H_0:μ_1-μ_2=d assuming heterogeneity t=((x ̅_1-x ̅_2 )-(μ_1-μ_2 ))/√((s_1^2)/n_1 +(s_2^2)/n_2 ) Student t(υ)
ν=((s_1^2)/n_1 +(s_2^2)/n_2 )^2/[((s_1^2)/n_1 )^2/(n_1-1)+((s_2^2)/n_2 )^2/(n_2-1)]
H_0:μ_1-μ_2=d assuming homogeneity t=((x ̅_1-x ̅_2 )-(μ_1-μ_2 ))/(s_p √(1/n_1 +1/n_2 )) Student t(υ) ν=n_1+n_2-2
H_0:μ_d=0 t=(d ̅-μ_d)/(s_d⁄√n) Student t(υ) ν=n-1
H_0:p_1-p_2=0 z=((p ̂_1-p ̂_2 )-(p_1-p_2 ))/√((p ̅(1-p ̅))/n_1 +(p ̅(1-p ̅))/n_2 ) N(0,1) NA
H_0:σ_1^2=σ_2^2 F=(s_Larger^2)/(s_Smaller^2 ) F_(ν_1,ν_2 ) ν_i=n_i
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that you have a large variance in the population and/or your sample size is too small
Adverse variances means unfavourable variance which is actual expenses are more than budgted variance.
Analysis of Variance (ANOVA) compares 3 or more means. The t-test would only compare 2 means.
It means that the variance remains the same across the range of values of the variable.
Homogeneity means that the statistical properties of the variable which is being studied remain the same across the population. Heterogeneity means that they do not: it could be that the mean changes between different subsets of the population or the variance does.