Question

Statistic

Q3(a) To estimate the mean amount of time children spend in physical activities daily, a researcher randomly selects 8 children and records the number of hours they spend in physical activities in one day. The numbers of hours are: 4.1, 1.2, 4.6, 2.4, 3.2, 1.8, 2.4, and 0.7. Obtain a 95% confidence interval of the mean number of hours that children spend in physical activities.

Q3(b) It is found that the average lifespan of 100 smart phones is 23.5 months with a standard deviation of 10.2 months. Construct the 99% confidence interval for the mean lifespan of all smart phones. State the assumption(s) made.

Q3(c) A machine produces DVD discs. The diameters of these DVD discs vary, and the standard deviation is 0.01 centimeter. How large a sample should be taken if we wish to have 95% confidence that our sample mean will not differ from the true mean by more than 0.001 centimeter?

Answer 1

a)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   1.3565
Sample Size ,   n =    8
Sample Mean,    x̅ = ΣX/n =    2.5500

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   7          
't value='   tα/2=   2.365   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   1.3565   / √   8   =   0.4796
margin of error , E=t*SE =   2.3646   *   0.4796   =   1.1340
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    2.55   -   1.134034   =   1.4160
Interval Upper Limit = x̅ + E =    2.55   -   1.134034   =   3.6840
95%   confidence interval is (   1.4160   < µ <   3.6840   )

b)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   99          
't value='   tα/2=   2.626   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   10.2000   / √   100   =   1.0200
margin of error , E=t*SE =   2.6264   *   1.0200   =   2.6789
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    23.50   -   2.678934   =   20.8211
Interval Upper Limit = x̅ + E =    23.50   -   2.678934   =   26.1789
99%   confidence interval is (   20.8211   < µ <   26.1789   )

c)

Standard Deviation ,   σ =    0.01                  
sampling error ,    E =   0.001                  
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   *   0.01   /   0.001   ) ² =   384.146
                          
                          
So,Sample Size needed=       385