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Chapter 13: Problem 13

Suppose that a single \(y\) observation is made at each of the \(x\) values \(5,10,15,20\), and 25 . a. If \(\sigma=4\), what is the standard deviation of the statistic \(b\) ? b. Now suppose that a second observation is made at every \(x\) value listed in Part (a) (for a total of 10 observations). Is the resulting value of \(\sigma_{b}\) half of what it was in Part (a)? c. How many observations at each \(x\) value in Part (a) are required to yield a \(\sigma_{b}\) value that is half the value calculated in Part (a)? Verify your conjecture.

Short Answer

Expert verified
a. Use the given values to calculate the standard deviation of \(b\). b. It will not be exactly half. It's because the standard deviation does not decrease linearly with increase in number of samples. c. The number of observations that should be made to halve the standard deviation value can be calculated by equating the formula of standard deviation to half of the original value and solving for number of samples.

Step by step solution

01

Find Standard Deviation for Single Observation

In the first part, we know the standard deviation \(\sigma = 4\) and the \(x\) values. We use the formula for standard deviation of \(b\) when there is one observation per \(x\) value, which is: \[\sigma_{b} = \frac{\sigma}{\sqrt{\sum (x - \overline{x})^2}}\] where \(\overline{x}\) is the mean of \(x\). Substitute the given values into the formula to calculate \(\sigma_{b}\).
02

Calculate Standard Deviation for Two Observations

In the second part, we have to understand what happens to the standard deviation when there are two observations at each \(x\) value instead of one. It requires the understanding of how increasing the sample size affects the standard deviation. It is known that as sample size increases, the standard deviation decreases but not necessarily by half. Use the formula for two observations at each \(x\) value and compare with the calculated value in Step 1.
03

Determine Number of Observations for Halved Standard Deviation

In the third part, we need to find how many observations should be made to halve the standard deviation value obtained in part (a). So, we need to equate the formula for standard deviation to half of the value obtained in part (a) and solve for the number of samples (\(n\)). Also verify the conjecture by substituting the calculated number of samples in to the formula.

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Key Concepts

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

Standard Deviation
In statistics, standard deviation is a measure of how spread out the numbers in a data set are. It indicates how much individual data points deviate from the mean. If you have a small standard deviation, your data points are close to the mean, while a large standard deviation indicates that the data points are spread out over a wider range. When calculating the standard deviation for the linear regression slope (\( b \)), we consider how the spread of data around the line affects this parameter. For an observation set, the formula for the standard deviation of the slope is:\[\sigma_{b} = \frac{\sigma}{\sqrt{\sum (x - \overline{x})^2}}\]This formula shows that the standard deviation of the slope depends on the spread of the \( x \)-values. Larger variability among the \( x \)-values means more precise estimates for the slope, resulting in a lower standard deviation. Understanding how to correctly use this concept allows us to anticipate how changes in data collection, like increasing observations, will affect results.
Sample Size
Sample size is a critical factor in statistical analysis, playing a significant role in the reliability and precision of statistical estimates. In our exercise, we explored how increasing the sample size affects the standard deviation of the regression slope (\( \sigma_{b} \)).As a rule, increasing the sample size generally decreases the standard deviation because more data points provide us a more stable mean and reduce error. However, the decrease in standard deviation doesn't progress linearly. For example:- Doubling the number of observations may not necessarily halve the standard deviation.- To halve the standard deviation, more than double the observations might be needed, depending on the data arrangement.By understanding these principles, students can predict how data collection will influence statistical outcomes. An informed estimation of the required sample size ensures precise and reliable analysis.
Linear Regression
Linear regression is a statistical method used to model the relationship between a dependent variable (\( y \)) and one or more independent variables (\( x \)). The primary goal is to determine the linear equation that best predicts the dependent variable based on the independent variables.The basic equation of a simple linear regression is:\[ y = mx + c \]In this context:- \( y \) is the dependent variable.- \( x \) is the independent variable.- \( m \) represents the slope of the regression line, indicating the change in \( y \) for a one-unit change in \( x \).- \( c \) is the y-intercept, representing the value of \( y \) when \( x \) is 0.In the exercise, linear regression helps us understand how \( y \) changes as \( x \) varies at different levels of observation. Crucially, by understanding both the slope and intercept's role in forming this relationship, students gain insight into how predictions can be made from data sets, further illustrated by changes in standard deviation and sample size.

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Most popular questions from this chapter

The accompanying data on \(x=\) treadmill run time to exhaustion (min) and \(y=20-\mathrm{km}\) ski time (min) were taken from the article "Physiological Characteristics and Performance of Top U.S. Biathletes" (Medicine and \(S c i\) ence in Sports and Exercise \([1995]: 1302-1310):\) \(\begin{array}{lrrrrrr}x & 7.7 & 8.4 & 8.7 & 9.0 & 9.6 & 9.6 \\ y & 71.0 & 71.4 & 65.0 & 68.7 & 64.4 & 69.4 \\ x & 10.0 & 10.2 & 10.4 & 11.0 & 11.7 & \\ y & 63.0 & 64.6 & 66.9 & 62.6 & 61.7 & \\ \sum x & =106.3 & \sum x^{2}=1040.95 & & \\ \sum y= & 728.70 & \sum x y=7009.91 & \sum y^{2}=48390.79\end{array}\) a. Does a scatterplot suggest that the simple linear regression model is appropriate? b. Determine the equation of the estimated regression line, and draw the line on your scatterplot. c. What is your estimate of the average change in ski time associated with a 1 -min increase in treadmill time? d. What would you predict ski time to be for an individual whose treadmill time is \(10 \mathrm{~min} ?\) e. Should the model be used as a basis for predicting ski time when treadmill time is \(15 \mathrm{~min}\) ? Explain. f. Calculate and interpret the value of \(r^{2}\). g. Calculate and interpret the value of \(s_{e}\)

An investigation of the relationship between traffic flow \(x\) (thousands of cars per \(24 \mathrm{hr}\) ) and lead content \(y\) of bark on trees near the highway (mg/g dry weight) yielded the accompanying data. A simple linear regression model was fit, and the resulting estimated regression line was \(\hat{y}=28.7+33.3 x .\) Both residuals and standardized residuals are also given. \(\begin{array}{lrrrrr}\text { iduals are also given. } & & & & \\ x & 8.3 & 8.3 & 12.1 & 12.1 & 17.0 \\ y & 227 & 312 & 362 & 521 & 640 \\ \text { Residual } & -78.1 & 6.9 & -69.6 & 89.4 & 45.3 \\ \text { St. resid. } & -0.99 & 0.09 & -0.81 & 1.04 & 0.51\end{array}\) \(\begin{array}{lrrrrr}x & 17.0 & 17.0 & 24.3 & 24.3 & 24.3 \\ y & 539 & 728 & 945 & 738 & 759 \\ \text { Residual } & -55.7 & 133.3 & 107.2 & -99.8 & -78.8 \\\ \text { St. resid. } & -0.63 & 1.51 & 1.35 & -1.25 & -0.99\end{array}\) a. Plot the \((x\), residual \()\) pairs. Does the resulting plot suggest that a simple linear regression model is an appropriate choice? Explain your reasoning. b. Construct a standardized residual plot. Does the plot differ significantly in general appearance from the plot in Part (a)?

In a study of bacterial concentration in surface and subsurface water ("Pb and Bacteria in a Surface Microlayer" Journal of Marine Research \([1982]: 1200-\) 1206 ), the accompanying data were obtained. \(\begin{aligned}&\text { Concentration }\left(\times 10^{6} / \mathrm{mL}\right) \\\&\text { Surface } & 48.6 & 24.3 & 15.9 & 8.29 & 5.75 \\\&\text { Subsurface } & 5.46 & 6.89 & 3.38 & 3.72 & 3.12 \\\&\text { Surface } & 10.8 & 4.71 & 8.26 & 9.41 & \\\&\text { Subsurface } & 3.39 & 4.17 & 4.06 & 5.16 & \end{aligned}\) Summary quantities are $$ \begin{aligned} &\sum x=136.02 \quad \sum y=39.35 \\ &\sum x^{2}=3602.65 \quad \sum y^{2}=184.27 \quad \sum x y=673.65 \end{aligned} $$ Using a significance level of \(.05\), determine whether the data support the hypothesis of a linear relationship between surface and subsurface concentration.

The accompanying summary quantities for \(x=\) particulate pollution \(\left(\mu \mathrm{g} / \mathrm{m}^{3}\right)\) and \(y=\) luminance \(\left(.01 \mathrm{~cd} / \mathrm{m}^{2}\right)\) were calculated from a representative sample of data that appeared in the article "Luminance and Polarization of the Sky Light at Seville (Spain) Measured in White Light" \((A t=\) mospheric Environment \([1988]: 595-599) .\) $$ a. Test to see whether there is a positive correlation between particulate pollution and luminance in the population from which the data were selected. b. What proportion of observed variation in luminance can be attributed to the approximate linear relationship between luminance and particulate pollution? \begin{aligned} &n=15 \quad \sum x=860 \quad \sum y=348 \\ &\sum x^{2}=56,700 \quad \sum y^{2}=8954 \quad \sum x y=22,265 \end{aligned} $$

The shelf life of packaged food depends on many factors. Dry cereal is considered to be a moisture-sensitive product (no one likes soggy cereal!) with the shelf life determined primarily by moisture content. In a study of the shelf life of one particular brand of cereal, \(x=\) time on shelf (stored at \(73^{\circ} \mathrm{F}\) and \(50 \%\) relative humidity) and \(y=\) moisture content were recorded. The resulting data are from "Computer Simulation Speeds Shelf Life Assessments" (Package Engineering [1983]: 72-73). \(\begin{array}{rrrrrrrr}x & 0 & 3 & 6 & 8 & 10 & 13 & 16 \\ y & 2.8 & 3.0 & 3.1 & 3.2 & 3.4 & 3.4 & 3.5 \\ x & 20 & 24 & 27 & 30 & 34 & 37 & 41 \\ y & 3.1 & 3.8 & 4.0 & 4.1 & 4.3 & 4.4 & 4.9\end{array}\) a. Summary quantities are $$ \begin{array}{ll} \sum x=269 & \sum y=51 \quad \sum x y=1081.5 \\ \sum y^{2}=7745 & \sum x^{2}=190.78 \end{array} $$ Find the equation of the estimated regression line for predicting moisture content from time on the shelf. b. Does the simple linear regression model provide useful information for predicting moisture content from knowledge of shelf time? c. Find a \(95 \%\) interval for the moisture content of an individual box of cereal that has been on the shelf 30 days. d. According to the article, taste tests indicate that this brand of cereal is unacceptably soggy when the moisture content exceeds 4.1. Based on your interval in Part (c), do you think that a box of cereal that has been on the shelf 30 days will be acceptable? Explain.
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