/*! 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 67 a. Let \(a\) and \(b\) be consta... [FREE SOLUTION] | 91Ó°ÊÓ

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a. Let \(a\) and \(b\) be constants and let \(y_{i}=a x_{i}+b\) for \(i=1,2, \ldots, n\). What are the relationships between \(\bar{x}\) and \(\bar{y}\) and between \(s_{x}^{2}\) and \(s_{y}^{2}\) ? b. The Australian army studied the effect of high temperatures and humidity on human body temperature (Neural Network Training on Human Body Core Temperature Data, Technical Report DSTO TN-0241, Combatant Protection Nutrition Branch, Aeronautical and Maritime Research Laboratory). They found that, at \(30^{\circ} \mathrm{C}\) and \(60 \%\) relative humidity, the sample average body temperature for nine soldiers was \(38.21^{\circ} \mathrm{C}\), with standard deviation \(.318^{\circ} \mathrm{C}\). What are the sample average and the standard deviation in \({ }^{\circ} \mathrm{F}\) ?

Short Answer

Expert verified
\(\bar{y} = a\bar{x} + b\), \(s_y^2 = a^2 s_x^2\); \(\bar{F} = 100.778\degree F\), \(s_F = 0.5724\degree F\).

Step by step solution

01

Understanding the Linear Transformation

Given the linear relationship, \(y_i = ax_i + b\), we can express \(\bar{y}\) in terms of \(\bar{x}\). The average \(\bar{y}\) is calculated as \(\bar{y} = \frac{1}{n} \sum y_i = \frac{1}{n} \sum (a x_i + b)\). This simplifies to \(a\bar{x} + b\) using properties of sums.
02

Determining the Relationship Between Variances

The variance \(s_y^2\) can be found from \(s_x^2\) since variance is affected by scaling but not by shifting. Given, \(y_i = ax_i + b\), the variance \(s_y^2 = a^2 s_x^2\). The constant \(b\) does not affect the variance.
03

Conversion from Celsius to Fahrenheit

For part b, use the conversion formula \(F = \frac{9}{5}C + 32\) to convert the average temperature \(38.21\degree C\) to Fahrenheit. Substitute to find \(\bar{F} = \frac{9}{5} \times 38.21 + 32 = 100.778\degree F\).
04

Calculating the Standard Deviation in Fahrenheit

Use the relationship between Celsius and Fahrenheit to convert the standard deviation. Since standard deviation measures spread and not location, conversion just involves multiplying by the slope (9/5) without adding the 32. Thus, \(s_F = \frac{9}{5} \times 0.318 = 0.5724\degree F\).

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Key Concepts

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

Linear Transformations
Linear transformations in statistics involve using a linear equation to convert one set of numbers to another. In this context, it's expressed as \(y_i = ax_i + b\), where \(x_i\) is your original variable, and \(y_i\) is the transformed variable.
This transformation can adjust data through multiplying by \(a\) (scaling) and adding \(b\) (shifting). Such transformations are useful in standardizing data or converting units.
  • Scaling (by \(a\)): Multiplies every data point, affecting the mean and variance.
  • Shifting (by \(b\)): Adds a constant to each data point, changing the mean but not the variance.
The effect on the mean after transformation is that \(\bar{y} = a\bar{x} + b\), demonstrating both scaling and shifting. Variance is affected only by scaling: \(s^2_y = a^2 s^2_x\).
Understanding these effects helps in predicting how data distributions change with transformations.
Sample Mean
The sample mean is a key measure of central tendency, representing the average of a set of data points. When a linear transformation \(y_i = ax_i + b\) is applied, the mean \(\bar{y}\) of the transformed data can be determined from the original mean \(\bar{x}\).
Using the linear transformation properties, we evaluate \(\bar{y}\) as:
  • \(\bar{y} = a\bar{x} + b\)
Here, scaling by \(a\) directly influences the magnitude of the mean, while adding \(b\) shifts the entire distribution. This is crucial for interpreting transformed data in practical scenarios like normalizing scores or adjusting monetary values.
By understanding the adjustments to the sample mean under different conditions, analysts can accurately interpret results regardless of transformation.
Variance Transformation
Variance measures data spread, and its behavior under a linear transformation \(y_i = ax_i + b\) is specific. When dealing with variability, it's essential to note that only the scaling factor \(a\) affects variance:
  • Transforming variance: \(s^2_y = a^2 s^2_x\)
Shifting by \(b\) does not alter the variance because variance is about spread, not about absolute position. Thus, all shifts affect only mean but leave variance intact.
Such understanding is particularly pertinent when data requires conversion for consistency, like converting units in physics or financial models. It assures that only true variability, not arbitrary adjustments, is considered in analyses.
Temperature Conversion
Transforming temperatures between scales, such as Celsius to Fahrenheit, is a practical application of linear transformations. The formula \(F = \frac{9}{5}C + 32\) adjusts measurements between the two widely used scales.
The conversion entails:
  • Scaling by \(\frac{9}{5}\): impacts spread, relevant for measures like standard deviation.
  • Shifting by 32: adjusts base, altering mean but not spread.
The mean temperature and variability (standard deviation) in \(^{\circ}F\) is calculated by:
  • Mean: \(\bar{F} = \frac{9}{5} \times 38.21 + 32 = 100.778^{\circ}F\)
  • Standard Deviation: \(s_F = \frac{9}{5} \times 0.318 = 0.5724^{\circ}F\)
These computations allow consistent interpretation across different units, vital in scientific and medical studies, to facilitate comparisons and conclusions.

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Most popular questions from this chapter

Fire load \(\left(\mathrm{MJ} / \mathrm{m}^{2}\right)\) is the heat energy that could be released per square meter of floor area by combustion of contents and the structure itself. The article "Fire Loads in Office Buildings" \((J .\) Struct. Engrg., 1997: \(365-368\) ) gave the following cumulative percentages (read from a graph) for fire loads in a sample of 388 rooms: \(\begin{array}{lccccc}\text { Value } & 0 & 150 & 300 & 450 & 600 \\ \text { Cumulative \% } & 0 & 19.3 & 37.6 & 62.7 & 77.5 \\ \text { Value } & 750 & 900 & 1050 & 1200 & 1350 \\ \text { Cumulative \% } & 87.2 & 93.8 & 95.7 & 98.6 & 99.1 \\ \text { Value } & 1500 & 1650 & 1800 & 1950 & \\ \text { Cumulative \% } & 99.5 & 99.6 & 99.8 & 100.0\end{array}\) a. Construct a relative frequency histogram and comment on interesting features. b. What proportion of fire loads are less than 600 ? At least 1200 ? c. What proportion of the loads are between 600 and 1200 ?

Every score in the following batch of exam scores is in the 60 's, 70 's, 80 's, or 90 's. A stem-and-leaf display with only the four stems \(6,7,8\), and 9 would not give a very detailed description of the distribution of scores. In such situations, it is desirable to use repeated stems. Here we could repeat the stem 6 twice, using 6L for scores in the low 60's (leaves 0, 1,2, 3 , and 4) and \(6 \mathrm{H}\) for scores in the high 60 's (leaves \(5,6,7,8\), and 9). Similarly, the other stems can be repeated twice to obtain a display consisting of eight rows. Construct such a display for the given scores. What feature of the data is highlighted by this display? \(\begin{array}{lllllllllllll}74 & 89 & 80 & 93 & 64 & 67 & 72 & 70 & 66 & 85 & 89 & 81 & 81 \\ 71 & 74 & 82 & 85 & 63 & 72 & 81 & 81 & 95 & 84 & 81 & 80 & 70 \\ 69 & 66 & 60 & 83 & 85 & 98 & 84 & 68 & 90 & 82 & 69 & 72 & 87 \\ 88 & & & & & & & & & & & & \end{array}\)

The accompanying data set consists of observations on shower-flow rate (L/min) for a sample of \(n=129\) houses in Perth, Australia ("An Application of Bayes Methodology to the Analysis of Diary Records in a Water Use Study," J. Amer. Statist. Assoc., 1987: 705-711): \(\begin{array}{rrrrrrrrrr}4.6 & 12.3 & 7.1 & 7.0 & 4.0 & 9.2 & 6.7 & 6.9 & 11.5 & 5.1 \\ 11.2 & 10.5 & 14.3 & 8.0 & 8.8 & 6.4 & 5.1 & 5.6 & 9.6 & 7.5 \\\ 7.5 & 6.2 & 5.8 & 2.3 & 3.4 & 10.4 & 9.8 & 6.6 & 3.7 & 6.4 \\ 8.3 & 6.5 & 7.6 & 9.3 & 9.2 & 7.3 & 5.0 & 6.3 & 13.8 & 6.2 \\ 5.4 & 4.8 & 7.5 & 6.0 & 6.9 & 10.8 & 7.5 & 6.6 & 5.0 & 3.3 \\ 7.6 & 3.9 & 11.9 & 2.2 & 15.0 & 7.2 & 6.1 & 15.3 & 18.9 & 7.2 \\ 5.4 & 5.5 & 4.3 & 9.0 & 12.7 & 11.3 & 7.4 & 5.0 & 3.5 & 8.2 \\ 8.4 & 7.3 & 10.3 & 11.9 & 6.0 & 5.6 & 9.5 & 9.3 & 10.4 & 9.7 \\ 5.1 & 6.7 & 10.2 & 6.2 & 8.4 & 7.0 & 4.8 & 5.6 & 10.5 & 14.6 \\ 10.8 & 15.5 & 7.5 & 6.4 & 3.4 & 5.5 & 6.6 & 5.9 & 15.0 & 9.6 \\ 7.8 & 7.0 & 6.9 & 4.1 & 3.6 & 11.9 & 3.7 & 5.7 & 6.8 & 11.3 \\ 9.3 & 9.6 & 10.4 & 9.3 & 6.9 & 9.8 & 9.1 & 10.6 & 4.5 & 6.2 \\ 8.3 & 3.2 & 4.9 & 5.0 & 6.0 & 8.2 & 6.3 & 3.8 & 6.0 & \end{array}\) a. Construct a stem-and-leaf display of the data. b. What is a typical, or representative, flow rate? c. Does the display appear to be highly concentrated or spread out? d. Does the distribution of values appear to be reasonably symmetric? If not, how would you describe the departure from symmetry? e. Would you describe any observation as being far from the rest of the data (an outlier)?

A transformation of data values by means of some mathematical function, such as \(\sqrt{x}\) or \(1 / x\), can often yield a set of numbers that has "nicer" statistical properties than the original data. In particular, it may be possible to find a function for which the histogram of transformed values is more symmetric (or, even better, more like a bell-shaped curve) than the original data. As an example, the article "Time Lapse Cinematographic Analysis of BerylliumLung Fibroblast Interactions" (Environ. Res., 1983: 34-43) reported the results of experiments designed to study the behavior of certain individual cells that had been exposed to beryllium. An important characteristic of such an individual cell is its interdivision time (IDT). IDTs were determined for a large number of cells both in exposed (treatment) and unexposed (control) conditions. The authors of the article used a logarithmic transformation, that is, transformed value \(=\log _{10}\) (original value). Consider the following representative IDT data: \(\begin{array}{lllllll}28.1 & 31.2 & 13.7 & 46.0 & 25.8 & 16.8 & 34.8 \\ 62.3 & 28.0 & 17.9 & 19.5 & 21.1 & 31.9 & 28.9 \\ 60.1 & 23.7 & 18.6 & 21.4 & 26.6 & 26.2 & 32.0 \\ 43.5 & 17.4 & 38.8 & 30.6 & 55.6 & 25.5 & 52.1 \\ 21.0 & 22.3 & 15.5 & 36.3 & 19.1 & 38.4 & 72.8 \\ 48.9 & 21.4 & 20.7 & 57.3 & 40.9 & & \end{array}\) Use class intervals \(10-20,20-30, \ldots\) to construct a histogram of the original data. Use intervals \(1.1-1.2\), \(1.2-1.3, \ldots\) to do the same for the transformed data. What is the effect of the transformation?

a. Give three different examples of concrete populations and three different examples of hypothetical populations. b. For one each of your concrete and your hypothetical populations, give an example of a probability question and an example of an inferential statistics question.

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