Question

2. The Edison Electric Institute has published figures on the number of kilowatt hours used annually by various home appliances. It is claimed that a vacuum cleaner uses an average of 46 kilowatt hours per year. A random sample of 12 homes included in a planned study indicates that vacuum cleaners use an average of 42 kilowatt hours per year. Assume the population of kilowatt hours to be normally distributed with a population standard devia- tion of 11.9 kilowatt hours. a. Construct an appropriate 95% confidence interval (C.I.) for u. Is the claim of the Edison Electric Institute reasonable? Explain. b. What sample size is required to obtain an interval with a width that is half of the width of the C.I. you constructed in (a)? c. Now, assume the population standard deviation is unknown. Use the sample standard deviation of 10 kilowatt hours to construct a 95% C.I. for u. Is the claim of the Edison Electric Institute reasonable? Explain. d. Use the sample standard deviation of 10 kilowatt hours to construct a 95% C.I. for 02. is the value of 11.9 reasonable for o?? Explain.

Keep in mind it says population standard deviation is 11.9 so it should be a Standard normal using Z tables and not T?

Answer 1

a ) Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   11.9000   / √   12   =   3.4352
margin of error, E=Z*SE =   1.9600   *   3.4352   =   6.7329
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    42.00   -   6.732935   =   35.2671
Interval Upper Limit = x̅ + E =    42.00   -   6.732935   =   48.7329
95%   confidence interval is (   35.27   < µ <   48.73   )

since, confidence interval contain 46, so, claim of 46 is reasonable

b)

Standard Deviation ,   σ =    11.9                  
sampling error ,    E =   3.366467559                  
Confidence Level ,   CL=   95%                  
                          
alpha =   1-CL =   5%                  
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.960   *   11.9   /   3.366467559   ) ² ≈ 48

c)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   11          
't value='   tα/2=   2.2010   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   10.0000   / √   12   =   2.8868
margin of error , E=t*SE =   2.2010   *   2.8868   =   6.3537
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    42.00   -   6.353697   =   35.6463
Interval Upper Limit = x̅ + E =    42.00   -   6.353697   =   48.3537
95%   confidence interval is (   35.65   < µ <   48.35   )

since, confidence interval contain 46, so, claim of 46 is reasonable  

d)

Sample Size,   n=   12
Sample Standard Deviation,   s=   10.0000
Confidence Level,   CL=   0.95
      
      
Degrees of Freedom,   DF=n-1 =    11
alpha,   α=1-CL=   0.05
alpha/2 ,   α/2=   0.025
Lower Chi-Square Value=   χ²1-α/2 =   3.816
Upper Chi-Square Value=   χ²α/2 =   21.920
      
95%   confidence interval for variance is  
lower bound= (n-1)s²/χ²α/2 =   11*10² / 21.92=   50.182
      
upper bound= (n-1)s²/χ²1-α/2 =   11*10² / 3.8157=   288.279

confidence interval for std dev is       
lower bound=   √(lower bound variance)=   7.084
      
      
upper bound=   √(upper bound of variance=   16.979

yes, 11.9 is reasonable for σ