Respuesta :
Answer:
[tex]\chi^2 = \frac{(83-90)^2}{90} +\frac{(68-72)^2}{72}+\frac{(85-78)^2}{78}+\frac{(64-60)^2}{60} = 1.66[/tex]
And the answer would be:
c. 1.66
Step-by-step explanation:
We need to conduct a chi square test
The statistic to check the hypothesis is given by:
[tex]\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
the observed values for this case are:
[tex] Freshemen=83[/tex]
[tex] Sophomores=68[/tex]
[tex] Juniors = 85[/tex]
[tex]Seniors= 64[/tex]
The expected values for this case are:
[tex] Fres_E =0.3*300= 90[/tex]
[tex] Sop_E =0.24*300= 72[/tex]
[tex]Jun_E = 0.26*300 =78[/tex]
[tex] Sen_E = 0.20*300= 60[/tex]
Then the statistic would be given by:
[tex]\chi^2 = \frac{(83-90)^2}{90} +\frac{(68-72)^2}{72}+\frac{(85-78)^2}{78}+\frac{(64-60)^2}{60} = 1.66[/tex]
And the answer would be:
c. 1.66
