Calculate the mean difference (male absences compared to female absences) within the population.
μd = μ1 - μ2 = 15 - 10 = 5
Determine the standard deviation of the difference.
σd = sqrt( σ12 / n1 + σ22 / n2 )
σd = sqrt(72/100 + 62/50) = sqrt(49/100 + 36/50) = sqrt(0.49 + 0.72) = sqrt(1.21) = 1.1
Calculate the z-score for when boys have three more absence days than girls. The difference between male and female absences is three in this situation. The corresponding z-score is
z = (x - μ)/σ = (3 - 5)/1.1 = -2/1.1 = -1.818
Determine the probability. We need to find the likelihood that the average absences from the male sample minus the average from the female sample is less than 3. Inputting the z-score (-1.818) into Stat Trek's Normal Distribution Calculator shows that the probability of a z-score being -1.818 or lower is approximately 0.035.
