/*! 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 6 The paper "Predicting Yolk Heigh... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The paper "Predicting Yolk Height, Yolk Width, Albumen Length, Eggshell Weight, Egg Shape Index, Eggshell Thickness, Egg Surface Area of Japanese Quails Using Various Egg Traits as Regressors" (International journal of Poultry Science [2008]: \(85-88\) ) suggests that the simple linear regression model is reasonable for describing the relationship between \(y=\) eggshell thickness (in micrometers) and \(x=\) egg length (mm) for quail eggs. Suppose that the population regression line is \(y=0.135+0.003 x\) and that \(\sigma=0.005 .\) Then, for a fixed \(x\) value, \(y\) has a normal distribution with mean \(0.135+0.003 x\) and standard deviation 0.005 . a. What is the mean eggshell thickness for quail eggs that are \(15 \mathrm{~mm}\) in length? For quail eggs that are \(17 \mathrm{~mm}\) in length? b. What is the probability that a quail egg with a length of \(15 \mathrm{~mm}\) will have a shell thickness that is greater than \(0.18 \mu \mathrm{m} ?\) c. Approximately what proportion of quail eggs of length \(14 \mathrm{~mm}\) have a shell thickness of greater than \(0.175 ?\) Less than \(0.178 ?\)

Short Answer

Expert verified
The mean eggshell thickness for 15mm and 17mm quail eggs is 0.18 micrometers and 0.186 micrometers, respectively. The probability that a quail egg with a length of 15mm will have a shell thickness greater than 0.18 micrometers is 50%. The proportion of 14mm quail eggs having shell thickness greater than 0.175 is approximately 34.46%, and less than 0.178 is approximately 57.93%.

Step by step solution

01

a. Mean eggshell thickness for 15mm and 17mm quail eggs

We use the regression equation \(y = 0.135 + 0.003x\) to find the mean thickness for quail eggs with lengths of 15mm and 17mm. 1. For eggs with length = 15mm: \(y = 0.135 + 0.003 (15) = 0.135 + 0.045 = 0.18\) micrometers. 2. For eggs with length = 17mm: \(y = 0.135 + 0.003 (17) = 0.135 + 0.051 = 0.186\) micrometers. The mean eggshell thickness for 15mm and 17mm quail eggs are 0.18 micrometers and 0.186 micrometers, respectively.
02

b. Probability of 15mm quail egg having a shell thickness greater than 0.18 micrometers

As we know, for a fixed x value, y has a normal distribution with mean \(0.135 + 0.003x\) and standard deviation 0.005. 1. For 15mm quail egg, the mean is 0.18, and standard deviation, \(\sigma = 0.005\). 2. We need to calculate the probability that the shell thickness will be greater than \(0.18\) micrometers. 3. To do this, find the z-score: \(z = \frac{y - (\mathrm{mean})}{\mathrm{standard deviation}} = \frac{0.18 - 0.18}{0.005} = 0\). 4. Now, we need to find the probability to the right of z=0 which is \(1 - P(z = 0) = 1 - 0.5 = 0.5\). The probability that a quail egg with a length of 15mm will have a shell thickness greater than 0.18 micrometers is 0.5 or 50%.
03

c. Proportion of quail eggs of length 14mm with shell thickness greater than 0.175 and less than 0.178

For 14mm quail egg, the mean is \(y=0.135+0.003(14)=0.177\) and the standard deviation, \(\sigma=0.005\). 1. To find proportion greater than 0.175: - Calculate z-score: \(z1=\frac{0.175-0.177}{0.005}=-0.4\). - Find probability: \(P(z>-0.4)=1-P(z=0.4)\). 2. To find proportion less than 0.178: - Calculate z-score: \(z2=\frac{0.178-0.177}{0.005}=0.2\). - Find probability: \(P(z<0.2)\). Using standard normal distribution table or calculator: - \(P(z=0.4)=0.6554\) - \(P(z<0.2)=0.5793\) So, the proportion of 14mm quail eggs having shell thickness: - Greater than 0.175 is approximately \(1-0.6554=0.3446\) or 34.46%. - Less than 0.178 is approximately 0.5793 or 57.93%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Regression Equation
The regression equation is central to linear regression analysis. It represents the relationship between a dependent variable and one or more independent variables. In our exercise, the regression equation is given as y = 0.135 + 0.003x, where y is the eggshell thickness and x is the egg length. This equation allows us to predict the mean value of the dependent variable for a given value of the independent variable. For example, when we substitute x with 15 mm for egg length, we find the predicted mean eggshell thickness is 0.18 micrometers.

Understanding and utilizing the regression equation is crucial for making predictions based on data. It encapsulates the slope of the line—signifying the change in y for a unit change in x—and the intercept, which is the expected value of y when x is zero. By analyzing the coefficients, one can interpret the strength and direction of the relationship between variables.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion of a set of values. In the context of our problem, the standard deviation of the eggshell thickness is given as 0.005 micrometers. A smaller standard deviation indicates that the values tend to be closer to the mean of the set, while a larger one indicates the values are spread out over a wider range. In the exercise, knowing the standard deviation is essential to assess the variability of eggshell thickness around the mean, which in turn is critical for calculating probabilities and z-scores for normal distributions.
Normal Distribution
A normal distribution, also known as the Gaussian distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In the example of eggshell thickness, the normal distribution depicts the probability of observing various thicknesses around the mean value. When we say that the eggshell thickness is normally distributed, we imply that, when plotted, the distribution of the thickness values forms the characteristic 'bell' shape. This property allows us to use standard statistical tools, such as z-scores, to make inferences about our population of interest.
Probability
Probability is the measure of the likelihood that an event will occur. It is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In our exercise, probability is used to express how likely it is for a quail egg to have an eggshell thickness greater than a certain value, given its length. Utilizing the normal distribution and the standard deviation, we can calculate the probability of randomly selecting a quail egg with an eggshell thickness above or below a specified measure. Learning to calculate probabilities is fundamental in statistics as it helps us quantify risk and make informed predictions.
Z-score
A z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, expressed in terms of standard deviations. It is calculated by subtracting the mean from the value and then dividing the result by the standard deviation. In the step-by-step solution, z-scores are used to find probabilities related to the eggshell thickness. For example, if an eggshell's thickness is equal to the mean thickness, its z-score would be 0, indicating that it is at the expected value. Z-scores allow us to translate individual data points into a standardized form which we can then compare to a normal distribution, facilitating the understanding of where a value stands in relation to others and how exceptional or ordinary it is.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Do taller adults make more money? The authors of the paper "Stature and Status: Height, Ability, and Labor Market Outcomes" (Journal of Political Economics [2008]: 499-532) investigated the association between height and earnings. They used the simple linear regression model to describe the relationship between \(x=\) height (in inches) and \(y=\) log(weekly gross earnings in dollars) in a very large sample of men. The logarithm of weekly gross earnings was used because this transformation resulted in a relationship that was approximately linear. The paper reported that the slope of the estimated regression line was \(b=0.023\) and the standard deviation of \(b\) was \(s_{b}=0.004\). Carry out a hypothesis test to decide if there is convincing evidence of a useful linear relationship between height and the logarithm of weekly earnings. You can assume that the basic assumptions of the simple linear regression model are met.

Consider a test of hypotheses about, \(\beta\) the population slope in a linear regression model. a. If you reject the null hypothesis, \(\beta=0\), what does this mean in terms of a linear relationship between \(x\) and \(y ?\) b. If you fail to reject the null hypothesis, \(\beta=0,\) what does this mean in terms of a linear relationship between \(x\) and \(y ?\)

Hormone replacement therapy (HRT) is thought to increase the risk of breast cancer. The accompanying data on \(x=\) percent of women using HRT and \(y=\) breast cancer incidence (cases per 100,000 women) for a region in Germany for 5 years appeared in the paper "Decline in Breast Cancer Incidence after Decrease in Utilisation of Hormone Replacement Therapy" (Epidemiology [2008]: \(427-430\) ). The authors of the paper used a simple linear regression model to describe the relationship between HRT use and breast cancer incidence. \begin{tabular}{|cc|} \hline HRT Use & Breast Cancer Incidence \\ \hline 46.30 & 103.3 \\ 40.60 & 105.0 \\ 39.50 & 100.0 \\ 36.60 & 93.8 \\ 30.00 & 83.5 \\ \hline \end{tabular} a. What is the equation of the estimated regression line? b. What is the estimated average change in breast cancer incidence associated with a 1 percentage point increase in HRT use? c. What would you predict the breast cancer incidence to be in a year when HRT use was \(40 \% ?\) d. Should you use this regression model to predict breast cancer incidence for a year when HRT use was \(20 \%\) ? Explain. e. Calculate and interpret the value of \(r^{2}\). f. Calculate and interpret the value of \(s_{e}\).

15.19 Acrylamide is a chemical that is sometimes found in cooked starchy foods and which is thought to increase the risk of certain kinds of cancer. The paper "A Statistical Regression Model for the Estimation of Acrylamide Concentrations in French Fries for Excess Lifetime Cancer Risk Assessment" (Food and Chemical Toxicology [2012]: \(3867-3876\) ) describes a study to investigate the effect of frying time (in seconds) and acrylamide concentration (in micrograms per kilogram) in french fries. The data in the accompanying table are approximate values read from a graph that appeared in the paper. \begin{tabular}{|cc|} \hline Frying Time & Acrylamide Concentration \\ \hline 150 & 155 \\ 240 & 120 \\ 240 & 190 \\ 270 & 185 \\ 300 & 140 \\ 300 & 270 \\ \hline \end{tabular} a. For these data, the estimated regression line for predicting \(y=\) acrylamide concentration based on \(x=\) frying time is \(y=87+0.359 x\). What is an estimate of the average change in acrylamide concentration associated with a 1-second increase in frying time? b. What would you predict for acrylamide concentration for a frying time of 250 seconds? c. Use the given Minitab output to decide if there is convincing evidence of a useful linear relationship between acrylamide concentration and frying time. You may assume that the necessary conditions have been met. R-sq \(\begin{array}{cc}\text { R-sq(adj) } & \text { R-sq(pred) } \\ 0.00 \% & 0.00 \%\end{array}\) \(\mathrm{S}\) 3 \(\mathrm{q}\) \(8 \%\) Coefficients \(\mathrm{K}-\mathrm{Sq}\) \(14.38 \%\) \(\begin{array}{lccccc}\text { Term } & \text { Coef } & \text { SE Coef } & \text { T-Value } & \text { P-Value } & \text { VIF } \\ \text { Constant } & 87 & 112 & 0.78 & 0.480 & \\ x & 0.359 & 0.438 & 0.82 & 0.459 & 1.00\end{array}\) Regression Equation \(y=87+0.359 x\)

A journalist is reporting about some research on appropriate amounts of sleep for people 9 to 19 years of age. In that research, a linear regression model is used to describe the relationship between alertness and number of hours of sleep the night before. The researchers reported a \(95 \%\) confidence interval, but newspapers usually report an estimate and a margin of error. Explain how the journalist could determine the margin of error from the reported confidence interval.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.