A New Approach for Comparing Means of Two Populations PowerPoint PPT Presentation

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Title: A New Approach for Comparing Means of Two Populations


1
A New Approach for Comparing Means of Two
Populations
  • By Brad Moss
  • And Sponsored by Dr. Chand Chauhan

2
The Problem
  • The research that is being presented is about a
    current problem in statistics.
  • The problem is how do you compare two means of
    different populations when the variance of each
    population is unknown and unequal.
  • There are many ways to deal with this problem.
    We will compare a new method to one of the other
    methods during this presentation.

3
The Idea!
  • After consulting with Dr. Chauhan and reading A
    Note on the Ratio of Two Normally Distributed
    Variables
  • It was decided that we could try to get around
    the unknown/unequal variance problem by taking a
    ratio of the estimated means which has not been
    done before.
  • We would approximate this ratio as a normal
    distribution and from that, decide whether or not
    the means are different.

4
A Context for the Idea
  • Lets say we are employees of Get Better Drug
    Company and that we are in the process of
    developing a drug for weight loss.
  • Our scientists have developed two drugs, A and B.
  • The company can only mass produce one of these
    drugs.

5
Things We Will Want to Know
  • Overall, what is the mean weight loss for people
    taking each drug?
  • Is one drug more effective than the other?

6
How To Answer
  •  

7
How this Works
  • A ratio (otherwise know as a fraction) of the two
    means should be one or close to one.
  • How close to 1 is close enough?
  • For this, we create what is know as a Confidence
    Interval using the estimated ratio and an
    estimate of the variance.
  • If 1 falls into this interval, we will say that
    the means are statistically the same.

8
The Ratio Can Be Approximated by a Normal
Distribution.
  • According to the paper mentioned on slide 3, a
    ratio of two normally distributed random
    variables can be approximated as a normal
    distribution.
  • This happens when the standard deviation of one
    of the random variables is significantly smaller
    than the other.
  • The ratio of the standard deviations should
    exceed 19 to 9 for this to work assuming the
    means are equal.

9
How Can We Use This
  •  

10
How Can We Use This
  •  

11
How Can We Use This
  •  

12
How Does that Change Things
  •  

13
Now Substituting
  •  

14
The Confidence Interval!
  • Now we need to determine how close to 1 do we
    have to be in order to say that the means are the
    same.
  • So we will create an interval around our
    approximate mean.
  • In statistics, we convert our normal values back
    to standard normal by subtracting the mean and
    then dividing by the standard deviation.

15
The Confidence Interval!
  •  

16
The Confidence Interval!
  •  

17
Doing Some Algebra
  •  

18
Results
  • So, given those two endpoints, if 1 is between
    them, then the means are statistically the same.
  • Otherwise the means are different.

19
Results
  • Using a program called Minitab, I ran 5000
    simulations for several cases and the results are
    on the following slides.

20
Case 1 Equal Sample Sizes
  • YN(100,10), XN(100,0.5), Sample Size 30
  • Confidence Interval of 94.04
  • Mean Length 0.0711821
  •  
  • YN(100,10), XN(100,2), Sample Size 30
  • Confidence Interval of94.42
  • Mean Length 0.0726839
  • YN(100,10), XN(100,5), Sample Size 30
  • Confidence Interval of 94.54
  • Mean Length 0.0795576

21
Case 2 Y has smaller sample size
  • YN(100,10) Sample Size 15, XN(100,0.5) Sample
    Size 20
  • Confidence Interval of 93.34
  • Mean Length 0.100127
  • YN(100,10) Sample size 15, XN(100,2) Sample
    Size 20
  • Confidence Interval of 92.88
  • Mean Length 0.101451
  • YN(100,10) Sample Size 15, XN(100,5) Sample
    Size 20
  • Confidence Interval of 93.68
  • Mean Length 0.108929

22
Case 3 X has smaller sample size
  • YN(100,10) Sample Size 20, XN(100,0.5) Sample
    Size 15
  • Confidence Interval of 93.8
  • Mean Length 0.0869802
  • YN(100,10) Sample Size 20, XN(100,2) Sample
    Size 15
  • Confidence Interval of 93.56
  • Mean Length .0891369
  • YN(100,10) Sample Size 20, XN(100,5) Sample
    Size 15
  • Confidence Interval of 93.84
  • Mean Length 0.100611

23
Sources
  • Jack Hayya, Donald Armstrong, and Nicolas
    Gressis. A Note on the Ratio of Two Normally
    Distributed Variables Management Science 21.11
    (1975). 14 Jan 2011 lt http//www.jstor.org/stabl
    e/2629897 gt
  • Roussas, George. An Introduction To Probability
    and Statistical Inference. San Diego Elsevier
    Science, 2003
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