Question

Assume the population distribution is normally distributed, the weights (in 1000 grams) of a random sample of 10 boxes of cereal are:

10.2 10.1 10.1 9.7 10.3 9.8 9.9 9.8 10.4 10.3

(a) Find a 99% confidence interval for the mean weight of each box of cereal.

(b) Assume the population standard deviation is 30 grams. Construct a 96% confidence interval for the mean weight of each box of cereal.

(c) Assume the population standard deviation is given. If one decreases the sample size while keeping the same confidence level, what is the effect on the length of a confidence interval of a population mean? Justify your answer.

(d) Assume the population standard deviation is given. If one increases the confidence level while keeping the same sample size, what is the effect on the length of a confidence interval of a population mean? Justify your answer.

Answer 1

a)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   0.2459
Sample Size ,   n =    10
Sample Mean,    x̅ = ΣX/n =    10.0600
Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   9          
't value='   tα/2=   3.2498   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   0.2459   / √   10   =   0.077746
margin of error , E=t*SE =   3.2498   *   0.07775   =   0.252662
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    10.06   -   0.252662   =   9.807338
Interval Upper Limit = x̅ + E =    10.06   -   0.252662   =   10.312662
99%   confidence interval is (   9.81   < µ <   10.31   )

.............

b)

population std dev ,    σ =    30.0000
Level of Significance ,    α =    0.04          
'   '   '          
z value=   z α/2=   2.0537   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   30.0000   / √   10   =   9.486833
margin of error, E=Z*SE =   2.0537   *   9.48683   =   19.483573
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    10.06   -   19.483573   =   -9.423573
Interval Upper Limit = x̅ + E =    10.06   -   19.483573   =   29.543573
96%   confidence interval is (   -9.42   < µ <   29.54   )
..................

if we decrease the sample size, length of a confidence interval will increase

..............

if we increase the confidence level, confidence interval will increase

////////////////////////////

THANKS

revert back for doubt

please upvote