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Eight measurements were made on the inside diameter of forged piston rings used in an automobile engine. The data (in millimeters) are 74.001,74.003,74.015,74.000 , \(74.005,74.002,74.005,\) and \(74.004 .\) Calculate the sample mean and sample standard deviation, construct a dot diagram, and comment on the data.

Short Answer

Expert verified
Mean: 74.0044 mm; Std Dev: 0.00366 mm; Data are consistent with low variability.

Step by step solution

01

Calculate the Sample Mean

The sample mean is calculated as the sum of all measurements divided by the number of measurements. The formula is:\[\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}\]Where \(x_i\) is each measurement and \(n\) is the number of measurements. For our data, \[\bar{x} = \frac{74.001 + 74.003 + 74.015 + 74.000 + 74.005 + 74.002 + 74.005 + 74.004}{8} = \frac{592.035}{8} = 74.004375\] mm.
02

Calculate the Sample Standard Deviation

The sample standard deviation is calculated using the formula: \[s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}}\]First, find the squared deviations from the mean, then sum them:\((74.001 - 74.004375)^2 + (74.003 - 74.004375)^2 + (74.015 - 74.004375)^2 + (74.000 - 74.004375)^2 + (74.005 - 74.004375)^2 + (74.002 - 74.004375)^2 + (74.005 - 74.004375)^2 + (74.004 - 74.004375)^2 = 0.00009375\). Divide this sum by \(n-1\) and take the square root:\[s = \sqrt{\frac{0.00009375}{8-1}} = \sqrt{0.00001339} \approx 0.00366\] mm.
03

Construct the Dot Diagram

To construct a dot diagram, plot each measurement on a number line. Each unique measurement gets a dot at its position on the line. Duplicate measurements are stacked vertically above their original position. For the data given, dots will be plotted at each measured value, with stacking for repeated values like 74.005.
04

Comment on the Data

The measurements are closely clustered around the mean of 74.004375 mm, indicating consistency. The standard deviation of approximately 0.00366 mm suggests a small spread, meaning the measurements do not vary significantly from each other.

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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 is a fundamental statistical measure used to determine the central tendency of a data set. When you have a series of numbers, like our piston ring measurements, the sample mean gives you a single value that represents the average of those numbers. To calculate it, add up all the values and then divide by the number of values.

Here's the formula for the sample mean:
  • Compute the sum of all measurements.
  • Divide by the number of measurements.
The result is called the sample mean, represented by \( \bar{x} \). In our case, the sample mean of piston ring diameters was calculated as follows:
  • Add: 74.001 + 74.003 + 74.015 + 74.000 + 74.005 + 74.002 + 74.005 + 74.004 = 592.035
  • Divide: \( \frac{592.035}{8} = 74.004375 \)
Thus, the sample mean is 74.004375 mm. It's a useful measure because it gives us a straightforward, simplified overview of the data's center.
Sample Standard Deviation
The sample standard deviation is another important statistical tool that measures the amount of variation or dispersion in a data set. In simple terms, it tells us how spread out the measurements are from their average value (the sample mean). A smaller standard deviation indicates that the measurements are closely clustered around the mean, while a larger one suggests more spread.

Here’s how to compute it:
  • Determine the deviation of each measurement from the sample mean.
  • Square each deviation to eliminate negative values.
  • Sum all squared deviations.
  • Divide this sum by one less than the number of measurements (\( n-1 \)), a step known as the degrees of freedom.
  • Take the square root of the result to get the standard deviation.
For our exercise, the squared deviations were calculated and summed to 0.00009375. Dividing by 7 (since there are 8 measurements, \( n-1 = 7 \)), we have:
  • \( s = \sqrt{\frac{0.00009375}{7}} \approx 0.00366 \) mm
Such a small standard deviation suggests that our piston ring measurements are very consistent, showing minimal variability among them.
Dot Diagram
A dot diagram, or dot plot, is a simple and effective way to visualize the frequency of each value in a data set. It consists of a simple number line where each measurement is represented by a dot above its value. When a value is repeated, dots are stacked vertically.

This visualization helps in quickly identifying clusters, gaps, peaks, and outliers. It’s particularly helpful for smaller data sets, as it gives a clear, visual representation of the distribution.

In our exercise with the piston ring diameters, you would:
  • Draw a horizontal number line covering the range of the data set.
  • Place a dot above the number line for each measurement. Repeat for duplicate values, like 74.005.
To comment on the data, you can see the values are clustered near the mean, with no extreme outliers, indicating stable manufacturing quality. The dot plot shows how often each specific measurement was reached, allowing a quick visual snapshot of data distribution.

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Most popular questions from this chapter

Cloud seeding, a process in which chemicals such as silver iodide and frozen carbon dioxide are introduced by aircraft into clouds to promote rainfall was widely used in the 20 th cen tury. Recent research has questioned its effectiveness [Journal of Atmospheric Research (2010, Vol. 97 (2), pp. 513-525)]. An experiment was performed by randomly assigning 52 clouds to be seeded or not. The amount of rain generated was then measured in acre-feet. Here are the data for the unseeded and seeded clouds: into clouds to promote rainfall was widely used in the 20 th century. Recent research has questioned its effectiveness [Journal of Atmospheric Research (2010, Vol. 97 (2), pp. 513-525)]. An experiment was performed by randomly assigning 52 clouds to be seeded or not. The amount of rain generated was then measured in acre-feet. Here are the data for the unseeded and seeded clouds: Unseeded: \(\begin{array}{lllllllll}81.2 & 26.1 & 95.0 & 41.1 & 28.6 & 21.7 & 11.5 & 68.5 & 345.5 & 321.2\end{array}\) \(\begin{array}{lllllllll}1202.6 & 1.0 & 4.9 & 163.0 & 372.4 & 244.3 & 47.3 & 87.0 & 26.3 & 24.4\end{array}\) \(\begin{array}{llllll}830.1 & 4.9 & 36.6 & 147.8 & 17.3 & 29.0\end{array}\) Seeded: \(\begin{array}{lllllll}274.7 & 302.8 & 242.5 & 255.0 & 17.5 & 115.3 & 31.4 & 703.4 & 334.1\end{array}\) \(\begin{array}{llllllll}1697.8 & 118.3 & 198.6 & 129.6 & 274.7 & 119.0 & 1656.0 & 7.7 & 430.0\end{array}\) \(\begin{array}{llllll}40.6 & 92.4 & 200.7 & 32.7 & 4.1 & 978.0 & 489.1 & 2745.6\end{array}\) Find the sample mean, sample standard deviation, and range of rainfall for (a) All 52 clouds (b) The unseeded clouds (c) The seeded clouds

For any set of data values, is it possible for the sample standard deviation to be larger than the sample mean? If so, give an example.

Suppose that the sample size \(n\) is such that the quantity \(n T / 100\) is not an integer. Develop a procedure for obtaining a trimmed mean in this case.

The \(\mathrm{pH}\) of a solution is measured eight times by one operator using the same instrument. She obtains the following data: \(7.15,7.20,7.18,7.19,7.21,7.20,7.16,\) and \(7.18 .\) Calculate the sample mean and sample standard deviation. Comment on potential major sources of variability in this experiment.

Consider the following two samples: Sample 1: 10,9,8,7,8,6,10,6 Sample 2: 10,6,10,6,8,10,8,6 (a) Calculate the sample range for both samples. Would you conclude that both samples exhibit the same variability? Explain. (b) Calculate the sample standard deviations for both samples. Do these quantities indicate that both samples have the same variability? Explain. (c) Write a short statement contrasting the sample range versus the sample standard deviation as a measure of variability.

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