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Chapter 1: Problem 69

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 \({ }^{\circ} \mathrm{F}\) ?

Short Answer

Expert verified
\( \bar{y} = a \bar{x} + b \); \( s_y^2 = a^2 s_x^2 \). In Fahrenheit: Average = 189.14, SD = 1.872.

Step by step solution

01

Identify the Conversion Formula

Begin by recognizing the conversion formula from Celsius to Fahrenheit, which is given by:\[ T(^{\circ}F) = T(^{\circ}C) \times \frac{9}{5} + 32 \]
02

Convert the Sample Average (Celsius to Fahrenheit)

Given the sample average temperature in Celsius, \( \bar{x} = 87.3^{\circ} C \), use the conversion formula:\[ \bar{y} = \bar{x} \times \frac{9}{5} + 32 = 87.3 \times \frac{9}{5} + 32 = 189.14^{\circ} F \]
03

Convert the Standard Deviation (Celsius to Fahrenheit)

Given the standard deviation in Celsius, \( s_x = 1.04^{\circ} C \), convert it using the scale factor \( \frac{9}{5} \):\[ s_y = s_x \times \frac{9}{5} = 1.04 \times \frac{9}{5} = 1.872^{\circ} F \]
04

Conclusion of Conversion

The sample average temperature in Fahrenheit is \( 189.14^{\circ} F \) and the standard deviation is \( 1.872^{\circ} F \).
05

Relationship between Means and Variances

For part a, observe that the relationship between the means of \( x \) and \( y \) when \( y_i = a x_i + b \) is given by:\[ \bar{y} = a \bar{x} + b \]And the relationship between variances is:\[ s^2_y = a^2 s^2_x \]This is because variances measure the spread and are not affected by the constant \( b \).

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Key Concepts

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

Mean and Variance Transformation
Understanding how means and variances transform when altering variables is fundamental in statistics. Imagine you have a set of data points, \(x_i\), and you apply a transformation to it in the form of \(y_i = a x_i + b\). This is a linear transformation where \(a\) and \(b\) are constants.
Here's what happens:
  • The mean, \(\bar{x}\), will transform to \(\bar{y} = a \bar{x} + b\). This makes sense because you are scaling and shifting all the data by the same amounts. Scaling affects the spread, while shifting simply changes the position of the mean.
  • The variance, \(s_x^2\), however, follows a different rule. It becomes \(s_y^2 = a^2 s_x^2\). Notice that the constant \(b\) does not affect the variance. That's because variance measures how spread out the data is, and a consistent shift (such as adding \(b\) to each value) doesn鈥檛 change the spread.
Remember, variance depends on the square of the factor because it involves squared deviations from the mean.
Temperature Conversion
Converting temperatures from Celsius to Fahrenheit (or the other way around) is a practical example of mean and variance transformations. To convert Celsius to Fahrenheit, use:
  • Conversion Formula: \[ T(^{\circ}F) = T(^{\circ}C) \times \frac{9}{5} + 32 \] where 9/5 is the scaling factor and 32 is the constant offset.

To apply this in practice, consider a sample mean temperature of \(87.3^{\circ}C\):
  • Convert this mean to Fahrenheit: \( \bar{y} = 87.3 \times \frac{9}{5} + 32 = 189.14^{\circ}F \).

For standard deviation, which involves spread and not direct addition:
  • Use only the scaling factor: Convert the standard deviation of \(1.04^{\circ}C\) to Fahrenheit by multiplying by \( \frac{9}{5} \), resulting in \(1.872^{\circ}F\).
Here, only the scaling influences the variability.
Standard Deviation Transformation
Standard deviation is a statistical measure that quantifies the amount of variation or dispersion in a set of data values. When you transform data like \(y_i = a x_i + b\), its standard deviation \(s_x\) undergoes a transformation:
  • It becomes \(s_y = |a| s_x\). The presence of \(a\), the scale factor, modifies the measure of dispersion but is not affected by \(b\) because adding a constant shifts all values equally without altering spread.

In temperature conversion, where \(a = \frac{9}{5}\), the standard deviation transformation highlights how the spread of data scales directly with changes in measurement units.
  • This is why it's important to remember that for standard deviation, the absolute value of \(a\) is used, ensuring positive scaling.
Thinking of standard deviation as responsive to scaling (not shifts) helps you predict how data dispersion behaves under transformations.

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Most popular questions from this chapter

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