Acrylic bone cement is commonly used in total joint replacement to secure the ar
ID: 3179711 • Letter: A
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. (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.) (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 mu_higher temperature - mu_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 =Explanation / Answer
37 degrees, dry
303.2
339.2
288.8
306.8
305.2
327.5
37 degrees, wet
363.9
376.5
327.7
331.9
338.1
394.6
Difference
-60.7
-37.3
-38.9
-25.1
-32.9
-67.1
Lower Limit = M1 - M2 -(tCL)(SM1-M2)
Upper Limit = M1 - M2 +(tCL)(SM1-M2)
where M1 - M2 is the difference between sample means, tCL is the t for the desired level of confidence, and is the estimated standard error of the difference between sample means.
MSE = (s12 + s22)/2 = (18.268932 + 27.103192)/2 = 534.1683558
SM1-M2 = sqrt(2*MSE/n)
= 13.34377
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
= 31703.98137 / (618.84223 +2997.844717) = 8. 766029 = 9
t value is 1.833 for df 9 and 90% confidence
M1 - M2 = 311.7833 – 355.45 = -43.6667
Lower Limit = -43.6667 - (1.833)(13.34377) = -68.1
Upper Limit = -43.6667 + (1.833)(13.34377) = -19.2
R code :
b)
37 degrees, dry
303.2
339.2
288.8
306.8
305.2
327.5
Mean = 311.7833
Std Dev = 18.26893
22 degrees, dry
100.9
141.8
194.8
118.4
176.1
213.1
Mean = 157.5167
Std Dev = 44.28875
Standard error, SE = sqrt[ (s12/n1) + (s22/n2) ] = sqrt[(18.268932/6) + (44.288752/6)] = 19.55866
t = [ (x1 - x2) - d ] / SE
d = mean of higher temp. – mean of lower temp. = 100
t = (154.2666 – 100)/19.55866 = 2.77
where x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (18.268932/6 + 44.288752/6)2/ {[(18.268932/6)2/5] + [(44.288752/6)2/5]} =
(55.62563 + 326.91556)2 / (618.84223 + 21374.75703) = 7(rounded off)
p value = 0.0137
reject null hypothesis. The result is significant at p < .10.
37 degrees, dry
303.2
339.2
288.8
306.8
305.2
327.5
37 degrees, wet
363.9
376.5
327.7
331.9
338.1
394.6
Difference
-60.7
-37.3
-38.9
-25.1
-32.9
-67.1
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