Question

A soda bottling plant fills cans labeled to contain 12 ounces of soda. The filling machine varies and does not fill each can with exactly 12 ounces. To determine if the filling machine needs adjustment, each day the quality control manager measures the amount of soda per can for a random sample of 50 cans. Experience shows that its filling machines have a known population standard deviation of 0.35 ounces.

In today's sample of 50 cans of soda, the sample average amount of soda per can is 12.1 ounces.

Construct and interpret a 90% confidence interval estimate for the true population average amount of soda contained in all cans filled today at this bottling plant. Use a 90% confidence level.

X =

population parameter: m =

random variable =

Answer 1

population std dev ,    σ =    0.3500
Sample Size ,   n =    50
Sample Mean,    x̅ =   12.10

Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.645   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   0.3500   / √   50   =   0.0495
margin of error, E=Z*SE =   1.6449   *   0.0495   =   0.0814
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    12.10   -   0.081416   =   12.0186
Interval Upper Limit = x̅ + E =    12.10   -   0.081416   =   12.1814
90%   confidence interval is (   12.0186   < µ <   12.1814   )