Let X denote the grades in a statistics course.
Hypothesis tested is,
H0: X ~ Poisson distribution against H1: X does not follow a Poisson distribution
We first compute the parameter "\\lambda=(\\displaystyle\\sum{(f*x)})\/n=716\/302=2.37(2 decimal places)."
We then compute respective probabilities for "x=1,2,3,4,5,6,7,8" using the Poisson distribution given by;
"P"x "=P(X=x)=(e^{-\\lambda}*\\lambda^x)\/x!"
Therefore,
P1=P(X=1)=e-2.37*(2.37)=0.22
P2=P(X=2)=e-2.37*(2.372)/2!=0.26
P3=P(X=3)=e-2.37*(2.373)/3!=0.21
P4=P(X=4)=e-2.37*(2.374)/4!=0.12
P5=P(X=5)=e-2.37*(2.375)/5!=0.06
P6=P(X=6)=e-2.37*(2.376)/6!=0.02
P7=P(X=7)=e-2.37*(2.377)/7!=0.008
P8=P(X=8)=e-2.37*(2.378)/8!=0.002
Next is to determine the expected frequencies, EX=n*PX
E1=n*p1=302*0.22=66.87
E2=n*p2=302*0.26=79.28
E3=n*p3=302*0.21=62.65
E4=n*p4=302*0.12=37.13
E5=n*p5=302*0.06=17.61
E6=n*p6=302*0.02=6.96
E7=n*p7=302*0.008=2.36
E8=n*p8=302*0.02=0.70
Since the expected frequencies for x=7 and x=8 are less than 5, they are both combined with the expected frequency for x=6 to get a total expected frequency of 10.01.
With these values, we can now find the chi square test statistic defined by,
"\\chi"2(calculated)="\\displaystyle\\sum_{i=1}^{6}(O_{i}-E_{i})^2\/E_i"
=(136-66.87)2/66.87+(60-79.28)2/79.28+(34-62.65)2/62.65+(12-37.13)2/37.13+(55-17.62)2/17.62+(5-10.01)2/10.01=188.17(2 decimal places).
"\\chi"2 table value with k-1= 6-1=5degrees of freedom at "\\alpha" =0.05=11.0705.
"\\chi^2"(calculated)=188.17 is greater than the table value of 11.0705 hence, we reject the null hypothesis that X follows a Poisson distribution and conclude that there is no sufficient evidence to show that grades in a statistics class follow a Poisson distribution.
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