About Chi square analysis and P value

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About Chi square analysis and P value
Message posted by Dr.Sankar (via 64.12.104.172) on November 8, 2001 at 6:18 PM (ET)

Iam a doctor.I need to know how chi square and P value can be calculated in two groups and the result is to be compared to find which one is significant?Is there any formula which related chi square and P value.After finding chi square how can we find P-value using that resultant of chi value.Is there any table to calculate P-value.Please explain me.

READERS RESPOND:
(In chronological order. Most recent at the bottom.)

Re: About Chi square analysis and P value
Message posted by Bill (via 192.231.71.108) on November 8, 2001 at 8:48 PM (ET)

There are numerous chisquare cal culators on the web that also provide the corresponding p value if you are sure that the chisquare is the appropriate test.

Re: About Chi square analysis and P value
Message posted by Dr.Sankar (via 152.163.201.56) on November 9, 2001 at 3:30 AM (ET)

I have a problem like this.
In a hospital during last year the total number of patients treated were 394.In that 30 patients are having complications and in the next year 192 patients were treated and 12 were having complications.We have to compare this and statistically determine both the years.can we calculate it using

1. chi square analysis
2.How to find P-value in this problem.Is there any table.
3.Is there any method to compare this two and find out which is significant.

Re: About Chi square analysis and P value
Message posted by Jack Tomsky (via 208.249.113.130) on November 9, 2001 at 5:28 PM (ET)

What you have here are two binomials,

x1 ~ Bin(394,p1)
x2 ~ Bin(192,p2).

We observe that x1 = 30 and x2 = 12. We want to test the null hypothesis H: p1 = p2 against the alternative hypothesis K: p1 \= p2.

The test statsitic is based on the hypergeometric distribution. Calculate

S = Sum C(394,x)*C(192, 42-x) / C(586,42),

where the sum is over x from 30 to 42 and C(m,n) = m!(m-n)!/m!.

The test statistic for the two-sided test is P = 2*min(S, 1-S). Reject the null hypothesis H if P > 0.05.

You can use Excel to calculate the hypergeometric probabilities.

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