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, 142.9, 194.8, 118.4, 176.1, 213.1 37 degrees Dry 303.3, 338.3, 288.8, 306.8, 305.2, 327.5 22 degrees Wet 386.4, 368.2, 322.6, 307.4, 357.9, 321.4 37 degrees Wet 363.6, 376.8, 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.) ( , ) (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 μhigher temperature − μlower 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)

Following is the output of descriptive statistics:

Descriptive statistics
	Dry, 37 degree	Wet, 37 degree
count	6	6
mean	311.650	355.450
sample standard deviation	17.991	27.132
sample variance	323.683	736.131
minimum	288.8	327.7
maximum	338.3	394.6
range	49.5	66.9

Here we have following information:

Here we have following information:

Since it is not given that variances are equal so degree of freedom of the test is

The critical value for 90% confidence interval is "=TINV(1-0.90,8)" is

And standard error for difference in mean will be

So required confidence interval is

So required confidence interval for will be

(b)

Hypotheses are:

Following is the output of independent sample t test:

Hypothesis Test: Independent Groups (t-test, unequal variance)
	higher	lower
	311.650	157.650	mean
	17.991	44.290	std. dev.
	6	6	n
		6	df
		154.0000	difference (higher - lower)
		19.5162	standard error of difference
		100	hypothesized difference
		2.77	t
		.0163	p-value (one-tailed, upper)
	F-test for equality of variance
		1,961.611	variance: lower
		323.683	variance: higher
		6.06	F
		.0699	p-value

The test statistics is

t = 2.77

Formula used:

Degree of freedom:

df = 6

The p-value is: 0.0163