# Test chi-square

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• Nov 6th 2012, 04:36 PM
martinj
Test chi-square
You cant solve this please!! In the telephone survey conducted last year by students results were very different, while some four scheduled interviews conducted others failed to complete any of them. The distribution of the number of interviews collected by the 57 students who participated in the project was the following:

 Numbres interview Numbers of student 0 6 1 16 2 24 3 9 4 2

… A confidence level of 90%. Does this mean that these differences have been due to chance? Or rather they are motivated by some other cause..((test chi-square))
• Nov 6th 2012, 05:34 PM
chiro
Re: Test chi-square
Hey martinj.

Did you calculate the test statistic for X^2? If so show us what you got and we can take it from there.
• Nov 7th 2012, 03:39 AM
martinj
Re: Test chi-square
---Expected frequency
Total stundents>> 57
If p=0.5 and q=0.5
binomial>>
P(x=0)=4C0x(0.5)^(0)x(0.5)^(4)=1/16 >> P(X=0)=57/16
p(x=1)............................................ .>>P(X=1)=57/4
p(x=2)............................................ .>>P(X=2)=171/8
p(x=3)............................................ .>>P(X=3)=57/4
p(x=4)............................................ .>>P(X=4)=57/16

Numbers interview Numbers of student
0 57/16
1 57/4
2 171/8
3 57/4
4 57/16

freedom degree>> v=k-1=4
chi-square(4,0.90)=7.77
chi-square-TS=4.82
7.77>4.82
Ho>> true
Hi>>false

Is it ok solved the problem?