Question

The weights of randomly selected 5 female students and 5 male students are given by the following:weights for male students:160,165,152,158,178weights for female students:169,154,158,156,162.Let the weight distributions of male and female students follow N(μ1,σ1) and N(μ2,σ2)respectively.•

TestH0:σ1=σ2vs.H1:σ16=σ2at 5% level of significance. UseF4,4,0.025= 0.10F4,4,0.975= 9.60.

Answer 1

   Sample 1:   
Sample Standard deviation,    s₁ =    9.788
Sample size,    n₁ =    5

   Sample 2:   
Sample Standard deviation,    s₂ =    5.933
Sample size,    n₂ =    5

Null and alternative hypothesis:
Hₒ : σ₁ = σ₂
H₁ : σ₁ ≠ σ₂

Test statistic:  
F = s₁² / s₂² = 9.788² / 5.933² =    2.722

Degree of freedom:  
df₁ = n₁-1 =    4
df₂ = n₂-1 =    4
Critical value(s):  
Lower tailed critical value, FL = 0.10
Upper tailed critical value, FU = 9.60

since, test stat = 2.72 < 9.60 , fail to reject Ho

hence, there is no enough evidence of significant difference between std dev of two population