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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 under two different temperature conditions and in two different mediums appear in the accompanying table. (These data are consistent with summary quantities appearing in the paper "Validation of the Small-Punch Test as a Technique for Characterizing the Mechanical Properties of Acrylic Bone Cement" (Journal of Engineering in Medicine [2006]: 11-21).) $$ \begin{array}{lcl} \text { Temperature } & \text { Medium } & \text { Data on Breaking Force } \\\ \hline \text { 22 degrees } & \text { Dry } & 100.8,141.9,194.8,118.4, \\ & & 176.1,213.1 \\ \text { 37 degrees } & \text { Dry } & 302.1,339.2,288.8,306.8, \\ & & 305.2,327.5 \\ \text { 22 degrees } & \text { Wet } & 385.3,368.3,322.6,307.4, \\ & & 357.9,321.4 \\ \text { 37 degrees } & \text { Wet } & 363.5,377.7,327.7,331.9, \\ & & 338.1,394.6 \\ \hline \end{array} $$ 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. 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 .10 .

Short Answer

Expert verified
The 90% confidence interval for the difference between the mean breaking force in a dry medium at 37 degrees and in a wet medium is (difference - error, difference + error). Next, for the hypothesis testing, the t-value is calculated as per step 5. If the calculated t is more than the critical t-value, we reject the null hypothesis and conclude that the mean breaking force in a dry medium at 37 degrees is greater than the mean breaking force in a dry medium at 22 degrees by more than 100 N with a significance level of 0.10. Otherwise, we fail to reject the null hypothesis.

Step by step solution

01

Calculate Mean Breaking Forces

First, calculate and denote \(M1\), \(M2\), \(M3\), \(M4\) which are the mean breaking forces for dry and wet mediums at each of the two temperatures. The mean is calculated as the sum of all values in a set divided by the number of values in that set.
02

Find the Difference in Means

Subtract the mean breaking force in a wet medium at 37 degrees \(M3\) from the mean breaking force in a dry medium at 37 degrees \(M1\). This gives the difference, \(D\) between the means.
03

Calculating the 90% Confidence Interval

Use the formula for a confidence interval of the difference of means which is \(D \pm z \times \sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2}}\) where \(D\) is the difference in means, \(z\) is the z-score corresponding to the desired confidence level (90% in this case), \(s^2\) is the variance of each dataset (calculated from the standard deviation of each dataset), and \(n\) is the size of each dataset.
04

Formulate Null and Alternative Hypotheses

Now let's use hypothesis testing to determine if 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. The null hypothesis \(H0: M1 - M2 \leq 100\) and the alternative hypothesis \(H1: M1 - M2 > 100\).
05

Conduct t-test

Finally, conduct a t-test to test these hypotheses. Use the formula for the test statistic in the hypotheses test for the difference of means, t = \((M1 - M2 - 100) / SE\), where SE is the standard error of the mean difference. The rejection region is \(t > t_{0.10,n1+n2-2}\), where \(t_{0.10,n1+n2-2}\) is the 0.10 critical value of the t-distribution with \(n1+n2-2\) degrees of freedom. We reject the null hypothesis if the calculated t statistic is in the rejection region.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Breaking Force
Understanding mean breaking force is essential for any study related to the structural integrity of materials, such as acrylic bone cement in joint replacement.

In technical terms, mean breaking force is the average amount of force needed to break or fracture a material. It's represented by the symbol 'M' in our context and is calculated by summing all the measured forces and dividing by the number of observations. Simple, right? Yet, grasping this concept is crucial, as it gives an insight into the material's reliability and performance under various conditions.

To demonstrate this, let's consider acrylic bone cement tested at different temperatures and mediums. Say the bone cement's mean breaking force at 37 degrees in dry conditions is denoted as M1. We'd find M1 by taking the individual breaking forces at this condition, adding them together and dividing by the number of these forces, thereby obtaining an average.

Similarly, mean forces for other conditions such as wet mediums, or at 22 degrees, can be worked out the same way. By comparing these mean forces, researchers, and by extension, our students, can discern how external factors like temperature and moisture affect the integrity of the bone cement—a fundamental step in ensuring the safety and effectiveness of joint replacements.
Hypothesis Testing
Diving into the domain of statistics, hypothesis testing is a methodological power tool that allows us to make inferences about population parameters based on sample statistics. It is a step-by-step procedure where we pose two opposing hypotheses and then determine which one the sample data supports.

The null hypothesis, typically symbolized as H0, represents a statement of 'no effect' or 'no difference'. On the other hand, the alternative hypothesis, symbolized as H1 or Ha, is what researchers genuinely believe to be the reality. When conducting a hypothesis test, the goal is to determine whether there is enough evidence from the sample to reject the null hypothesis in favor of the alternative hypothesis.

Applying Hypothesis Testing

In our exercise, let's say we want to assess whether the mean breaking force at a higher temperature and in a dry condition significantly exceeds that of a lower temperature. The null hypothesis might state that the difference in mean breaking forces is less than or equal to 100 Newtons. The alternative, conversely, posits that the difference is more than 100 Newtons.

To test this, the t-test comes into play. If our calculated statistic exceeds the critical value from t-distribution tables at a chosen significance level (such as 0.10), we have grounds to reject the null hypothesis. This insinuates that there is a statistically significant difference in mean breaking forces, precisely what we sought to discover.
T-test
Now let's focus on the t-test—a statistical procedure used to determine whether there's a significant difference between the means of two groups. It basically helps us understand if the differences in sample statistics are likely to reflect actual differences in the populations or if they're just due to sampling variability.

There are different types of t-tests, but in the context of our exercise, we'd be working with an independent samples t-test for comparing the means from two different groups under different conditions or treatments.

Calculating the T-test

The t-statistic is calculated by taking the difference between sample means, subtracting any hypothesized difference (in our case, 100 Newtons), and dividing by the standard error of the difference. This gives us a ratio which tells us how many standard errors our observed difference is away from the hypothesized difference.

Upon determining the t-statistic, we compare it with a critical value from the t-distribution table (often at the back of statistics textbooks), which corresponds to our desired confidence level and degrees of freedom. If the absolute value of our t-statistic is larger than this critical value, we have reason to reject our null hypothesis, implying that our observed difference is statistically significant and not due to chance.

Mastering the t-test requires not just computing it correctly, but also understanding the implications of the results—a critical thinking skill that can be applied far beyond statistics to broader scientific inquiry.

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Most popular questions from this chapter

"Doctors Praise Device That Aids Ailing Hearts" (Associated Press, November 9,2004 ) is the headline of an article that describes the results of a study of the effectiveness of a fabric device that acts like a support stocking for a weak or damaged heart. In the study, 107 people who consented to treatment were assigned at random to either a standard treatment consisting of drugs or the experimental treatment that consisted of drugs plus surgery to install the stocking. After two years, \(38 \%\) of the 57 patients receiving the stocking had improved and \(27 \%\) of the patients receiving the standard treatment had improved. Do these data provide convincing evidence that the proportion of patients who improve is higher for the experimental treatment than for the standard treatment? Test the relevant hypotheses using a significance level of \(.05 .\)

The paper "If It's Hard to Read, It's Hard to Do" (Psychological Science [2008]\(: 986-988)\) described an interesting study of how people perceive the effort required to do certain tasks. Each of 20 students was randomly assigned to one of two groups. One group was given instructions for an exercise routine that were printed in an easy-to-read font (Arial). The other group received the same set of instructions, but printed in a font that is considered difficult to read (Brush). After reading the instructions, subjects estimated the time (in minutes) they thought it would take to complete the exercise routine. Summary statistics are given below. $$ \begin{array}{ccc} & \text { Easy font } & \text { Difficult font } \\ \hline n & 10 & 10 \\ \bar{x} & 8.23 & 15.10 \\ s & 5.61 & 9.28 \\ \hline \end{array} $$ The authors of the paper used these data to carry out a two-sample \(t\) test, and concluded that at the .10 significance level, there was convincing evidence that the mean estimated time to complete the exercise routine was less when the instructions were printed in an easy-to-read font than when printed in a difficult-to-read font. Discuss the appropriateness of using a two-sample \(t\) test in this situation.

The paper "The Observed Effects of Teenage Passengers on the Risky Driving Behavior of Teenage Drivers" (Accident Analysis and Prevention [2005]: 973-982) investigated the driving behavior of teenagers by observing their vehicles as they left a high school parking lot and then again at a site approximately \(\frac{1}{2}\) mile from the school. Assume that it is reasonable to regard the teen drivers in this study as representative of the population of teen drivers. Use a .01 level of significance for any hypothesis tests. a. Data consistent with summary quantities appearing in the paper are given in the accompanying table. The measurements represent the difference between the observed vehicle speed and the posted speed limit (in miles per hour) for a sample of male teenage drivers and a sample of female teenage drivers. Do these data provide convincing support for the claim that, on average, male teenage drivers exceed the speed limit by more than do female teenage drivers? $$ \begin{array}{cc} \hline \text { Male Driver } & \text { Female Driver } \\ \hline 1.3 & -0.2 \\ 1.3 & 0.5 \\ 0.9 & 1.1 \\ 2.1 & 0.7 \\ 0.7 & 1.1 \\ 1.3 & 1.2 \\ 3 & 0.1 \\ 1.3 & 0.9 \\ 0.6 & 0.5 \\ 2.1 & 0.5 \\ \hline \end{array} $$ b. Consider the average miles per hour over the speed limit for teenage drivers with passengers shown in the table at the top of the following page. For purposes of this exercise, suppose that each driver-passenger combination mean is based on a sample of size \(n=40\) and that all sample standard deviations are equal to .8 . $$ \begin{array}{l|cc} & \text { Male Passenger } & \text { Female Passenger } \\ \hline \text { Male Driver } & 5.2 & .3 \\ \text { Female Driver } & 2.3 & .6 \\ \hline \end{array} $$ i. Is there sufficient evidence to conclude that the average number of miles per hour over the speed limit is greater for male drivers with male passengers than it is for male drivers with female passengers? ii. Is there sufficient evidence to conclude that the average number of miles per hour over the speed limit is greater for female drivers with male passengers than it is for female drivers with female passengers? iii. Is there sufficient evidence to conclude that the average number of miles per hour over the speed limit is smaller for male drivers with female passengers than it is for female drivers with male passengers? c. Write a few sentences commenting on the effects of gender on teenagers driving with passengers.

In December 2001 , the Department of Veterans Affairs announced that it would begin paying benefits to soldiers suffering from Lou Gehrig's disease who had served in the Gulf War (The New york Times, December 11,2001 ). This decision was based on an analysis in which the Lou Gehrig's disease incidence rate (the proportion developing the disease) for the approximately 700,000 soldiers sent to the Gulf between August 1990 and July 1991 was compared to the incidence rate for the approximately 1.8 million other soldiers who were not in the Gulf during this time period. Based on these data, explain why it is not appropriate to perform a formal inference procedure (such as the two-sample \(z\) test) and yet it is still reasonable to conclude that the incidence rate is higher for Gulf War veterans than for those who did not serve in the Gulf War.

The director of the Kaiser Family Foundation's Program for the Study of Entertainment Media and Health said, "It's not just teenagers who are wired up and tuned in, its babies in diapers as well." A study by Kaiser Foundation provided one of the first looks at media use among the very youngest children \(-\) those from 6 months to 6 years of age (Kaiser Family Foundation, \(2003,\) www .kff.org). Because previous research indicated that children who have a TV in their bedroom spend less time reading than other children, the authors of the Foundation study were interested in learning about the proportion of kids who have a TV in their bedroom. They collected data from two independent random samples of parents. One sample consisted of parents of children age 6 months to 3 years old. The second sample consisted of parents of children age 3 to 6 years old. They found that the proportion of children who had a TV in their bedroom was . 30 for the sample of children age 6 months to 3 years and .43 for the sample of children age 3 to 6 years old. Suppose that the two sample sizes were each 100 . a. Construct and interpret a \(95 \%\) confidence interval for the proportion of children age 6 months to 3 years who have a TV in their bedroom. Hint: This is a one-sample confidence interval. b. Construct and interpret a \(95 \%\) confidence interval for the proportion of children age 3 to 6 years who have a TV in their bedroom. c. Do the confidence intervals from Parts (a) and (b) overlap? What does this suggest about the two population proportions? d. Construct and interpret a \(95 \%\) confidence interval for the difference in the proportion that have TVs in the bedroom for children age 6 months to 3 years and for children age 3 to 6 years. e. Is the interval in Part (d) consistent with your answer in Part (c)? Explain.

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