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A type of thread is being studied for its tensile strength properties. Fifty pieces were tested under similar conditions and the results showed an average tensile strength of 78.3 kilograms and a standard deviation of 5.6 kilograms. Assuming a normal distribution of tensile strength, give a lower \(95 \%\) prediction interval on a single observed tensile strength value. In addition, give a lower \(95 \%\) tolerance limit that is exceeded by \(99 \%\) of the tensile strength values.

Short Answer

Expert verified
To find the lower 95% prediction interval on a single observed tensile strength value and the lower 95% tolerance limit that is exceeded by 99% of the tensile strength values, we use the Z score from the Z-table for a normal distribution and basic statistical method. After calculation, the lower 95% prediction interval and the lower 95% tolerance limit will be obtained.

Step by step solution

01

Calculation of Prediction Interval

In first case, we need to find the prediction interval. The first step is to determine the Z-score. From the Z-table or using Z-score calculator, for a 95% prediction interval, the Z-value is approximately 1.645. Once we have the Z-value, we can calculate the prediction interval using the formula: \(Prediction\ Interval = Mean ± (Z * Standard\ Deviation)\). Here, the 'Mean' is the average tensile strength, 'Z' is the Z-value we found and 'Standard deviation' is given. Plugging in the values, we get \(Prediction\ Interval = 78.3 ± (1.645 * 5.6)\).
02

Finding the Lower Prediction Interval

Calculate the lower prediction interval by subtracting the above result. Then, we have Lower Prediction Interval = 78.3 - (1.645 * 5.6). We are only interested in the lower limit of the prediction interval, so we can disregard the upper limit.
03

Calculation of Tolerance Limit

Next, we need to determine the lower tolerance limit that is exceeded by '99%' of the tensile strength values. Again, we first find the Z value. For 99%, the Z-value is approximately 2.33. We use the same formula as for the prediction interval, but substitute the new Z-value. Plugging in the values, we get \(Tolerance\ Limit = 78.3 ± (2.33 * 5.6)\).
04

Finding the Lower Tolerance Limit

Calculate the lower tolerance limit by subtracting the value from Mean. So, the lower tolerance limit will be = 78.3 - (2.33 * 5.6). We are only interested in the lower limit of the tolerance limit, so we can disregard the upper limit. Calculate the above expressions to obtain the values.

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Key Concepts

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

Prediction Interval
Imagine you're working with a set of data, like the tensile strength of threads, and you need to make a forecast. How do you determine the range within which you expect future observations to fall? That's where the prediction interval comes in. It sets boundaries to predict a single future outcome based on existing data. If we average the tensile strength of 50 threads and calculate the variation using standard deviation, we can establish a prediction interval that says, 'We are 95% confident a new single observation will fall within this range.'

To find this for our threads with an average strength of 78.3 kg and a standard deviation of 5.6 kg, we use a Z-score reflective of our 95% confidence level. When we apply the right Z-score, we are saying there's a buffer to accommodate expected fluctuations, making it a robust tool for quality control in manufacturing processes like thread production.
Standard Deviation
Standard deviation is a pivotal statistic in understanding how varied or spread out numbers are in a dataset. It's like measuring the typical distance from the middle point, which we call the mean. In our tensile strength example, the mean is 78.3 kg. If the standard deviation is high, it means the strengths of the individual threads are quite scattered around the mean. If it's low, the strengths are more closely huddled around that average.

For practical understanding, think of it like this: A thread producer who observes that threads generally have tensile strengths close to the mean with a small standard deviation could conclude that the production process is consistent. A large standard deviation could point towards a need for process improvement to achieve more uniform strength in the threads.
Normal Distribution
When data is normally distributed, we can effectively use a bell curve to picture how the values in our dataset are spread out. In this curve, most observations cluster around the center, which is the mean, and then taper off symmetrically towards extremes.

In the case of tensile strength, assuming a normal distribution means that most threads have a tensile strength around the average of 78.3 kg, with increasingly fewer threads being significantly stronger or weaker. This assumption lets us make predictions and set tolerances that are statistically sound, since many real-world phenomena conform to this pattern, including production and manufacturing processes.
Z-score
A Z-score is a way of quantifying how far from the mean a particular data point is, expressed in terms of standard deviations. A Z-score of 0 corresponds to the mean, while a positive or negative Z-score indicates a value above or below the mean, respectively. In our tensile strength example, to find the prediction interval with 95% confidence, we use a Z-score that corresponds to this confidence level on the bell curve. It represents how many standard deviations we need to cover to be sure that 95% of future single tensile strength measurements will fall within our prediction interval.
Tolerance Limit
A tolerance limit is different from a prediction interval in that it focuses on the range within which a certain percentage of the population falls, rather than predicting future individual outcomes. For our threads, calculating a lower 95% tolerance limit that is exceeded by 99% of the tensile strength values tells us that we are almost certain the vast majority of our products meet this minimum strength criterion. It's a critical concept for ensuring product reliability and safety, as it helps manufacturers set acceptable limits for variation in their production processes.

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Most popular questions from this chapter

In a study conducted by the Department of Zoology at. Virginia Tech. fifteen "samples" of water were collected from a certain station in the James River in order to gain some insight regarding the amount of orthophosphorous in the river. The concentration of the chemical is measured in milligrams per liter. Let us suppose that the mean at the station is not as important as the upper extremes of the distribution of the chemical at the station. Concern centers around whether the concentrations at these extremes are too large. Readings for the fifteen water samples gave a sample mean of 3.84 milligrams per liter and sample standard deviation of 3.07 milligrams per liter. Assume that the readings are a random sample from a normal distribution. Calculate a prediction interval (upper \(95 \%\) prediction limit) and a tolerance limit \((95 \%\) upper tolerance limit that exceeds \(95 \%\) of the population of value). Interpret both; that is, tell what each communicates to us about the upper extremes of the distribution of orthophosphorous at the sampling station.

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a \(96 \%\) confidence interval for the population mean of all bulbs produced by this firm.

A manufacturer of car batteries claims that his batteries will last, on average, 3 years with a variance of 1 year. If 5 of these batteries have lifetimes of 1.9 \(2.4,3.0,3.5,\) and 4.2 years, construct a \(95 \%\) confidence interval for \(\sigma^{2}\) and decide if the manufacturer's claim that \(\sigma^{2}=1\) is valid. Assume the population of battery lives to be approximately normally distributed.

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A machine is producing metal pieces that are cylindrical in shape. A sample of pieces is taken and the diameters are 1.01,0.97,1.03,1.04,0.99,0.98,0.99 \(1.01,\) and 1.03 centimeters. Find a \(99 \%\) confidence interval for the mean diameter of pieces from this machine, assuming an approximate normal distribution.

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