Question

In: Statistics and Probability

Do larger universities tend to have more property crime? University crime statistics are affected by a...

Do larger universities tend to have more property crime? University crime statistics are affected by a variety of factors. The surrounding community, accessibility given to outside visitors, and many other factors influence crime rate. Let x be a variable that represents student enrollment (in thousands) on a university campus, and let y be a variable that represents the number of burglaries in a year on the university campus. A random sample of n = 8 universities in California gave the following information about enrollments and annual burglary incidents.

x 14.5 30.2 24.5 14.3 7.5 27.7 16.2 20.1
y 25 69 39 23 15 30 15 25

(a)

Make a scatter diagram of the data. Then visualize the line you think best fits the data. (Submit a file with a maximum size of 1 MB.)

CrimeonCampus.xlsx

This answer has not been graded yet.

(b)

Use a calculator to verify that Σ(x) = 155.0, Σ(x2) = 3417.02, Σ(y) = 241, Σ(y2) = 9411 and Σ(x y) = 5419.7.

Compute r. (Enter a number. Round to 3 decimal places.)

As x increases, does the value of r imply that y should tend to increase or decrease? Explain your answer.

Given our value of r, y should tend to decrease as x increases.Given our value of r, y should tend to remain constant as x increases.    Given our value of r, y should tend to increase as x increases.Given our value of r, we can not draw any conclusions for the behavior of y as x increases.

(b)

Verify the given sums Σx, Σy, Σx2, Σy2, Σx y, and the value of the sample correlation coefficient r. (For each answer, enter a number. Round your value for r to three decimal places.)

Σx =

Σy =

Σx2 =

Σy2 =

Σx y =

r =

(c)

Find , and . Then find the equation of the least-squares line  = a + b x. (For each answer, enter a number. Round your answers for  and  to two decimal places. Round your answers for a and b to three decimal places.)

= x bar =

= y bar =

= value of a coefficient + value of b coefficient x

(d)

Graph the least-squares line. Be sure to plot the point (, ) as a point on the line. (Select the correct graph.)

(e)

Find the value of the coefficient of determination r2. What percentage of the variation in y can be explained by the corresponding variation in x and the least-squares line? What percentage is unexplained? (For each answer, enter a number. Round your answer for r2 to three decimal places. Round your answers for the percentages to one decimal place.)

r2 =

explained =  %

unexplained =  %

(f)

The calves you want to buy are 25 weeks old. What does the least-squares line predict for a healthy weight (in kg)? (Enter a number. Round your answer to two decimal places.)
kg

Solutions

Expert Solution

(a) Using excel:

We notice an upward trend from the above plot:

(b) Using excel we may verify Σ(x) = 155.0, Σ(x2) = 3417.02, Σ(y) = 241, Σ(y2) = 9411 and Σ(x y) = 5419.7 as follows:

x y x2 y2 xy
14.5 25 210.25 625 362.5
30.2 69 912.04 4761 2083.8
24.5 39 600.25 1521 955.5
14.3 23 204.49 529 328.9
7.5 15 56.25 225 112.5
27.7 30 767.29 900 831
16.2 15 262.44 225 243
20.1 25 404.01 625 502.5
155 241 3417.02 9411 5419.7

The Pearson's correlation r can be computed using the formula:

  

r = 0.795

The correlation coefficient r measures the strength and direction of the linear relationship between x and y. It ranges from -1 to 1; negative and positive values indicating a negative and positive linear relationship respectively. Values close to unity, depicts a strong linear relationship and those close to zero implies weak or no linear relationship.Here , R =0.795. This may be interpreted as: There is a moderately strong positive linear relationship between the two variables.

This implies that given our value of r, y should tend to increase as x increases

(b) Verifying r: Using CORREL function in excel,

We get r = 0.795

(c) Estimating the intercept (a) and Slope (b) coefficients :

Substituting the values,

= 1.813

a = 30.125 - (1.813)(19.375)

= -4.999

The fitted regression equation can be expressed as:

y = -4.999 + 1.813x

(d) Fitting a least square line:

(e) Also, here, r2 = (0.795)2 = 0.632

About 63.2% of the variation in y can be explained by the corresponding variation in x and the least-squares line. And about 1 - 0.632 = 36.8% percentage of the variation in y is unexplained.

r2 = 0.632

Explained = 63.2%

Unexplained = 36.8%


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