Question

In order to determine the load capacity of a lift, a random sample of 15lifts are stress tested and the maximum weight that each life can bear before malfunctioning is recordedas the “Maximum Load”. Suppose thatthe mean of the maximum load is1279 pounds and the standard deviation of the maximum loadis 123 pounds.

a.Verify that the conditions for constructing confidence intervals for the mean are met.

b.Compute and interpret a 99% confidence interval for the mean of the maximum weight the lifts can bear.

c.Compute and interpret a 99% confidence interval for the standard deviation of the maximum weight the lifts can bear.

d.A safety mechanism built into the lift will detect the weight currently on the lift and prevent it from activating if it exceeds a certain threshold. The manufacturers want the mechanism to activate below the maximum weight of nearly all of the liftsthey produce. Currently, the mechanism will activate if it detects 1200 pounds or more on the lift.

Based off of your answers to parts b and c, should the threshold for the mechanism be changed? Justify your answer.

Answer 1

a)

sample is random and normalyy distirbuted

......

b)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   14          
't value='   tα/2=   2.9768   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   123.0000   / √   15   =   31.758463
margin of error , E=t*SE =   2.9768   *   31.75846   =   94.539951
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    1279.00   -   94.539951   =   1184.460049
Interval Upper Limit = x̅ + E =    1279.00   -   94.539951   =   1373.539951
99%   confidence interval is (   1184.46   < µ <   1373.54   )

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

c)

Sample Size,   n=   15  
Sample Standard Deviation,   s=   123.0000  
Confidence Level,   CL=   0.99  
          
          
Degrees of Freedom,   DF=n-1 =    14  
alpha,   α=1-CL=   0.01  
alpha/2 ,   α/2=   0.005  
Lower Chi-Square Value=   χ²1-α/2 =   4.0747  
Upper Chi-Square Value=   χ²α/2 =   31.3193  
          

lower bound= (n-1)s²/χ²α/2 =   14*123² / 31.3193=   6762.784  
          
upper bound= (n-1)s²/χ²1-α/2 =   14*123² / 4.0747=   51981.079  

confidence interval for std dev is       
lower bound=   √(lower bound variance)=   82.236
      
      
upper bound=   √(upper bound of variance=   227.994

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