Fwd: Fw: Chi Square question

[Heidi Voordeckers]

---------- Forwarded message ----------
From: Heidi Voordeckers <hvoord...@wbcity.org>
Date: Tue, Dec 10, 2013 at 7:41 AM
Subject: Fw: Chi Square question
To: "Heidi Voordeckers (wbc....@gmail.com)" <wbc....@gmail.com>

 

 

Sent from Windows Mail

 

From: Heidi Voordeckers
Sent: Monday, December 9, 2013 8:24 PM
To: Bev Uipi
Cc: UteCohorts

 

Hi Bev,

 

I worked on this for two hours last night and realized that the problem is with the expected outcomes he presented Friday night. He used the percentage table to calculate expected results instead of the observed table ... big difference!

 

Here is a link that I found helpful in explaining how to get the expected values:

 

http://mathforum.org/library/drmath/view/52756.html

 

 and a calculator that also shows you the expected values it uses:

 

http://www.socscistatistics.com/tests/chisquare/Default2.aspx

 

Hope this helps!

 

Heidi

 

Sent from Windows Mail

 

From: Bev Uipi
Sent: Monday, December 9, 2013 7:57 PM
To: d_mon42
Cc: UteCo...@googlegroups.com

 

All,

Paula found this youtube on chi square that made the most sense to me. My numbers aren't at all what Simon came up with in class this weekend. 

http://m.youtube.com/watch?v=WXPBoFDqNVk&desktop_uri=%2Fwatch%3Fv%3DWXPBoFDqNVk

Good luck,

Bev

On Mon, Dec 9, 2013 at 6:47 PM, d_mon42 <d_m...@hotmail.com> wrote:

Hey all, I wanted some extra help with the chi square piece of the final exam and asked Simon for some direction. 

He sent me this but I haven't dug into it yet.   Hopefully it will help some of you too

Sent from my Samsung Galaxy S®4

-------- Original message --------
From: CHRISTOPHER A SIMON
Date:12/09/2013 1:12 AM (GMT-07:00)
To: u0312540
Subject: RE: Chi Square question

I copied and pasted another example below.  Does this help?

 

Chi Square Goodness of Fit (One Sample Test)

This test allows us to compare a collection of categorical data with some theoretical expected distribution. This test is often used in genetics to compare the results of a cross with the theoretical distribution based on genetic theory. Suppose you preformed a simple monohybrid cross between two individuals that were heterozygous for the trait of interest.

Aa x Aa

The results of your cross are shown in Table 4.

 

Table 4. Results of a monohybrid cross between two heterozygotes for the 'a' gene.

	A	a	Totals
A	10	42	52
a	33	15	48
Totals	43	57	100

The penotypic ratio 85 of the A type and 15 of the a-type (homozygous recessive). In a monohybrid cross between two heterozygotes, however, we would have predicted a 3:1 ratio of phenotypes. In other words, we would have expected to get 75 A-type and 25 a-type. Are or resuls different?

Calculate the chi square statistic x² by completing the following steps:

1.     For each observed number in the table subtract the corresponding expected number (O — E).

2.     Square the difference [ (O —E)² ].

3.     Divide the squares obtained for each cell in the table by the expected number for that cell [ (O - E)² / E ].

4.     Sum all the values for (O - E)² / E. This is the chi square statistic.

For our example, the calculation would be:

	Observed	Expected	(O — E)	(O — E)2	(O — E)2/ E
A-type	85	75	10	100	1.33
a-type	15	25	10	100	4.0
Total	100	100			5.33

x² = 5.33

We now have our chi square statistic (x² = 5.33), our predetermined alpha level of significalnce (0.05), and our degrees of freedom (df =1). Entering the Chi square distribution table with 1 degree of freedom and reading along the row we find our value of x² 5.33) lies between 3.841 and 5.412. The corresponding probability is 0.05<P<0.02. This is smaller than the conventionally accepted significance level of 0.05 or 5%, so the null hypothesis that the two distributions are the same is rejected. In other words, when the computed x² statistic exceeds the critical value in the table for a 0.05 probability level, then we can reject the null hypothesis of equal distributions. Since our x² statistic (5.33) exceeded the critical value for 0.05 probability level (3.841) we can reject the null hypothesis that the observed values of our cross are the same as the theoretical distribution of a 3:1 ratio.

Table 3. Chi Square distribution table.

probability level (alpha)

Df	0.5	0.10	0.05	0.02	0.01	0.001
1	0.455	2.706	3.841	5.412	6.635	10.827
2	1.386	4.605	5.991	7.824	9.210	13.815
3	2.366	6.251	7.815	9.837	11.345	16.268
4	3.357	7.779	9.488	11.668	13.277	18.465
5	4.351	9.236	11.070	13.388	15.086	20.517

To put this into context, it means that we do not have a 3:1 ratio of A_ to aa offspring.

 

 

From: DEE NORTON [mailto:d_m...@hotmail.com]
Sent: Sunday, December 08, 2013 9:05 PM
To: CHRISTOPHER A SIMON
Subject: Chi Square question

 

Professor Simon,

I am having some trouble understanding the use of the Percentage Points of the Chi-Square Distribution sheet you sent to the class.

For the final exam, I have calculated the final Chi-Square for the mining town example, and my final score was 0.1599.  For one degree of freedom, and looking at the column for 0.05 (5% limit for accuracy?), the score to compare against is 3.84.  Does this mean that my score of 0.1599 is well below the 3.84 and therefore does not have any statistical significance?

I thought I had this leaving class yesterday, but find myself second guessing what I thought I took from it.

Any guidance you can give would be great.

Thanks
Dee Norton

 

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