/*! 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 22 The following data are the measu... [FREE SOLUTION] | 91Ó°ÊÓ

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The following data are the measures of the diameters of 36 rivet heads in \(1 / 100\) of an inch. $$ \begin{array}{llllllll} 6.72 & 6.77 & 6.82 & 6.70 & 6.78 & 6.70 & 6.62 & 6.75 \\ 6.66 & 6.66 & 6.64 & 6.76 & 6.73 & 6.80 & 6.72 & 6.76 \\ 6.76 & 6.68 & 6.66 & 6.62 & 6.72 & 6.76 & 6.70 & 6.78 \\ 6.76 & 6.67 & 6.70 & 6.72 & 6.74 & 6.81 & 6.79 & 6.78 \\ 6.66 & 6.76 & 6.76 & 6.72 & & & & \end{array} $$ (a) Compute the sample mean and sample standard deviation. (b) Construct a relative frequency histogram of the data. (c) Comment on whether there is any clear indication or not that the sample came from a population that depicts a bell-shaped distribution.

Short Answer

Expert verified
The sample mean and standard deviation need computation and the result will depend on these calculations. Constructing the relative frequency histogram involves determining intervals for the data, tallying the data falling into those intervals, and plotting the resulting relative frequencies. Visual inspection of the histogram gives insight on whether the sample data are likely to have come from a population with a normal distribution.

Step by step solution

01

Calculation of Sample Mean

First, sum up all the 36 data points and then divide the sum by 36. This calculation yields the sample mean.
02

Calculation of Sample Standard Deviation

The formula for calculating the standard deviation of a sample is \(\sqrt{\frac{1}{N-1} \sum (x_i - \overline{x})^2}\), where \(x_i\) represents the dataset values, \(\overline{x}\) represents the mean of the dataset, N stands for the numbers in the dataset. After calculating the squared deviations, sum them up, divide by \(N-1\) (which is 35) and then take the square root.
03

Construction of a Relative Frequency Histogram

Begin by deciding on the intervals for the bars in the histogram. Then count the number of data points that fall within each interval. These counts are converted into relative frequencies by dividing by the number of data points (36 in this case). The histogram is then built by setting the intervals on the x-axis and the relative frequencies on the y-axis. The counts in each interval determine the height of the corresponding bar in the histogram.
04

Analysis of the Distribution Type

Observe the shape of the relative frequency histogram constructed in Step 3. If the data appears to be distributed symmetrically around the mean, the sample has a bell-shaped or normal distribution. If there is a skew to the left or right, the sample does not have a normal distribution. Keep in mind that a single sample may not depict the true distribution of the population accurately.

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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 measure of central tendency, which reveals the average value of a dataset. To compute the sample mean, you first add up all the data points. In our case with 36 rivet heads, you would sum all their diameters together. Once you have this sum, you divide by the number of data points, which is 36 in this exercise.

The formula for the sample mean, denoted as \( \overline{x} \), is:
  • \( \overline{x} = \frac{1}{N} \sum_{i=1}^{N} x_i \)
Where \( N \) is the number of observations (36), and \( x_i \) represents each individual data point.

This mean helps in determining the average size of the rivet heads. It is crucial for quality control as it provides a single value summary of the entire dataset.
Sample Standard Deviation
Sample standard deviation measures how spread out the numbers in a dataset are. It gives insight into the variability of the data. In simpler terms, it tells us how much the individual measurements differ from the average one.

To calculate the sample standard deviation \( s \), use the following formula:
  • \( s = \sqrt{\frac{1}{N-1} \sum (x_i - \overline{x})^2} \)
Where:
  • \( x_i \) is each individual data point
  • \( \overline{x} \) is the sample mean
  • \( N \) is the total number of data points

The expression \( (x_i - \overline{x})^2 \) is the squared deviation of each data point from the mean.

Standard deviation is valuable for assessing the consistency of the rivet head sizes. A smaller standard deviation indicates that the sizes are closely packed near the mean, while a larger one suggests more variation.
Frequency Histogram
A frequency histogram provides a graphical representation of how often different values occur in a dataset. It is an effective tool for visualizing the distribution of data.

To create a frequency histogram, you must first determine suitable intervals or "bins" for the data values. You then count how many values fall into each bin. This count is divided by the total number of observations to get the relative frequency, which is used to construct the histogram bars.

The histogram has:
  • Intervals on the x-axis
  • Relative frequencies on the y-axis
This visual tool allows you to quickly see which ranges of rivet head sizes are most common, making it convenient for detecting patterns or anomalies in the data.
Bell-Shaped Distribution
A bell-shaped distribution, often referred to as a normal distribution, is a common probability distribution in statistics, characterized by its symmetric shape. In a bell-shaped distribution, most of the observations cluster around the central peak, which is the mean. As you move away from the mean, the frequency of observations decreases.

Identifying a bell-shaped distribution in your dataset is essential because it helps in making probabilistic interpretations and statistical inferences.

To determine if the rivet head sizes follow a bell-shaped distribution, you should examine the frequency histogram you've plotted. If the histogram shows symmetry around the mean with most of the data near the central peak, followed by a gradual decrease on either side, then the distribution is likely bell-shaped.

However, note that a single sample might not perfectly exhibit a bell shape due to sample variability, emphasizing the importance of larger sample sizes for more accurate analysis.

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Most popular questions from this chapter

The nicotine contents, in milligrams, for 40 cigarettes of a certain brand were recorded as follows: $$ \begin{array}{rrrrr} 1.09 & 1.92 & 2.31 & 1.79 & 2.28 \\ 1.74 & 1.47 & 1.97 & 0.85 & 1.24 \\ 1.58 & 2.03 & 1.70 & 2.17 & 2.55 \\ 2.11 & 1.86 & 1.90 & 1.68 & 1.51 \\ 1.64 & 0.72 & 1.69 & 1.85 & 1.82 \\ 1.79 & 2.46 & 1.88 & 2.08 & 1.67 \\ 1.37 & 1.93 & 1.40 & 1.64 & 2.09 \\ 1.75 & 1.63 & 2.37 & 1.75 & 1.69 \end{array} $$ (a) Find the sample mean and sample median. (b) Find the sample standard deviation.

The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint. $$ \begin{array}{lllll} 3.4 & 2.5 & 4.8 & 2.9 & 3.6 \\ 2.8 & 3.3 & 5.6 & 3.7 & 2.8 \\ 4.4 & 4.0 & 5.2 & 3.0 & 4.8 \end{array} $$ Assume that the measurements are a simple random sample. (a) What is the sample size for the above sample? (b) Calculate the sample mean for this data. (c) Calculate the sample median. (d) Plot the data by way of a dot plot. (e) Compute the \(20 \%\) trimmed mean for the above data set.

The following data set is related to that in Exercise \(1.24 .\) It gives the percent of the families that are in the upper income level at the same individual schools in the same order as in Exercise 1.24 . $$ \begin{array}{rrrrrrrr} 72.2 & 31.9 & 26.5 & 29.1 & 27.3 & 8.6 & 22.3 & 26.5 \\ 20.4 & 12.8 & 25.1 & 19.2 & \mathbf{2 4 . 1} & 58.2 & 68.1 & 89.2 \\ 55.1 & 9.4 & 14.5 & 13.9 & 20.7 & 17.9 & 8.5 & 55.4 \\ 38.1 & 54.2 & 21.5 & 26.2 & 59.1 & 43.3 & & \end{array} $$ (a) Calculate the sample mean. (b) Calculate the sample median. (c) Construct a relative frequency histogram of the data. (d) Compute the \(10 \%\) trimmed mean. Compare with the results in (a) and (b) and comment.

Show that the \(n\) pieces of information in \(\sum_{j=1}^{N}\left(x_{i}-\right.\) \(\bar{x})^{2}\) are not independent; that is, show that, \(\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)=\mathbf{0}\).

A certain polymer is used for evacuation systems for aircraft. It is important that the polymer be resistant to the aging process. Twenty specimens of the polymer were used in an experiment. Ten were assigned randomly to be exposed to the accelerated batch aging process that involved exposure to high temperatures for 10 days. Measurements of tensile strength of the specimens were made and the following data were recorded on tensile strength in psi. $$ \begin{array}{llllll} \text { No aging: } & 227 & 222 & 218 & 217 & 225 \\ & 218 & 216 & 229 & 228 & 221 \\ \text { Aging: } & 219 & 214 & 215 & 211 & 209 \\ & 218 & 203 & 204 & 201 & 205 \end{array} $$ (a) Do a dot plot of the data. (b) From your plot, does it appear as if the aging process has had an effect on the tensile strength of this polymer? Explain. (c) Calculate the sample mean tensile strength of the two samples. (d) Calculate the median for both. Discuss the similarity or lack of similarity between the mean and median of each group.

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