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	100.7, 142.8, 194.8, 118.4, 176.1, 213.1
37 degrees	Dry	302.1, 338.1, 288.8, 306.8, 305.2, 327.5
22 degrees	Wet	386.3, 367.3, 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.)
( ¹ , ² )

(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	=	3
df	=	4
P	=	5

????

Answer 1

a) Use the Minitab software to determine confidence interval with the help of following steps: Enterthe data in to the Minitab work sheet, 1. Select stat from the Minitab menu bar 2. Choose stat->basic statistics->2 sample t test. 4. 3. In the 2 sample t test dialog box, click on summarized data, and enter the values of sample size, sample mean, sample standard deviations. 5. Choose options-level ofsignificance (0.10->alternativehypothesis (). 6. Click ok The required output is shown below: Two-Sample T-Test and Cl: Dry, Wet Two-sample T for Dry vs Wet N Mean StDev SE Mean Dry 6 311.4 18.0 7.4 Wet 6 355.4 27.1 Difference -u (Dry) - H (Wet) Estimate for difference: -44.0 90% CI for difference : (-68.8, -19.3) T-Test of difference o (vs ·) : T-value-一3.31 P-value- 0.011 DE-8

From the output, 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 is (-68.8,-19.3) b) Null hypothesis: Ho : μί-μ,-100 Alternativehypothesis: H, -2>100 Level ofsignificance: α = 0, 10 Use the Minitab software to find the test statistic for the given data with the help of following steps: 1. Enterthe data in to the Minitab work sheet, 2. Select stat from the Minitab menu bar 3. Choose stat->basic statistics->2 sample t test. 4. In the2 sample t test dialog box, click on summarized data, and enter the values of sample size, sample mean, sample standard deviations. 5. Chooseoptions-level of significance (0.10 alternativehypothesis (>) and hypothesized difference as 100. 6. Click ok.

The obtained output as follows: wo-Sample T-Test and Cl: High temperature, Lower temperature sample T for High temperature vs Lower temperature N Mean StDev SE Mean igh temperature 6311.4 18.0 r temperature 6 157.7 44.3 18 ifference -μ (High temperature) -μ (Lower temperature) stimate for difference: 153.8 90% lower bound for difference : 125.7 -Test of difference= 100 (vs>): T-value=2.75 P-Value= 0.017 DF= 6 From the output The test statistic value is 2.75 The p-value is 0.017 The degree of freedom is, 6 Conclusion Since the p value is less than the significance level, reject the null hypothesis at 1 090 igni1ica nce level Hence, conclude that there is sufficient evidence to support the claim 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