The fill weights of 6pound boxes of a particular brand of la

The fill weights of 6-pound boxes of a particular brand of laundry soap have historically been normally distributed, with a mean of 6.13 lbs. with a standard deviation of 0.095 lbs. New procedures were put into place, with the goal of reducing the variation. After these procedures were implemented, a random sample of n = 20 containers yielded y = 6.10 and s = 0.065. Using = 0.05, can the company conclude that the new procedures were successful?

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   sigma   >=   0.095  
Ha:    sigma   <   0.095  
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical chi^2, as alpha =    0.05   ,      
alpha =    0.05          
df = N - 1 =    19          
chi^2 (crit) =    10.11701306        
              
Getting the test statistic, as              
s = sample standard deviation =    0.065          
sigmao = hypothesized standard deviation =    0.095          
n = sample size =    20          
              
              
Thus, chi^2 = (N - 1)(s/sigmao)^2 =    8.894736842          
              
As chi^2 < chi^2(crit), then we REJECT THE NULL HYPOTHESIS.              

Thus, there is significant evidence that the new procedure has less variation; thus, it is successful. [CONCLUSION]