Question

Acrylic bone cement is commonly used in total joint replacement to secure the artificial joint. Data on the force (measured in Newtons, N) required to break a cement bond was determined under two different temperature conditions and in two different mediums appear in the following table.

Temperature	Medium	Data on Breaking Force
22 degrees	Dry	100.6, 142.7, 194.8, 118.4, 176.1, 213.1
37 degrees	Dry	302.3, 339.3, 288.8, 306.8, 305.2, 327.5
22 degrees	Wet	385.3, 368.2, 322.6, 307.4, 357.9, 321.4
37 degrees	Wet	363.9, 376.5, 327.7, 331.9, 338.1, 394.6

(a) Estimate the difference between the mean breaking force in a dry medium at 37 degrees and the mean breaking force at the same temperature in a wet medium using a 90% confidence interval. (Round your answers to one decimal place.)
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(b) Is there sufficient evidence to conclude that the mean breaking force in a dry medium at the higher temperature is greater than the mean breaking force at the lower temperature by more than 100 N? Test the relevant hypotheses using a significance level of 0.10. (Use μ_{higher temperature} − μ_{lower temperature}. Round your test statistic to two decimal places. Round your degrees of freedom to the nearest whole number. Round your p-value to three decimal places.)

t	=
df	=
P	=

Answer 1

a)

Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   311.65                  
standard deviation of sample 1,   s1 =    18.39                  
size of sample 1,    n1=   6                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   355.45                  
standard deviation of sample 2,   s2 =    27.10                  
size of sample 2,    n2=   6                  
                          
difference in sample means =    x̅1-x̅2 =    311.6500   -   355.5   =   -43.80  
Degree of freedom, DF=   n1+n2-2 =    10              
t-critical value =    t α/2 =    1.8125   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    23.1589              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    13.3708              
margin of error, E = t*SE =    1.8125   *   13.3708   =   24.2340  
                      
difference of means =    x̅1-x̅2 =    311.6500   -   355.450   =   -43.8000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -43.8000   -   24.2340   =   -68.0340
Interval Upper Limit=   (x̅1-x̅2) + E =    -43.8000   +   24.2340   =   -19.5660
CI (-68.0 , -19.6)

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B)

Ho :   µ1 - µ2 =   100                  
Ha :   µ1-µ2 >   100                  
                          
Level of Significance ,    α =    0.1                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   311.65                  
standard deviation of sample 1,   s1 =    18.39                  
size of sample 1,    n1=   6                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   157.62                  
standard deviation of sample 2,   s2 =    44.30                  
size of sample 2,    n2=   6                  
                          
difference in sample means =    x̅1-x̅2 =    311.6500   -   157.6   =   154.03  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    33.9181                  
std error , SE =    Sp*√(1/n1+1/n2) =    19.5826                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   154.0333   -   100   ) /    19.58   =   2.76
                          
Degree of freedom, DF=   n1+n2-2 =    10                  

p-value =        0.010 [excel function: =T.DIST.RT(t stat,df) ]             
Conclusion:     p-value <α , Reject null hypothesis                      
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