculate
a confidence interval. Explain why this is either correct or not correct. Calculate and interpret an appropriate confidence interval.

(d) It was asserted that had the raw data been available, a paired t-test would be the appropriate test for analyzing the study results. Discuss.

(e) Before a study such as this one can be carried out, the researchers need to decide on a sample size for the study. Discuss the issues that need to be considered in arriving at a suitable sample size for
this type of study.

Homework Equations





The Attempt at a Solution



(a) I used F = (s1)^2/(s2)^2
This gave (16.5)^2/(10.9)^2 = 2.29

The null hypotheses, Ho: s1=s2

Is this correct?

I don't know how to get the significance level or critical value.


(b) I used t = (x1 - x2)/root[(s1^2 - s2^2)/n)
which gave me 3.49

Ho: u1 = u2

11-1=10 degrees of freedom

Is it okay to do a 95% confidence interval? That's what all the examples do, but I don't know why...


(c) Using (x1 - x2)+-t*(root[(s1^2 - s2^2)/n)

I need to get t* from the previous part.


For (d) and (e) any advice would be greatly appreciated as I'm lost.

For part (a): if the null hypothesis [itex] \sigma_1 = \sigma_2 [/itex] holds then both s1^2/s2^2 and s2^2/s1^2 should have the so-called F(10,10) distribution. (Here, the '10s' are the "degrees of freedom" (n-1) used in computing the s^2 values.) In other words, we want to know how likely it would be to see a value of s1^2/s2^2 as large as 2.29, or a value of s2^2/s1^2 as small as 1/2.29 = 0.437, from a random variable with distribution F(10,10). You need to look at F-tables (or use an on-line calculator, or spreadsheet, or whatever) to look at the corresponding probabilities.

Note: you may be confusing yourself by using sloppy notation: s1 is a sample standard deviation, not a true "population" standard deviation. We are not testing whether s1 = s2 because this is obviously not true: one of them is about 1.5 times the other. However, the underlying standard deviations [itex] \sigma_1 [/itex] and [itex] \sigma_2 [/itex] may, or may not be equal, or nearly so. That is what we are trying to find out.

RGV
 
  • #6
I like Serena said:
An F-test has 2 parameters that are both degrees of freedom (df).
They are called the df of the nominator and the df of the denominator.
I believe in this case you need 10 and 10, which comes out at alpha=5% as F*=2.97.

So what is your conclusion for the F-test?

My book says to double F* obtaining 6. (in my tables the value for F(10,10) is 3 so I'm just going to use that.)

This is greater than the 5% level of significance, so is significant in our test.
So the null hypothesis fails?
 
  • #7
Maybe_Memorie said:
My book says to double F* obtaining 6. (in my tables the value for F(10,10) is 3 so I'm just going to use that.)

This is greater than the 5% level of significance, so is significant in our test.
So the null hypothesis fails?

Hmm, doubling F*?
I've never heard of that.
Aren't you mixing it up with the t-test?
There's doubling involved in the t-test... but we'll get to that later.

Either way, your F was less then 3 (I assume this is a rounded value), so whatever the doubling, the F is less than the critical F*.

Now concerning the null hypothesis.
Suppose the null hypothesis was perfectly true, what F-value would you get?
And if the null hypothesis would be very untrue, what kind of F-value would you have then?
 
  • #8
I like Serena said:
Now concerning the null hypothesis.
Suppose the null hypothesis was perfectly true, what F-value would you get?
And if the null hypothesis would be very untrue, what kind of F-value would you have then?

If it was perfectly true, F would equal 1, if it was very untrue, F would be very large.
Since my F is smaller than 5, can I take it that the variances are equal and then in (b) use
t = (x1 - x2)/root[(s1^2 + s2^2)/n) ?
 
  • #9
Maybe_Memorie said:
If it was perfectly true, F would equal 1, if it was very untrue, F would be very large.
Since my F is smaller than 5, can I take it that the variances are equal and then in (b) use
t = (x1 - x2)/root[(s1^2 + s2^2)/n) ?

Yes. :smile:
To be a little more accurate, according to the F-test you do not have enough proof that you can reject the null hypothesis, so we'll assume equal variances.

And so yes, you can use the formula for the independent two-sample t-test for equal variances and equal sample sizes. :wink:
But usually, before you do that, you should first state the null hypothesis and the alternative hypothesis.
In particular you need to decide whether you're testing 1-sided or 2-sided...
 
  • #10
I like Serena said:
Yes. :smile:
To be a little more accurate, according to the F-test you do not have enough proof that you can reject the null hypothesis, so we'll assume equal variances.

And so yes, you can use the formula for the independent two-sample t-test for equal variances and equal sample sizes. :wink:
But usually, before you do that, you should first state the null hypothesis and the alternative hypothesis.
In particular you need to decide whether you're testing 1-sided or 2-sided...

The null hypothesis is that u1 = u2
The alternative is that u1 =/= u2.

My book only mentions and uses the 1-sided test, I can't find a mention of a 2-sided test...
 
  • #11
Maybe_Memorie said:
Since my F is smaller than 5

Btw, I noticed that now you are referring to 5.
Any particular reason?

Actually, I was not entirely sure about the degrees of freedom being 10 and 10.
They might well be 2-1 resp 11-1, in which case you would get 4.9647...
Pity I do not have a good reference book available. :(
 
  • #12
Maybe_Memorie said:
The null hypothesis is that u1 = u2
The alternative is that u1 =/= u2.

My book only mentions and uses the 1-sided test, I can't find a mention of a 2-sided test...

Well, you just defined a 2-sided test!

Basically you have three choices for the alternative hypothesis Ha.
1. u1 < u2
2. u1 > u2
3. u1 ≠ u2

The first and second are 1-sided, the third is 2-sided.

You should always carefully read the problem statement to find out what the alternative hypothesis should be.
Did you?
Anyway, let's assume a 2-sided test, which is what you correctly selected! :wink:
 
  • #13
I like Serena said:
Btw, I noticed that now you are referring to 5.
Any particular reason?

Actually, I was not entirely sure about the degrees of freedom being 10 and 10.
They might well be 2-1 resp 11-1, in which case you would get 4.9647...
Pity I do not have a good reference book available. :(

When I said 5 I was referring to my alpha.
I'm assuming this is wrong?

My book said when the populations have n1 and n2 members, then my degrees of freedom are n1 - 1 and n2 -1
 
  • #14
Maybe_Memorie said:
When I said 5 I was referring to my alpha.
I'm assuming this is wrong?

You are wrong if you were comparing your F-value with the significance level alpha of 5%.
In all these statistical theories you're dealing each and every time with two types of variables: test-statistics (like t-values and F-values) and probabilities (like significance level alpha and p-values).
Never mix these two types!


Maybe_Memorie said:
My book said when the populations have n1 and n2 members, then my degrees of freedom are n1 - 1 and n2 -1

Okay, then we were and are on track! :smile:
 
  • #15
I like Serena said:
Well, you just defined a 2-sided test!

Basically you have three choices for the alternative hypothesis Ha.
1. u1 < u2
2. u1 > u2
3. u1 ≠ u2

The first and second are 1-sided, the third is 2-sided.

You should always carefully read the problem statement to find out what the alternative hypothesis should be.
Did you?
Anyway, let's assume a 2-sided test, which is what you correctly selected! :wink:

Oh right I see! :smile:
So I should be using 2-sided because the question said "compare the sample means", not investigate which is bigger?
 
  • #16
I like Serena said:
You are wrong if you were comparing your F-value with the significance level alpha of 5%.
In all these statistical theories you're dealing each and every time with two types of variables: test-statistics (like t-values and F-values) and probabilities (like significance level alpha and p-values).
Never mix these two types!

Right... so what should I have been comparing my F-value to in order to not be able to reject Ho?
 
  • #17
Maybe_Memorie said:
Oh right I see! :smile:
So I should be using 2-sided because the question said "compare the sample means", not investigate which is bigger?

Almost! :smile:
As you formulate it, they would both be 2-sided tests.
"Investigate which is bigger" means you have no clue which one it is.

What you're looking for, is for instance a test to find out if people become more intelligent after taking a course.
That would be typically 1-sided.
Or for instance if the weight of rats becomes less after taking a diet.
That is, if the problem statement uses the word "more" or "less". :wink:
 
  • #18
Maybe_Memorie said:
Right... so what should I have been comparing my F-value to in order to not be able to reject Ho?

You should compare your F-value with the critical F-value F*, which you found was 3 (or 6?!?).

Alternatively you can look up your F-value in a table to find a so called p-value.
In that case you would compare your p-value with the significance level alpha of 5%.
 
  • #19
Oh! Right!

Sorry I got a little lost there. :redface:

Okay, I understand the F test. :smile:

So in my t-test, I'm getting t = 3.49.

My critical value is 2.26, using 10 degrees of freedom and alpha = 5%

Because my t is greater than t* I can reject the null hypothesis.

Is this right?
 
  • #20
Maybe_Memorie said:
Oh! Right!

Sorry I got a little lost there. :redface:

Okay, I understand the F test. :smile:

Good! :smile:


Maybe_Memorie said:
My critical value is 2.26, using 10 degrees of freedom and alpha = 5%

Wait! Stop! :smile:

Where did you find 2.26?
I can't find it.

Btw, this is the point where there is possible doubling or halving in the t-test.
For these calculations a 1-sided alpha of 2.5% corresponds to a 2-sided alpha of 5%.
Do you have references to that in your table?

Then the degrees of freedom.
I think you should not use 10 degrees of freedom.
Does your book say anything about the degrees of freedom in a t-test with equal variances assumed?
(I think it should be 20 degrees of freedom.)
 
  • #21
I like Serena said:
Good! :smile:




Wait! Stop! :smile:

Where did you find 2.26?
I can't find it.

Btw, this is the point where there is possible doubling or halving in the t-test.
For these calculations a 1-sided alpha of 2.5% corresponds to a 2-sided alpha of 5%.
Do you have references to that in your table?

Then the degrees of freedom.
I think you should not use 10 degrees of freedom.
Does your book say anything about the degrees of freedom in a t-test with equal variances assumed?
(I think it should be 20 degrees of freedom.)

Sorry, should have said 2.23.

Yes, in my tables I have a row for alpha and a column for degrees of freedom.

I can't find anything about that in my book, but if I used df=20 I would get 2.09 according to my tables..
 
  • #22
Maybe_Memorie said:
Sorry, should have said 2.23.

Yes, in my tables I have a row for alpha and a column for degrees of freedom.

I can't find anything about that in my book, but if I used df=20 I would get 2.09 according to my tables..

Ah well, I'm relying on wikipedia here:
http://en.wikipedia.org/wiki/T-value

It says for an Independent two-sample t-test with Equal sample sizes, equal variance:
"For significance testing, the degrees of freedom for this test is 2n − 2 where n is the number of participants in each group."

I hope it's reliable.
SPSS would know. :wink:

But yes, I also find 2.09.
Maybe_Memorie said:
Because my t is greater than t* I can reject the null hypothesis.

Is this right?

And yes, this is right! :smile:
 
  • #23
Okay, so for my confidence interval,

x1-x2 +- t*[root(s1^2 + s2^2)/n)

giving 20.8 +- (2.09)(5.96)

giving my confidence interval (8.34, 33.26)

However I don't know why it would be better to do so or what the result means. :confused:
 
  • #24
Let's start with the definition of a confidence interval.
From wikipedia:
"In statistics, a confidence interval (CI) is a particular kind of interval estimate of a population parameter and is used to indicate the reliability of an estimate. It is an observed interval [...]".

Which "population parameter" did you "observe"?
What is the chance that the actual population parameter is within this interval?
How "significant" is this interval? Or rather, what is the significance of it?
 
  • #25
I like Serena said:
Which "population parameter" did you "observe"?
What is the chance that the actual population parameter is within this interval?
How "significant" is this interval? Or rather, what is the significance of it?

The population parameter was the mean, yes?
And since we used alpha = 0.05 there's a 95% chance the actual mean is in this interval.
 
  • #26
Maybe_Memorie said:
The population parameter was the mean, yes?
And since we used alpha = 0.05 there's a 95% chance the actual mean is in this interval.

Nope.
The population parameter is the difference in means.
And yes, there's a 95% chance the actual mean is in this interval.

So how can you tell from this confidence interval whether it is significant?
Actually, you already know that it is significant from the previous question.
But how would it look if it were not significant?
 
  • #27
I like Serena said:
So how can you tell from this confidence interval whether it is significant?
Actually, you already know that it is significant from the previous question.
But how would it look if it were not significant?

If it were not significant the boundaries of the interval would be very close?
Or even 2(standard deviation) would be the difference in the boundaries?
 
  • #28
Maybe_Memorie said:
If it were not significant the boundaries of the interval would be very close?
Or even 2(standard deviation) would be the difference in the boundaries?

Nope.
Now you have the interval (8.34, 33.26) for the difference in population means.
How would this interval look if the population means were the same?
 
  • #29
I like Serena said:
Nope.
Now you have the interval (8.34, 33.26) for the difference in population means.
How would this interval look if the population means were the same?

If the means were the same the difference should be zero?
 
  • #30
Maybe_Memorie said:
If the means were the same the difference should be zero?

Yes...
 
  • #31
I like Serena said:
Yes...

So the interval would be close to zero in the plus and minus direction, something like
(-3, 3)

Like that yeah?
 
  • #32
Yes! :smile:

So how can you see from your interval whether it is significant or not?
 
  • #33
Because my interval has both boundaries greater than zero, we're 95% certain the difference is between these boundaries, so that means we're 95% certain the difference is greater than zero. This is significant and forces us to reject the null hypothesis?
 
  • #34
Yep! :smile:

When you use a CI in a test to compare the means of two samples, the criterion is whether the CI contains zero.
 
  • #35
Ah I see! :smile:

So it would be better to do a confidence interval instead of a t-test?
 

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