The following exercises are based on the following sample data consisting of numbers of enrolled students (in thousands) and numbers of burglaries for randomly selected large colleges in a recent year (based on data from the New York Times).

Exercise 1 stated that ris found to be 0.499. Does that value change if the actual enrollment values of 53,000, 28,000, 27,000, 36,000, and 42,000 are used instead of 53, 28, 27, 36, and 42?

Short Answer

Expert verified

The value of the correlation coefficient remains the same at 0.499.

Step by step solution

01

Given information

The table representing the number of enrolled students (in thousands) and the number of burglaries for randomly selected large colleges in recent years is provided.

\(r = 0.499\)

02

State the formula for the correlation coefficient

The formula for the correlation coefficient is

\(r = \frac{{n\left( {\sum {xy} } \right)--\left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right)--{{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right)--{{\left( {\sum y } \right)}^2}} \right)} }}\).

The measure is computed as 0.499 using the original data set given in Exercise 1.

03

Discuss the change in the measure

All the values for variable x are multiplied by a fixed constant of 1000.

Change observation x as 1000x in the formula.

\(\begin{array}{c}{r_{new}} = \frac{{n\left( {\sum {1000xy} } \right)--\left( {\sum {1000x} } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{{\left( {1000x} \right)}^2}} } \right)--{{\left( {\sum {1000x} } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right)--{{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{1000\left( {n\left( {\sum {xy} } \right)--\left( {\sum x } \right)\left( {\sum y } \right)} \right)}}{{1000\sqrt {\left( {\left( {n\sum {{{\left( x \right)}^2}} } \right)--{{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right)--{{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{n\left( {\sum {xy} } \right)--\left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{{\left( x \right)}^2}} } \right)--{{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right)--{{\left( {\sum y } \right)}^2}} \right)} }}\\ = r\end{array}\)

Therefore, there will be no change in the value of the correlation coefficient.

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 Vaia!

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

Different hotels on Las Vegas Boulevard (“the strip”) in Las Vegas are randomly selected, and their ratings and prices were obtained from Travelocity. Using technology, with xrepresenting the ratings and yrepresenting price, we find that the regression equation has a slope of 130 and a y-intercept of -368.

a. What is the equation of the regression line?

b. What does the symbol\(\hat y\)represent?

The following exercises are based on the following sample data consisting of numbers of enrolled students (in thousands) and numbers of burglaries for randomly selected large colleges in a recent year (based on data from the New York Times).

Enrollment (thousands)

53

28

27

36

42

Burglaries

86

57

32

131

157

True or false: If the sample data lead us to the conclusion that there is sufficient evidence to support the claim of a linear correlation between enrollment and number of burglaries, then we could also conclude that higher enrollments cause increases in numbers of burglaries.

Interpreting\({R^2}\)For the multiple regression equation given in Exercise 1, we get \({R^2}\)= 0.928. What does that value tell us?

Stocks and Sunspots. Listed below are annual high values of the Dow Jones Industrial Average (DJIA) and annual mean sunspot numbers for eight recent years. Use the data for Exercises 1–5. A sunspot number is a measure of sunspots or groups of sunspots on the surface of the sun. The DJIA is a commonly used index that is a weighted mean calculated from different stock values.

DJIA

14,198

13,338

10,606

11,625

12,929

13,589

16,577

18,054

Sunspot

Number

7.5

2.9

3.1

16.5

55.7

57.6

64.7

79.3

Confidence Interval Construct a 95% confidence interval estimate of the mean sunspot number. Write a brief statement interpreting the confidence interval.

let the predictor variable x be the first variable given. Use the given data to find the regression equation and the best predicted value of the response variable. Be sure to follow the prediction procedure summarized in Figure 10-5 on page 493. Use a 0.05 significance level.

For 50 randomly selected speed dates, attractiveness ratings by males of their

female date partners (x) are recorded along with the attractiveness ratings by females of their male date partners (y); the ratings are from Data Set 18 “Speed Dating” in Appendix B. The 50 paired ratings yield\(\bar x = 6.5\),\(\bar y = 5.9\), r= -0.277, P-value = 0.051, and\(\hat y = 8.18 - 0.345x\). Find the best predicted value of\(\hat y\)(attractiveness rating by female of male) for a date in which the attractiveness rating by the male of the female is x= 8.

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.

Sign-up for free