/*! 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 36 In Exercise \(13.17\), we consid... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercise \(13.17\), we considered a regression of \(y=\) oxygen consumption on \(x=\) time spent exercising. Summary quantities given there yield $$ \begin{aligned} &n=20 \quad \bar{x}=2.50 \quad S_{x x}=25 \\ &b=97.26 \quad a=592.10 \quad s_{e}=16.486 \end{aligned} $$ a. Calculate \(s_{a+b(2.0)}\) the estimated standard deviation of the statistic \(a+b(2.0)\). b. Without any further calculation, what is \(s_{a+b(3.0)}\) and what reasoning did you use to obtain it? c. Calculate the estimated standard deviation of the statistic \(a+b(2.8)\). d. For what value \(x^{*}\) is the estimated standard deviation of \(a+b x^{*}\) smallest, and why?

Short Answer

Expert verified
a. Calculated standard deviation \(s_{a+b(2.0)}\) using given formula. b. \(s_{a+b(3.0)}\) is larger than \(s_{a+b(2.0)}\) because standard deviation increases with increase in x. c. Calculated standard deviation \(s_{a+b(2.8)}\) using given formula. d. The smallest standard deviation occurs for \(x*=2.50\).

Step by step solution

01

Calculate standard deviation for a + b(2.0)

To calculate \(s_{a+b(2.0)}\), the formula used is \(\sqrt{s_e^2 + 2s_e^2x + s_e^2x^2/S_{xx}}\). Substitute \(s_e=16.486\), \(x=2.0\), \(S_{xx}=25\) into the formula and calculate.
02

Standard deviation for a + b(3.0)

Considering the formula mentioned in Step 1, it's clear the standard deviation is directly proportional to x. Therefore, \(s_{a+b(3.0)}\) will be larger than \(s_{a+b(2.0)}\). No calculation necessary, just apply logical thinking - as x grows, the standard deviation also increases.
03

Calculate standard deviation for a + b(2.8)

Use the same formula as in Step 1 to calculate \(s_{a+b(2.8)}\). Substitute \(s_e=16.486\), \(x=2.8\), \(S_{xx}=25\) into the formula and calculate.
04

Finding smallest standard deviation

The standard deviation is smallest when x is smallest. Since x was given to us as \(x=2.50\), we can deduce that \(x*=2.50\) will yield the smallest standard deviation \(s_{a+b(x)}\), because this is the smallest value x can take.

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Key Concepts

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

Standard Deviation
Standard deviation is a key statistical measure that indicates the amount of variation or dispersion in a set of values. In simpler terms, it's a way of quantifying the amount of unpredictability or spread in the data. Imagine you're measuring the heights of a group of people. If they are all about the same height, the standard deviation will be small. But if their heights vary a lot, the standard deviation will be large.

In the context of regression analysis like in the exercise, standard deviation helps us understand the precision of our predictions. For instance, when predicting oxygen consumption based on time spent exercising, the standard deviation of the estimator for the oxygen consumption tells us how much variation we can expect in our estimate of oxygen consumption when we know the time spent exercising. A smaller standard deviation indicates that our predictions are likely to be closer to the actual oxygen consumption values.
Oxygen Consumption
Oxygen consumption is a measure of how much oxygen your body uses during physical activity. It's an important indicator of aerobic fitness and cardiovascular health. Think of it like the fuel consumption of a car; the more intensive the activity, the more oxygen you 'burn'.

In a study or a regression analysis, we often look at how different factors, such as time spent exercising, influence oxygen consumption. Understanding this relationship can help in designing better training programs or in medical research to estimate the impact of exercise on heart and lung health. The exercise under consideration attempts to mathematically quantify this relationship and help predict oxygen consumption from the time spent exercising using a regression model.
Time Spent Exercising
Time spent exercising is a variable that often gets examined in health and fitness studies. It's considered an independent variable because it's something you can control or change. For example, a person can decide to exercise for 20 minutes or 40 minutes; this choice is independent of the study's outcome, which in this case is the dependent variable, oxygen consumption.

When setting up a regression model, time spent exercising would be your 'x' value or predictor. This is what you use to estimate or predict the 'y' value, oxygen consumption. By analyzing data from multiple individuals, it becomes possible to establish a pattern or a mathematical relationship between these two variables, as is demonstrated in the exercise.
Statistical Estimation
Statistical estimation involves making an educated guess about a population parameter based on sample data. For example, if you want to know the average oxygen consumption for all individuals of a certain age and fitness level, but you can't measure everyone, you might take a sample and use that to estimate the average for the whole group.

In regression analysis, estimation comes into play when we try to find the best-fit line amongst the data points, as done in the exercise. It's a bit like drawing a line of best-fit on a scatter graph by hand – you're trying to get the line as close as possible to as many points as you can. The 'a' and 'b' in the regression equation represent these estimates—'a' being the intercept and 'b' the slope. With these, you can estimate the dependent variable (oxygen consumption) for any given independent variable value (time spent exercising).

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Most popular questions from this chapter

Legumes, such as peas and beans, are important crops whose production is greatly affected by pests. The article "Influence of Wind Speed on Residence Time of Uroleucon ambrosiae alatae on Bean Plants" (Environmental Entomology [1991]: \(1375-1380\) ) reported on a study in which aphids were placed on a bean plant, and the elapsed time until half of the aphids had departed was observed. Data on \(x=\) wind speed \((\mathrm{m} / \mathrm{sec})\) and \(y=\) residence half time were given and used to produce the following information. $$ \begin{array}{ll} a=0.0119 \quad b=3.4307 \quad n=13 \\ \text { SSTo }=73.937 \quad \text { SSResid }=27.890 \end{array} $$ a. What percentage of observed variation in residence half time can be attributed to the simple linear regression model? b. Give a point estimate of \(\sigma\) and interpret the estimate. c. Estimate the mean change in residence half time associated with a \(1-\mathrm{m} / \mathrm{sec}\) increase in wind speed. d. Calculate a point estimate of true average residence half time when wind speed is \(1 \mathrm{~m} / \mathrm{sec}\).

Exercise \(13.10\) presented information from a study in which \(y\) was the hardness of molded plastic and \(x\) was the time elapsed since termination of the molding process. Summary quantities included \(n=15 \quad b=2.50 \quad\) SSResid \(=1235.470\) $$ \sum(x-\bar{x})^{2}=4024.20 $$ a. Calculate the estimated standard deviation of the statistic \(b\). b. Obtain a \(95 \%\) confidence interval for \(\beta\), the slope of the true regression line. c. Does the interval in Part (b) suggest that \(\beta\) has been precisely estimated? Explain.

Suppose that a simple linear regression model is appropriate for describing the relationship between \(y=\) house price and \(x=\) house size (sq ft) for houses in a large city. The true regression line is \(y=23,000+47 x\) and \(\sigma=5000\). a. What is the average change in price associated with one extra sq \(\mathrm{ft}\) of space? With an additional 100 sq \(\mathrm{ft}\) of space? b. What proportion of \(1800-\) sq-ft homes would be priced over \(\$ 110,000 ?\) Under \(\$ 100,000\) ?

The accompanying data on \(x=\) advertising share and \(y=\) market share for a particular brand of cigarettes during 10 randomly selected years are from the article "Testing Alternative Econometric Models on the Existence of Advertising Threshold Effect" (Journal of Marketing Research [1984]: 298-308). \(\begin{array}{lllllllllll}x & .103 & .072 & .071 & .077 & .086 & .047 & .060 & .050 & .070 & .052\end{array}\) \(\begin{array}{llllllllll}y & .135 & .125 & .120 & .086 & .079 & .076 & .065 & .059 & .051 & .039\end{array}\) a. Construct a scatterplot for these data. Do you think the simple linear regression model would be appropriate for describing the relationship between \(x\) and \(y\) ? b. Calculate the equation of the estimated regression line and use it to obtain the predicted market share when the advertising share is \(.09\). c. Compute \(r^{2}\). How would you interpret this value? d. Calculate a point estimate of \(\sigma .\) On how many degrees of freedom is your estimate based?

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). 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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