/*! 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 41 According to the size of a femal... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

According to the size of a female salamander's snout is correlated with the number of eggs in her clutch. The following data are consistent with summary quantities reported in the article. Partial Minitab output is also included. \(\begin{array}{lrrrrr}\text { Snout-Vent Length } & 32 & 53 & 53 & 53 & 54 \\\ \text { Clutch Size } & 45 & 215 & 160 & 170 & 190 \\ \text { Snout-Vent Length } & 57 & 57 & 58 & 58 & 59 \\ \text { Clutch Size } & 200 & 270 & 175 & 245 & 215 \\ \text { Snout-Vent Length } & 63 & 63 & 64 & 67 & \\ \text { Clutch Size } & 170 & 240 & 245 & 280 & \end{array}\) 2 The regression equation is \(\begin{aligned}&Y=-133+5.92 x \\\&\text { Predictor } & \text { Coef } & \text { StDev } & T & P \\\&\text { Constant } & -133.02 & 64.30 & 2.07 & 0.061 \\\&x & 5.919 & 1.127 & 5.25 & 0.000 \\\&s=33.90 & \text { R-Sq }=69.7 \% & R-S q(a d j)=67.2 \%\end{aligned}\) Additional summary statistics are \(n=14 \quad \bar{x}=56.5 \quad \bar{y}=201.4\) \(\sum x^{2}=45,958 \quad \sum y^{2}=613,550 \quad \sum x y=164,969\) a. What is the equation of the regression line for predicting clutch size based on snout-vent length? b. What is the value of the estimated standard deviation of \(b\) ? c. Is there sufficient evidence to conclude that the slope of the population line is positive? d. Predict the clutch size for a salamander with a snoutvent length of 65 using a \(95 \%\) interval. e. Predict the clutch size for a salamander with a snoutvent length of 105 using a \(90 \%\) interval.

Short Answer

Expert verified
a. The equation of the regression line is \(Y = -133 + 5.92x\). b. The estimated standard deviation of 'b' is 1.127. c. Yes, there is sufficient evidence to conclude that the slope of the population line is positive. d. The predicted clutch size for a salamander with a snout-vent length of 65 is 251.4 (95% interval cannot be calculated without further data). e. The predicted clutch size for a salamander with a snout-vent length of 105 is 489.6 (90% interval cannot be calculated without further data).

Step by step solution

01

Equation of the Regression Line

The equation of the regression line, also known as the best fit line, is given by \(Y = -133 + 5.92x\). Therefore, this is the equation for predicting clutch size based on snout-vent length. In this equation, \(x\) represents the snout-vent length, and \(Y\) represents the predicted clutch size.
02

Value of Estimated Standard Deviation

The estimated standard deviation of 'b' (slope), denoted as \(s_b\), is the standard error of the regression slope. Looking at the given summary statistics, we see for x (snout-vent length) the standard deviation value is 1.127.
03

Slope of the Population Line

To assess whether there is sufficient evidence to conclude that the slope of the population line is positive, we look at the P-value associated with the slope. The P-value of x is given as 0.000. This P-value is less than the common alpha level of 0.05, so we can reject the null hypothesis that the slope is zero (or negative). This provides sufficient evidence to conclude that the slope of the population line is positive.
04

Predicting Clutch Size (65 snout-vent length)

To predict the clutch size for a salamander with a snout-vent length of 65 using a 95% interval, we substitute \(x = 65\) into the regression equation: \(Y = -133 + 5.92(65) = 251.4\). The 95% interval calculation would involve t-statistics and standard error, which are not provided, so cannot be obtained here.
05

Predicting Clutch Size (105 snout-vent length)

Similarly, to predict the clutch size for a salamander with a snout-vent length of 105 using a 90% interval, we substitute \(x = 105\) into the regression equation: \(Y = -133 + 5.92(105) = 489.6\). The 90% interval calculation would involve t-statistics and standard error, which are not provided, so cannot be obtained here.

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.

Predictor Variables
When it comes to understanding regression analysis, the term predictor variables plays a key role. These are the independent variables in a regression equation that are used to predict the value of the dependent variable. In our exercise, the snout-vent length of a salamander is the predictor variable because it is used to anticipate the clutch size, which is the dependent variable.

In regression analysis, the relationship between the predictor and the outcome is quantified by the regression coefficients. These coefficients tell us the amount of change we can expect in the dependent variable for a one-unit change in the predictor variable, assuming all other variables remain constant.
Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range. In the context of the exercise, the standard deviation value of 1.127 for the predictor variable's coefficient signifies the dispersion of snout-vent length values around the estimated regression line.

Understanding standard deviation in the scope of regression helps us assess the reliability of the predictor. If the standard deviation of the estimated coefficient is high, the reliability of the prediction decreases since there's more variability in the data points.
Null Hypothesis
The null hypothesis is a fundamental concept in hypothesis testing used as a starting point to test if there is a significant effect or relationship between variables. It proposes that there is no effect or no difference, and any observed effect is due to random chance.

In regression analysis, the null hypothesis often states that there is no relationship between the predictor and dependent variable – meaning, the regression coefficient is zero. This hypothesis is tested against an alternative hypothesis, which proposes that there is indeed an effect or relationship. Rejecting the null hypothesis implies that there is statistically significant evidence to support the existence of a relationship.
P-value
The P-value is the probability of obtaining test results at least as extreme as the observed data, assuming that the null hypothesis is true. In simpler terms, it's a measure used to determine the statistical significance of the findings from a study or an experiment.

In the salamander exercise, the P-value associated with the slope is 0.000, which is less than the commonly used significance level of 0.05. This extremely low P-value indicates that the probability of finding such a result, if in fact, the true slope is zero, is virtually non-existent. Therefore, we have sufficient evidence to reject the null hypothesis and conclude that there is a positive relationship between snout-vent length and clutch size.
Confidence Interval
A confidence interval is a range of values that's used to estimate the true value of an unknown population parameter. The confidence level represents the proportion of times that the interval estimate would contain the parameter if you repeated the study multiple times.

For instance, a 95% confidence interval means that if we repeated the study 100 times, we would expect the interval to contain the true parameter 95 times out of 100. It's important to note that the confidence interval includes an element of uncertainty. In regression, confidence intervals for predictions account for both the uncertainty in estimating the population mean and the random variation around this mean.

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

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

Give a brief answer, comment, or explanation for each of the following. a. What is the difference between \(e_{1}, e_{2}, \ldots, e_{n}\) and the \(n\) residuals? b. The simple linear regression model states that \(y=\alpha+\beta x\). c. Does it make sense to test hypotheses about \(b\) ? d. SSResid is always positive. e. A student reported that a data set consisting of \(n=6\) observations yielded residuals \(2,0,5,3,0\), and 1 from the least-squares line. f. A research report included the following summary quantities obtained from a simple linear regression analysis: \(\sum(y-\bar{y})^{2}=615 \quad \sum(y-\hat{y})^{2}=731\)

The accompanying data were read from a plot (and are a subset of the complete data set) given in the article . The data represent the mean response times for a group of individuals with closed-head injury (CHI) and a matched control group without head injury on 10 different tasks. Each observation was based on a different study, and used different subjects, so it is reasonable to assume that the observations are independent. \begin{tabular}{ccc} & \multicolumn{2}{l} { Mean Response Time } \\ \cline { 2 - 3 } Study & Control & CHI \\ \hline 1 & 250 & 303 \\ 2 & 360 & 491 \\ 3 & 475 & 659 \\ 4 & 525 & 683 \\ 5 & 610 & 922 \\ 6 & 740 & 1044 \\ 7 & 880 & 1421 \\ 8 & 920 & 1329 \\ 9 & 1010 & 1481 \\ 10 & 1200 & 1815 \\ \hline \end{tabular} a. Fit a linear regression model that would allow you to predict the mean response time for those suffering a closed-head injury from the mean response time on the same task for individuals with no head injury. b. Do the sample data support the hypothesis that there is a useful linear relationship between the mean response time for individuals with no head injury and the mean response time for individuals with CHI? Test the appropriate hypotheses using \(\alpha=.05\). c. It is also possible to test hypotheses about the \(y\) intercept in a linear regression model. For these data, the null hypothesis \(H_{0}: \alpha=0\) cannot be rejected at the \(.05\) significance level, suggesting that a model with a \(y\) intercept of 0 might be an appropriate model. Fitting such a model results in an estimated regression equation of \(\mathrm{CHI}=1.48(\) Control \()\) Interpret the estimated slope of \(1.48\).

The authors of the article used a simple linear regression model to describe the relationship between \(y=\) vigor (average width in centimeters of the last two annual rings) and \(x=\) stem density (stems/ \(\mathrm{m}^{2}\) ). The estimated model was based on the following data. Also given are the standardized residuals. \(\begin{array}{lrrrrr}x & 4 & 5 & 6 & 9 & 14 \\ y & 0.75 & 1.20 & 0.55 & 0.60 & 0.65 \\ \text { St. resid. } & -0.28 & 1.92 & -0.90 & -0.28 & 0.54 \\ x & 15 & 15 & 19 & 21 & 22 \\ y & 0.55 & 0.00 & 0.35 & 0.45 & 0.40 \\ \text { St. resid. } & 0.24 & -2.05 & -0.12 & 0.60 & 0.52\end{array}\) a. What assumptions are required for the simple linear regression model to be appropriate? b. Construct a normal probability plot of the standardized residuals. Does the assumption that the random deviation distribution is normal appear to be reasonable? Explain. c. Construct a standardized residual plot. Are there any unusually large residuals? d. Is there anything about the standardized residual plot that would cause you to question the use of the simple linear regression model to describe the relationship between \(x\) and \(y\) ?

A random sample of \(n=347\) students was selected, and each one was asked to complete several questionnaires, from which a Coping Humor Scale value \(x\) and a Depression Scale value \(y\) were determined. The resulting value of the sample correlation coefficient was \(-.18\). a. The investigators reported that \(P\) -value \(<.05 .\) Do you agree? b. Is the sign of \(r\) consistent with your intuition? Explain. (Higher scale values correspond to more developed sense of humor and greater extent of depression.) c. Would the simple linear regression model give accurate predictions? Why or why not?

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.