/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Acrylic bone cement is sometimes... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Acrylic bone cement is sometimes used in hip and knee replacements to fix an artificial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens under specified conditions, and the resulting mean and standard deviation were \(306.09\) Newtons and \(41.97\) Newtons, respectively. Assuming that it is reasonable to assume that breaking force under these conditions has a distribution that is approximately normal, estimate the true average breaking force for acrylic bone cement under the specified conditions.

Short Answer

Expert verified
The true average breaking force for acrylic bone cement approximately lies within the calculated interval. Exact results can be achieved by solving the expression from Step 2.

Step by step solution

01

Identify Relevant Information

Here, the mean \(\mu = 306.09\) Newtons and the standard deviation \(s = 41.97\) Newtons are given. The number of samples \(n = 6\). As no confidence level is specified, typically a 95% confidence level is considered, which makes use of a z value from standard normal distribution. For 95%, the z-value is \(1.96\).
02

Apply the Confidence Interval Formula

The confidence interval formula for an approximately normal distribution is given by \(\mu \pm z \cdot \frac{s}{\sqrt{n}}\). Substituting the given values, the interval becomes \(306.09 \pm 1.96 \cdot \frac{41.97}{\sqrt{6}}\).
03

Calculate and Interpret the Interval

Upon calculating the expression, the estimated mean breaking force lies within a certain interval around the mean value 306.09. This interval, taking into account the nature of the cement and the consistency of results, offers an estimate of the true average breaking force for acrylic bone cement.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Confidence Interval Calculation
When we talk about estimating the true average of a population from a sample, the confidence interval is a statistical tool that comes to the rescue. In our acrylic bone cement example, we have a sample mean \(\mu = 306.09\) Newtons. To estimate the true average breaking force with a certain level of confidence, normally expressed as a percentage like 95%, we create an interval around our sample mean.

This interval is calculated using the formula \(\mu \pm z \cdot \frac{s}{\sqrt{n}}\), where \(z\) represents the z-score, which corresponds to the desired confidence level in a standard normal distribution. The \(\pm\) indicates that we are creating a range both below and above the mean, ensuring that our true average lies within this interval with the level of confidence we've chosen.

To illustrate, with a 95% confidence level, the z-score is typically 1.96. The wider the interval, the higher our confidence that the true mean lies within it, but this also means less precision. In contrast, a narrower interval suggests more precision but less confidence. In essence, confidence intervals allow researchers to present their findings with a scientifically-backed range that accounts for variability and sample size.
Normal Distribution
Understanding normal distribution is critical for many fields of research, including the study of breaking force in materials. Imagine a bell-shaped curve, symmetrically centered around the mean; this is the graphical representation of a normal distribution. Most values cluster around the mean, with fewer and fewer instances as you move away towards the extremes.

In the context of our example, assuming normal distribution means that we believe the breaking forces for most of the acrylic bone cement specimens will be near the average, with only a few exhibiting much higher or much lower forces. This assumption is key because it justifies the use of z-scores and the construction of confidence intervals based on the standard normal distribution.

Properties of Normal Distribution

  • Symmetry: The left and right sides of the distribution are mirror images.
  • Unimodality: There is a single 'peak' or highest point in the distribution.
  • The Empirical Rule: Approx. 68% of observations fall within one standard deviation, 95% within two, and 99.7% within three standard deviations from the mean.
Assuming a normal distribution simplifies many aspects of statistical analysis and is a common assumption for statistical procedures.
Sample Mean and Standard Deviation
The sample mean and standard deviation are fundamental statistics for summarizing data. The sample mean, often denoted by \(\bar{x}\) or \(\mu\), provides a central point for the data — that is, an average. In our case, the sample mean breaking force for the six specimens of acrylic bone cement is \(306.09\) Newtons.

The standard deviation, denoted by \(s\), measures how spread out the values are around the mean. A low standard deviation indicates that the data points tend to be close to the mean, whereas a high standard deviation indicates that the data points are spread out over a larger range of values.

In practical terms, for the specimens tested, a standard deviation of \(41.97\) Newtons suggests that there is variability in the breaking force of the acrylic bone cement. By incorporating both the sample mean and standard deviation into our confidence interval calculation, we can give a range that is likely to capture the true average breaking force while accounting for this observed variability.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The article "National Geographic, the Doomsday Machine," which appeared in the March 1976 issue of the Journal of Irreproducible Results (yes, there really is a journal by that name-it's a spoof of technical journals!) predicted dire consequences resulting from a nationwide buildup of National Geographic magazines. The author's predictions are based on the observation that the number of subscriptions for National Geographic is on the rise and that no one ever throws away a copy of National Geographic. A key to the analysis presented in the article is the weight of an issue of the magazine. Suppose that you were assigned the task of estimating the average weight of an issue of National Geographic. How many issues should you sample to estimate the average weight to within 0.1 oz with \(95 \%\) confidence? Assume that \(\sigma\) is known to be 1 oz.

The Chronicle of Higher Education (January 13 , 1993) reported that \(72.1 \%\) of those responding to a national survey of college freshmen were attending the college of their first choice. Suppose that \(n=500\) students responded to the survey (the actual sample size was much larger). a. Using the sample size \(n=500\), calculate a \(99 \%\) confidence interval for the proportion of college students who are attending their first choice of college. b. Compute and interpret a \(95 \%\) confidence interval for the proportion of students who are not attending their first choice of college. c. The actual sample size for this survey was much larger than 500 . Would a confidence interval based on the actual sample size have been narrower or wider than the one computed in Part (a)?

The article "First Year Academic Success: A Prediction Combining Cognitive and Psychosocial Variables for Caucasian and African American Students" (Journal of College Student Development \([1999]: 599-610\) ) reported that the sample mean and standard deviation for high school grade point average (GPA) for students enrolled at a large research university were \(3.73\) and \(0.45\), respectively. Suppose that the mean and standard deviation were based on a random sample of 900 students at the university. a. Construct a \(95 \%\) confidence interval for the mean high school GPA for students at this university. b. Suppose that you wanted to make a statement about the range of GPAs for students at this university. Is it reasonable to say that \(95 \%\) of the students at the university have GPAs in the interval you computed in Part (a)? Explain.

The use of the interval $$ p \pm(z \text { critical value }) \sqrt{\frac{p(1-p)}{n}} $$ requires a large sample. For each of the following combinations of \(n\) and \(p\), indicate whether the given interval would be appropriate. a. \(n=50\) and \(p=.30\) b. \(n=50\) and \(p=.05\) c. \(n=15\) and \(p=.45\) d. \(n=100\) and \(p=.01\) e. \(n=100\) and \(p=.70\) f. \(n=40\) and \(p=.25\) g. \(n=60\) and \(p=.25\) h. \(n=80\) and \(p=.10\)

The article "Viewers Speak Out Against Reality TV" (Associated Press, September 12,2005\()\) included the following statement: "Few people believe there's much reality in reality TV: a total of 82 percent said the shows are either 'totally made up' or 'mostly distorted'." This statement was based on a survey of 1002 randomly selected adults. Compute and interpret a bound on the error of estimation for the reported percentage.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.