/*! 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 31 Acrylic bone cement is sometimes... [FREE SOLUTION] | 91Ó°ÊÓ

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Acrylic bone cement is sometimes used in hip and knee replacements to secure an artificial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens, and the resulting mean and standard deviation were 306.09 Newtons and 41.97 Newtons, respectively. Assuming that it is reasonable to believe that breaking force has a distribution that is approximately normal, use a confidence interval to estimate the mean breaking force for acrylic bone cement.

Short Answer

Expert verified
We can be 95% confident that the true mean breaking force for acrylic bone cement in the population lies between 261.02 Newtons and 351.16 Newtons.

Step by step solution

01

Identify the given information

We know the following information: - Sample mean (\(\bar{x}\)) = 306.09 Newtons - Sample standard deviation (s) = 41.97 Newtons - Sample size (n) = 6 specimens - Confidence level = 95%
02

Find the t-value corresponding to the confidence level

Since we are working with a small sample size (n < 30), we will use a t-distribution. First, find the degrees of freedom: df = n - 1 = 6 - 1 = 5. Next, find the t-value corresponding to a 95% confidence interval and 5 degrees of freedom. You can use a t-table or a calculator with a t-distribution function. The t-value for a 95% confidence interval with 5 degrees of freedom is approximately 2.571.
03

Calculate the margin of error

Margin of error (E) is calculated as follows: \[E = t \times \frac{s}{\sqrt{n}}\] Plug in the values, and we have: \[E = 2.571 \times \frac{41.97}{\sqrt{6}}\] Calculate the margin of error: \[E ≈ 45.07\]
04

Calculate the confidence interval

Now, we can calculate the 95% confidence interval for the population mean (μ) using our sample mean and margin of error: \[(\bar{x} - E, \bar{x} + E)\] Substitute the values: \[(306.09 - 45.07, 306.09 + 45.07)\] Calculate the confidence interval: \[(261.02, 351.16)\]
05

Interpret the results

We can be 95% confident that the true mean breaking force for acrylic bone_cement in the population lies between 261.02 Newtons and 351.16 Newtons.

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Key Concepts

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

T-Distribution
When dealing with small sample sizes, especially less than 30, and we don't know the population standard deviation, the t-distribution comes into play. It is similar to the normal distribution but has thicker tails, allowing for more variability in the data. This extra allowance accounts for the uncertainty that comes with smaller samples. The shape of the t-distribution depends on the degrees of freedom, which is calculated as the sample size minus one. In this exercise, with a sample size of 6, the degrees of freedom is 5. A t-table or calculator can help us find the t-value needed for the confidence interval estimation. These characteristics make the t-distribution particularly useful for small sample statistics, ensuring our confidence intervals are accurate and reliable even with limited data.
Sample Mean
The sample mean, denoted as \(\bar{x}\), represents the average of all data points in a sample. It provides a single value that summarizes the center of the data set. In our context, the sample mean of the breaking force for the acrylic bone cement is 306.09 Newtons. This value is crucial as it serves as the point estimate for the population mean we're interested in. While the sample mean gives us a point estimate, it doesn't tell us how much this estimate might vary from the true mean. That's why it's used in conjunction with other statistical tools, like confidence intervals, to give a fuller picture of where the true population mean might lie.
Margin of Error
The margin of error reflects how confident we are about the sample statistic representing the true population parameter. It's computed as the product of the t-value and the standard error of the sample mean. The formula used here is \(E = t \times \frac{s}{\sqrt{n}}\). In this exercise, the margin of error is approximately 45.07 Newtons. This means our sample mean could differ from the true population mean by about 45.07 Newtons. By adding and subtracting this margin from the sample mean, we form an interval that we can say with 95% confidence contains the true mean. The margin of error accounts for possible discrepancies due to random sampling variability.
Sample Size
The sample size, denoted \( n \), is the number of observations in a sample. It plays a critical role in statistical analysis as it impacts the reliability of the estimates and the width of the confidence interval. A smaller sample size, like the one in this exercise with only 6 specimens, typically leads to wider confidence intervals, reflecting greater uncertainty about the population parameter. The sample size affects the degrees of freedom in a t-distribution. As the sample size increases, the t-distribution approaches a normal distribution, making the estimates more precise. Thus, when planning experiments, researchers aim for larger sample sizes to obtain more precise estimates and narrower confidence intervals.
Standard Deviation
The standard deviation, symbolized as \( s \), quantifies the amount of variation or dispersion in a set of data values. In this exercise, the standard deviation of the breaking force is 41.97 Newtons. It measures how much each data point in the sample differs from the sample mean. A smaller standard deviation indicates that the data points are closer to the mean, whereas a larger one suggests more spread out data. The standard deviation is pivotal in the calculation of the margin of error, as it directly influences the standard error of the mean. Knowing the standard deviation is crucial because it impacts the confidence interval's width; the greater the uncertainty in your data, the wider your confidence interval will need to be to maintain the same level of confidence.

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Most popular questions from this chapter

The report "Majoring in Money: How American College Students Manage Their Finances" (SallieMae, \(2016,\) www.news/salliemae.com, retrieved December 24,2106\()\) includes data from a survey of college students. Each person in a representative sample of 793 college students was asked if they had one or more credit cards and if so, whether they paid their balance in full each month. There were 500 who paid in full each month. For this sample of 500 students, the sample mean credit card balance was reported to be \(\$ 825 .\) The sample standard deviation of the credit card balances for these 500 students was not reported, but for purposes of this exercises, suppose that it was \(\$ 200\). Is there convincing evidence that college students who pay their credit card balance in full each month have mean balance that is lower than \(\$ 906\), the value reported for all college students with credit cards? Carry out a hypothesis test using a significance level of 0.01 .

USA TODAY reported that the average amount of money spent on coffee drinks each month is \(\$ 78.00\) (USA Snapshot, November 4, 2016). a. Suppose that this estimate was based on a representative sample of 20 adult Americans. Would you recommend using the one-sample \(t\) confidence interval to estimate the population mean amount spent on coffee for the population of all adult Americans? Explain why or why not. b. If the sample size had been 200 , would you recommend using the one-sample \(t\) confidence interval to estimate the population mean amount spent on coffee for the population of all adult Americans? Explain why or why not.

A sign in the elevator of a college library indicates a limit of 16 persons. In addition, there is a weight limit of 2500 pounds. Assume that the average weight of students, faculty, and staff at this college is 150 pounds, that the standard deviation is 27 pounds, and that the distribution of weights of individuals on campus is approximately normal. A random sample of 16 persons from the campus will be selected. a. What is the mean of the sampling distribution of \(\bar{x} ?\) b. What is the standard deviation of the sampling distribution of \(\bar{x} ?\) c. What average weights for a sample of 16 people will result in the total weight exceeding the weight limit of 2500 pounds? d. What is the probability that a random sample of 16 people will exceed the weight limit?

A student organization uses the proceeds from a soft drink vending machine to finance its activities. The price per can was \(\$ 0.75\) for a long time, and the mean daily revenue during that period was \(\$ 75.00\). The price was recently increased to \(\$ 1.00\) per can. A random sample of \(n=20\) days after the price increase yielded a sample mean daily revenue and sample standard deviation of \(\$ 70.00\) and \(\$ 4.20\), respectively. Does this information suggest that the mean daily revenue has decreased from its value before the price increase? Test the appropriate hypotheses using \(\alpha=\) \(0.05 .\)

A random sample is selected from a population with mean \(\mu=200\) and standard deviation \(\sigma=15\). Determine the mean and standard deviation of the sampling distribution of \(\bar{x}\) for each of the following sample sizes: a. \(n=12\) d. \(n=40\) b. \(n=20\) e. \(n=90\) c. \(n=25\) f. \(n=300\)

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