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))

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.

---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?