Acrylic bone cement is commonly used in total joint replacement to secure the ar

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 higher temperature 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.)

ConclusionYes, there is sufficient evidence.

No, there is not sufficient evidence.    

Temperature Medium Data on Breaking Force 22 degrees Dry 100.6, 141.9, 194.8, 118.4, 176.1, 213.1 37 degrees Dry 303.2, 339.3, 288.8, 306.8, 305.2, 327.5 22 degrees Wet 385.3, 368.3, 322.6, 307.4, 357.9, 321.4 37 degrees Wet 363.5, 377.7, 327.7, 331.9, 338.1, 394.6

Explanation / Answer

a) x1 = 311.8 , x2 = 355.58 , s1 = 18.298 , s2 = 27.269 , n1 = n2 = 6

t value at 90% CI = +/- 1.812

CI = (x1 - x2) + /- t * sqrt ( s1^2 / n1 + s2^2 / n2)

= ( 311.8 - 355.58) + / -1.812 ( 18.298^2/6 + 27.269^2/6)

= (68.1 , -19.5)

b )

Null hypothesis: 1 - 2 <= 100
Alternative hypothesis: 1 - 2 >100

Test statistic:

x1 = 311.8 , x2 = 157.48 , s1 = 18.298 , s2 = 44.358

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt [(18.298^2 / 6) + (44.358^2/6)]

= 19.589

t = [ (x1 - x2) - d ] / SE

= [(311.8- 157.48) - 100] / 19.589

= 2.77

df = 6 + 6 - 2 = 10

p value is calculated using t = 2.77 , df =10

p value = .009894.

We reject the null hypothesis. p value is less than 0.10

there is sufficient evidence.

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