In May the fill weights of 6pound boxes of laundry soap had

In May the fill weights of \"6-pound\" boxes of laundry soap had a mean of 6.13 pounds with a standard deviation of 0.095. The goal was to decrease the standard deviation. The company decided to adjust the filling machines so that the standard deviation will be reduced. In June a random sample of 20 boxes yielded mean of 6.10 and a standard deviation of 0.065.

At a 0.05 significance level, was the company successful?

Set up a 99% confidence interval for s².

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   sigma   <=   0.095  
Ha:    sigma   >   0.095  
              
As we can see, this is a    right   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          
              
Comparing chi^2 < chi^2(crit), we              
              
FAIL TO REJECT THE NULL HYPOTHESIS.   

There is no significant evidence that the that the company was successful in reducing the variability of their filling machines. [conclusion]   
*********************************************              
              
As              
              
df = n - 1 =    19          
alpha = (1 - confidence level)/2 =    0.005          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    38.58225655          
chi^2(alpha/2) =    6.843971445          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    0.00208062          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    0.011729301          
              
Thus, the confidence interval for the variance is              
              
(   0.00208062   ,   0.011729301   ) [ANSWER, CONFIDENCE INTERVAL]