Acrylic bone cement is commonly used in total joint replacement to secure the ar
ID: 3393350 • 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. (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.)
Conclusion
Yes, there is sufficient evidence. OR No, there is not sufficient evidence.
Explanation / Answer
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.
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.
Let X be breaking force in a dry medium at 37 degrees.
Y be reaking force at the same temperature in a wet medium.
mean breaking force in a dry medium at 37 degrees is µ1.
mean breaking force in a wet medium at 37 degrees is µ2.
We have to calculate confidence interval for µ1 - µ2.
The 90% confidence interval for µ1 - µ2 is,
(X1bar - X2bar) - E < µ1 - µ2 < (X1bar - X2bar) + E
where X1bar = mean breaking force in a dry medium at 37 degrees = 311.75
X2bar = mean breaking force in a wet medium at 37 degrees = 355.6167
E is the margin of error.
E = tc * s sqrt [1/n1 + 1/n2]
Considering variances are equal.
s = sqrt [ ( (n1-1)*s12 + (n2-1)*s22 ) / n1 + n2 - 2 ]
tc is critical value for confidence level.
c = confidence level = 90% = 0.90
n1 = number of observations for breaking force in a dry medium at 37 degrees = 6
n2 = number of observations for breaking force in a wet medium at 37 degrees = 6
d.f. = n1 + n2 - 2 = 6+6-2 = 10
tc = 1.8125
s = sqrt [ ((6-1)*333.003 + (6-1)*744.8377) / 6 + 6 -2 ] = 23.21466
E = tc * s * sqrt [1/n1 + 1/n2]
= 1.8125 * 23.21466 * sqrt [ 1/6 + 1/6 ]
0E = 24.2924
lower limit = (X1bar - X2bar) - E = (311.75 - 355.6167) - 24.2924 = -68.2
upper limit = (X1bar - X2bar) + E = (311.75 - 355.6167) + 24.2924 = -19.6
90% confidence interval for difference of mean is (-68.2,-19.6)
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. (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.)
H0 : µ1 - µ2 = 0 Vs H1 : µ1 - µ2 > 0
µ1 = mean breaking force in a dry medium at the higher temperature
µ2 = mean breaking force at the lower temperature
Here we consider the same medium at different temperature.
First we have to test the hypothesis about variances.
H0 : variances are equal.
H1 : variances are not equal.
X1bar = 157.4833
X2bar = 311.75
s1^2 = 1962.758
s2^2 = 333.003
Test statistic is,
F = larger variance / smaller variance = 1962.758 / 333.003 = 5.8941
P-value = 0.03696
alpha = 0.1
P-value < alpha
Reject H0 at 0.1 level significance level.
variances are not equal.
Here we use unpooled variance.
S = sqrt [ s1^2/n1 + s2^2/n2 ]
S = sqrt [ 1962.758/6 + 333.003/6 ] = 19.56085
t = (X1bar - X2bar) / s * sqrt[1/n1 + 1/n2]
t = (157.4833-311.75) / 19.56085*sqrt[1/6 + 1/6 ]
t = -13.7
P-value = 4.16397E-08
P-value < alpha (0.1)
Reject H0 at 0.1 level of significance.
Conclusion : 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
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