Vacuum CleanerWeights Upright vacuumcleaners have either a hard body type or a soft bodytype. Shown are the weights in pounds of a sample ofeach type. At a=0.05, can the claim that there is adifference in the variances of the weights of the twotypes be substantiated?Hard body 21 17 17 20 16 17 15 20 23 16 17 1713 15 16 1818Soft bodied types24,13,11,13,12,15
Hard body : Arithmetic mean : Xbar = ΣX / N = 296 / 17 = 17.4 Standard deviation : s = √ [Σ(X - Xbar)^2 / N] = √ [96.12 / 17] = 2.38 Variance : s^2 = 5.66 Degrees of freedom = 17 - 1 = 16 -------------------- Soft body : Xbar = 88 / 6 = 14.7 s = √ [113.34 / 6] = 4.35 s^2 = 18.92 Degrees of freedom = 6 - 1 = 5 -------------------- Using the F distribution : F = (larger variance) / (smaller variance) = 18.92 / 5.66 = 3.34 (with v1 = 5 and v2 = 16) This is a two-tailed test because the null hypothesis is that there is no difference between the variances. With a 5% probability limit, each tail area = 0.025. Looking up a table of the F distribution for v1 = 5 and v2 = 16, the value given for 0.025 is 3.50. The observed value of 3.34 is less than 3.50, so it lies between the 5% probability limits. Thus, the null hypothesis is accepted, so the claim that there is a difference in the variances of the weights of the two types CANNOT be substantiated, at the 0.05 level of confidence. (My apologies to statistical experts if that is incorrect)
