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A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress of \(8.48 \mathrm{MPa}\) and a sample standard deviation of \(79 \mathrm{MPa}\) ("6haracterization of Bearing Strength Factors in Pegged Timber Connections," J. of Structural Engr., 1997: 326-332). a. Calculate and interpret a \(95 \%\) lower confidence bound for the true average proportional limit stress of all such joints. What, if any, assumptions did you make about the distribution of proportional limit stress? b. Calculate and interpret a \(95 \%\) lower prediction bound for the proportional limit stress of a single joint of this type.

Short Answer

Expert verified
The 95% lower confidence bound is greater than -28.866 MPa, and the lower prediction bound is greater than -137.08 MPa, assuming normality.

Step by step solution

01

Understand the Problem

We need to calculate two bounds: a 95% lower confidence bound for the true average proportional limit stress and a 95% lower prediction bound for the stress of a single joint. This involves dealing with sample mean, standard deviation, and understanding assumptions about the distribution.
02

Identify Relevant Formulas

For the confidence bound, we use the formula for the one-sided lower confidence interval:\[ \bar{x} - t \left(\frac{s}{\sqrt{n}}\right) \]For the prediction bound, the formula is:\[ \bar{x} - t\sqrt{s^2\left(1 + \frac{1}{n}\right)} \]where \(\bar{x}\) is the sample mean, \(s\) is the standard deviation, \(n\) is the sample size, and \(t\) is the t-value from the t-distribution for the desired confidence level and degrees of freedom \(n-1\).
03

Calculate the t-value

For \(n=14\) and a 95% confidence level (one-tailed), we need to find the t-value for 13 degrees of freedom. This is usually found in statistical tables or using software. The t-value is approximately 1.771.
04

Calculate the 95% Lower Confidence Bound

Plug the values into the confidence interval formula:\[ 8.48 - 1.771 \left(\frac{79}{\sqrt{14}}\right) \]Calculate:- \(\frac{79}{\sqrt{14}} \approx 21.099\)- \(1.771 \times 21.099 \approx 37.346\)- The lower confidence bound: \(8.48 - 37.346 \approx -28.866\)Interpreted, this means the true average proportional limit stress is greater than \(-28.866 \text{ MPa}\) with 95% confidence. This negative result suggests a reconsideration of data reasonability and assumptions.
05

Calculate the 95% Lower Prediction Bound

For the prediction bound:\[ 8.48 - 1.771\sqrt{79^2\left(1+\frac{1}{14}\right)} \]Calculate:- \(79^2\left(1+\frac{1}{14}\right) \approx 6753.143\)- \(\sqrt{6753.143} \approx 82.179\)- \(1.771 \times 82.179 \approx 145.56\)- The lower prediction bound: \(8.48 - 145.56 \approx -137.08\)This means the stress for a single joint is greater than \(-137.08 \text{ MPa}\) with 95% confidence.
06

Assumptions and Interpretation

We assumed that the distribution of proportional limit stress is approximately normal. The negative results imply assumptions may not hold, or data should be re-evaluated. However, mathematically, they indicate confidence and prediction bounds under given assumptions.

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Key Concepts

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

Sample Mean
The sample mean, often represented as \( \bar{x} \), is a crucial measure in statistics that gives us the average value of a sample set. For instance, in our given exercise, the sample mean of 14 joint specimens is 8.48 MPa. Calculating the sample mean involves adding up all the values in your sample and then dividing by the number of values. It acts as an unbiased estimator for the population mean if the sample is randomly drawn from the population.
This makes it a central component in calculating confidence and prediction intervals, as it represents our best estimate of the population's average based on the data we have collected. Understanding the sample mean is vital because it sets the foundation for more complex analysis involving variations from this central tendency.
T-Distribution
The t-distribution, or Student's t-distribution, is a fundamental concept when dealing with small sample sizes or when the population standard deviation is unknown. Unlike the standard normal distribution, the t-distribution accounts for the additional variability expected when working with smaller samples.
It is characterized by degrees of freedom, which generally equals the sample size minus one \( (n-1) \). In our exercise, with a sample size of 14, we have 13 degrees of freedom. The t-distribution helps determine the critical t-value used in calculating confidence and prediction intervals. This critical value adjusts the intervals to reflect the uncertainty inherent in a smaller sample, which is why the t-distribution features heavier tails compared to the normal distribution. This reflects the increased likelihood of observing results further from the mean when dealing with small samples.
Prediction Interval
A prediction interval provides a range within which we expect a future observation to fall. Unlike a confidence interval, which estimates a parameter like the mean, a prediction interval anticipates the variability of individual outcomes.
In our scenario, the prediction interval helps us understand the range for the proportional limit stress of a single joint specimen. The formula for the lower prediction bound in the exercise includes the sample standard deviation and accounts for variability in the data set.
The resulting calculation anticipates where one might expect future single observations to land, considering both the typical variation captured in our sample and additional variability because we're predicting a single event. It's essential, particularly in quality control and reliability engineering, where estimating future single outcomes is critical for safety assessments and operational planning.
Standard Deviation
Standard deviation is a key indicator of data dispersion around the mean. It tells us how much the individual data points in a sample deviate from the sample mean. In our exercise, the standard deviation of 79 MPa reflects the variability within our 14 joint specimens. Calculating it involves finding the square root of the variance, which is the average of the squared deviations from the mean.
Understanding standard deviation is crucial as it forms the basis for determining the width of our confidence and prediction intervals. It measures the degree of spread in our data, influencing the intervals we calculate and, consequently, the precision of our estimates. High standard deviation indicates that data points are spread out over a wider range of values, whereas a low standard deviation indicates they are closer to the mean. This distinction plays a critical role in assessing data reliability and predicting future observations.

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