Question

A sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms. We know from past testing that the population standard deviation is 0.35 ohms.

1. Determine a 95% confidence interval for the true mean resistance of the population.

2. In part 1 above, do you need any assumptions, if yes what, if no why.

Answer 1

1)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   0.3500   / √   11   =   0.1055
margin of error, E=Z*SE =   1.9600   *   0.1055   =   0.2068
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    2.20   -   0.206833   =   1.993
Interval Upper Limit = x̅ + E =    2.20   -   0.206833   =   2.407
95%   confidence interval is (   1.99   < µ <   2.41   )

2)

no, because population from which sample is taken is normally distributed.

and sample are taken randomly