/*! 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 26 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. A sample of temperatures for initiating a certain chemical reaction yielded a sample average \(\left({ }^{\circ} \mathrm{C}\right)\) of \(87.3\) and a sample standard deviation of \(1.04\). What are the sample average and standard deviation measured in "F? [Hint: \(F=\frac{2}{5} C+32\).]

Short Answer

Expert verified
\(\bar{y} = a\bar{x} + b\); \(s_y^2 = a^2s_x^2\). \(\bar{F} = 189.14\)°F; \(s_F = 1.872\)°F.

Step by step solution

01

Understanding the Relationship between Means

Given that the relation between each pair of \((x_i, y_i)\) is defined as \(y_i = ax_i + b\), we need to find the relationship between \(\bar{x}\) and \(\bar{y}\). The mean of \(y\) can be calculated as \(\bar{y} = \frac{1}{n} \sum_{i=1}^n y_i\). Substituting \(y_i = ax_i + b\), we get \(\bar{y} = \frac{1}{n} \sum_{i=1}^n (ax_i + b) = a\bar{x} + b\). This shows that \(\bar{y} = a\bar{x} + b\).
02

Understanding the Relationship between Variances

The variance of \(y\), \(s_y^2\), is \(\frac{1}{n-1} \sum_{i=1}^n (y_i - \bar{y})^2\). Substituting \(y_i = ax_i + b\), we derive \(s_y^2 = \frac{1}{n-1} \sum_{i=1}^n (ax_i + b - a\bar{x} - b)^2 = a^2 s_x^2\). Therefore, the relationship between the variances is \(s_y^2 = a^2 s_x^2\).
03

Converting Celsius Mean to Fahrenheit Mean

The temperature in Fahrenheit, \(F\), can be converted from Celsius using \(F = \frac{9}{5}C + 32\). Given \(\bar{C} = 87.3\), we substitute it into the formula: \(\bar{F} = \frac{9}{5} \times 87.3 + 32 = 189.14\). The sample average in Fahrenheit is therefore 189.14°F.
04

Converting Celsius Standard Deviation to Fahrenheit Standard Deviation

To find the standard deviation in Fahrenheit, we use the linear transformation. Since \(F = \frac{9}{5}C + 32\), the standard deviation in Fahrenheit is scaled by \(\frac{9}{5}\) only (constant addition doesn't affect the spread): \(s_F = \frac{9}{5} \times 1.04 = 1.872\). The sample standard deviation in Fahrenheit is therefore 1.872°F.

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Key Concepts

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

Mean Transformation
When dealing with linear transformations in statistics, understanding how the mean changes is crucial. The process involves transforming the data linearly using an equation such as \( y_i = ax_i + b \). Here, \( a \) and \( b \) are constants. The original data has a mean \( \bar{x} \), which represents the central tendency.
The transformed mean \( \bar{y} \) is obtained using the same linear equation applied to individual data points. It results in the equation \( \bar{y} = a \bar{x} + b \). This means the new mean is a scaled version of the original mean, amplified by \( a \), and adjusted by adding \( b \). This change reflects how the shift and scale affect the central location of the data.
Key points to remember:
  • Scaling the data by \( a \) changes the spread.
  • Adding \( b \) affects the shift of the mean.
Variance Transformation
Variance is a measure of the spread of data values around the mean. When a linear transformation is applied to data, its variance also changes. The transformation \( y_i = ax_i + b \) affects the variance of the dataset, which is denoted \( s_y^2 \) for the transformed data.
The transformation impacts variance through scaling, where \( s_y^2 = a^2 s_x^2 \). Here, \( s_x^2 \) is the original variance before transformation. Notice that only the scaling factor \( a^2 \) impacts variance, while the addition of \( b \) does not.
This is because variance, being a measure of dispersion, is unaffected by uniform shifts:
  • Larger values of \( a \) increase variance, making the data more spread out.
  • Smaller values of \( a \) decrease variance, reducing the spread.
  • Addition constant \( b \) has no effect on the variance.
Standard Deviation Conversion
When converting the standard deviation through a linear transformation, you modify how dispersed the data is around the mean. Standard deviation, denoted \( s \), is the square root of variance. When transforming data using the formula \( y_i = ax_i + b \), the conversion impacts standard deviation as \( s_y = |a| s_x \).
This transformation shows that the standard deviation scales linearly with \( |a| \), while the addition of \( b \) does not alter it. It's crucial to remember that although standard deviation is a measure of spread, transformations only affect it through scaling. Here's why that matters:
  • Multiplying the data by \( a \) scales the deviation proportionally.
  • Adding a constant \( b \) shifts the data but doesn't impact the deviation.
Temperature Conversion in Statistics
Converting temperatures between scales, such as Celsius and Fahrenheit, involves linear transformations that apply to both means and standard deviations. For Celsius to Fahrenheit conversions, the formula used is \( F = \frac{9}{5}C + 32 \). This represents a linear transformation.
To convert the mean temperature, apply the formula to the mean in Celsius. If the mean Celsius temperature is \( \bar{C} \), the Fahrenheit mean \( \bar{F} \) becomes \( \frac{9}{5} \times \bar{C} + 32 \). This transformation affects the central figure but keeps the relational dynamics of temperature consistent.
For standard deviation, the focus is only on the scaling factor of \( \frac{9}{5} \). This is because a shift (constant addition like \( +32 \)) does not affect deviation. Thus, the Fahrenheit standard deviation \( s_F \) is \( \frac{9}{5} \times s_C \), where \( s_C \) is the initial Celsius deviation.
  • Transforming mean emphasizes shifts and scaling in temperature.
  • Converting deviation highlights the effect of scaling without shifts.

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Most popular questions from this chapter

A study carried out to investigate the distribution of total braking time (reaction time plus accelerator-to-brake movement time, in ms) during real driving conditions at \(60 \mathrm{~km} /\) hr gave the following summary information on the distribution of times ("A Field Study on Braking Responses During Driving," Ergonomics, 1995: 1903–1910): mean \(=535\) median \(=500 \quad\) mode \(=500\) \(\mathrm{sd}=96 \quad\) minimum \(=220 \quad\) maximum \(=925\) 5 th percentile \(=400 \quad 10\) th percentile \(=430\) 90 th percentile \(=640 \quad 95\) th percentile \(=720\) What can you conclude about the shape of a histogram of this data? Explain your reasoning.

For each of the following hypothetical populations, give a plausible sample of size 4 : a. All distances that might result when you throw a football b. Page lengths of books published 5 years from now c. All possible earthquake-strength measurements (Richter scale) that might be recorded in California during the next year d. All possible yields (in grams) from a certain chemical reaction carried out in a laboratory

Consider a sample \(x_{1}, x_{2}, \ldots, x_{n}\) and suppose that the values of \(\bar{x}, s^{2}\), and \(s\) have been calculated. a. Let \(y_{i}=x_{i}-\bar{x}\) for \(i=1, \ldots, n\). How do the values of \(s^{2}\) and \(s\) for the \(y_{i}\) 's compare to the corresponding values for the \(x_{i}\) 's? Explain. b. Let \(z_{i}=\left(x_{i}-\bar{x}\right) / s\) for \(i=1, \ldots, n\). What are the values of the sample variance and sample standard deviation for the \(z_{i} s\) ?

The article "A Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankease Lubricants" (Lubric. Engr., 1984: 75-83) reported the following data on oxidation-induction time (min) for various commercial oils: \(\begin{array}{lllllllllll}87 & 103 & 130 & 160 & 180 & 195 & 132 & 145 & 211 & 105 & 145\end{array}\) \(\begin{array}{llllllll}153 & 152 & 138 & 87 & 99 & 93 & 119 & 129\end{array}\) a. Calculate the sample variance and standard deviation. b. If the observations were reexpressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the reexpression.

The article "Effects of Short-Term Warming on Low and High Latitude Forest Ant Communities" (Ecoshpere, May 2011, Article 62) described an experiment in which observations on various characteristics were made using minichambers of three different types: (1) cooler (PVC frames covered with shade cloth), (2) control (PVC frames only), and (3) warmer (PVC frames covered with plastic). One of the article's authors kindly supplied the accompanying data on the difference between air and soil temperatures \(\left({ }^{\circ} \mathrm{C}\right)\). \(\begin{array}{ccc}\text { Cooler } & \text { Control } & \text { Warmer } \\\ 1.59 & 1.92 & 2.57 \\ 1.43 & 2.00 & 2.60 \\ 1.88 & 2.19 & 1.93 \\ 1.26 & 1.12 & 1.58 \\ 1.91 & 1.78 & 2.30 \\ 1.86 & 1.84 & 0.84 \\ 1.90 & 2.45 & 2.65 \\ 1.57 & 2.03 & 0.12 \\ 1.79 & 1.52 & 2.74 \\ 1.72 & 0.53 & 2.53 \\\ 2.41 & 1.90 & 2.13 \\ 2.34 & & 2.86 \\ 0.83 & & 2.31 \\ 1.34 & & 1.91 \\\ 1.76 & & \end{array}\) a. Compare measures of center for the three different samples. b. Calculate, interpret, and compare the standard deviations for the three different samples. c. Do the fourth spreads for the three samples convey the same message as do the standard deviations about relative variability? d. Construct a comparative boxplot (which was included in the cited article) and comment on any interesting features.

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