/*! 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} Q18BSC Exercises 13鈥28 use the same d... [FREE SOLUTION] | 91影视

91影视

Exercises 13鈥28 use the same data sets as Exercises 13鈥28 in Section 10-1. In each case, find the regression equation, letting the first variable be the predictor (x) variable. Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5 on page 493.

Use the foot lengths and heights to find the best predicted height of a male

who has a foot length of 28 cm. Would the result be helpful to police crime scene investigators in trying to describe the male?

Short Answer

Expert verified

The regression equation is\(\hat y = 125 + 1.73x\).

The best-predicted value is the mean height of 177 cm. It will not be helpful to the police in trying to obtain a description of the male.

Step by step solution

01

Given information

The given data provides the information of the shoe print (in cm) and the height (in cm), as follows.

02

State the equation for the regression linea

The formula for the estimated regression line is

\(y = {b_0} + {b_1}x\).

Here,

\({b_0}\)is the Y-intercept,

\({b_1}\)is the slope,

\(x\)is the explanatory variable, and

\(\hat y\)is the response variable (predicted value).

Let X denote the foot length (in cm) and Y denote the height (in cm) of the male.

03

Compute the slope and intercept

The calculations required to compute the slope and intercept are as follows.

The sample size is \(\left( n \right) = 5\).

The slope is computed as

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{5 \times 23209.27 - 130.8 \times 886.5}}{{5 \times 3426.96 - {{130.8}^2}}}\\ = 3.5226\end{array}\).

The intercept is computed as

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{886.5 \times 3426.96 - 130.8 \times 23209.27}}{{5 \times 3426.96 - {{130.8}^2}}}\\ = 85.15\end{array}\).

The estimated regression equation is

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = 85.15 + 3.5226x\end{array}\).

04

Checking the model

Refer to exercise 18 of section 10-1 for the following result.

1) The scatter plot does not show an approximate linear relationship between the variables.

2) The P-value is 0.085.

As the P-value is greater than the level of significance (0.05), the null hypothesis is failed to be rejected.

Therefore, the correlation is not significant.

Referring to figure 10-5, the criteria for a good regression model are not satisfied.

The best-predicted value of a variable is simply its sample mean.

05

Compute the prediction 

The best-predicted height of a male who has a foot length of 28 cm is obtained as the mean of the sample responses.

The sample mean is computed as

\(\begin{array}{c}\bar y = \frac{{\sum\limits_{i = }^n {{y_i}} }}{n}\\ = \frac{{\left( {175.3 + 177.8 + ... + 172.7} \right)}}{5}\\ = 177.3\end{array}\).

Therefore, the best-predicted height of the male who has a foot length of 28 cm will be 177 cm. As the best-predicted value is the mean height (177 cm), it will not help the police to describe the male.

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 91影视!

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

In Exercises 5鈥8, use a significance level of A = 0.05 and refer to the

accompanying displays.

Casino Size and Revenue The New York Times published the sizes (square feet) and revenues (dollars) of seven different casinos in Atlantic City. Is there sufficient evidence to support the claim that there is a linear correlation between size and revenue? Do the results suggest that a casino can increase its revenue by expanding its size?

Prediction Interval Using the heights and weights described in Exercise 1, a height of 180 cm is used to find that the predicted weight is 91.3 kg, and the 95% prediction interval is (59.0 kg, 123.6 kg). Write a statement that interprets that prediction interval. What is the major advantage of using a prediction interval instead of simply using the predicted weight of 91.3 kg? Why is the terminology of prediction interval used instead of confidence interval?

In Exercises 5鈥8, use a significance level of A = 0.05 and refer to theaccompanying displays.Garbage Data Set 31 鈥淕arbage Weight鈥 in Appendix B includes weights of garbage discarded in one week from 62 different households. The paired weights of paper and glass were used to obtain the XLSTAT results shown here. Is there sufficient evidence to support the claim that there is a linear correlation between weights of discarded paper and glass?

Best Multiple Regression Equation For the regression equation given in Exercise 1, the P-value is 0.000 and the adjusted \({R^2}\)value is 0.925. If we were to include an additional predictor variable of neck size (in.), the P-value becomes 0.000 and the adjusted\({R^2}\)becomes 0.933. Given that the adjusted \({R^2}\)value of 0.933 is larger than 0.925, is it better to use the regression equation with the three predictor variables of length, chest size, and neck size? Explain.

Testing for a Linear Correlation. In Exercises 13鈥28, construct a scatterplot, and find the value of the linear correlation coefficient r. Also find the P-value or the critical values of r from Table A-6. Use a significance level of A = 0.05. Determine whether there is sufficient evidence to support a claim of a linear correlation between the two variables. (Save your work because the same data sets will be used in Section 10-2 exercises.)

POTUS Media periodically discuss the issue of heights of winning presidential candidates and heights of their main opponents. Listed below are those heights (cm) from severalrecent presidential elections (from Data Set 15 鈥淧residents鈥 in Appendix B). Is there sufficient evidence to conclude that there is a linear correlation between heights of winning presidential candidates and heights of their main opponents? Should there be such a correlation?

President

178

182

188

175

179

183

192

182

177

185

188

188

183

188

Opponent

180

180

182

173

178

182

180

180

183

177

173

188

185

175

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.