1792. On the exam for a given subject, the examiner asks the student only one question from one of the four parts of the course. Out of 100 students, 26 received a question from the first part, 32 from the second, 17 from the third, and the rest from the fourth. Can we accept the hypothesis that a student coming to the exam has an equal probability of receiving a question from any of the four parts? (Take $\alpha=0.05$).
Solution. In this case: $m_{1}=26, m_{2}=32, m_{3}=17$, $m_{4}=25, p_{i}=0.25, \quad n=100, \quad n p_{i}=25(i=1,2,3,4)$. We find $\chi_{0}^{2}: \chi_{0}^{2}=\frac{(26-25)^{2}}{25}+\frac{(32-25)^{2}}{25}+\frac{(17-25)^{2}}{25}+\frac{(25-25)^{2}}{25}=4.56$. Since none of the parameters of the assumed distribution were estimated from the sample, $s=0$ and the number of degrees of freedom is $k-s-1=4-0-1=3$. From tables (appendix 4) we find the boundary of the critical region. For $\alpha=0.05$ it is 7.815. Since $4.56<7.815$, the hypothesis is confirmed.