## Question

x y

2 10.32

3 11.78

4 15.84

5 20.1

6 23.96

7 24.52

8 28.58

9 30.24

10 32.2

11 35.66

2.

Based on the data shown below, calculate the regression line (each value to two decimal places)

y = x +

x y

1 11.14

2 11.48

3 13.02

4 12.16

5 12.3

6 11.74

7 10.78

8 13.32

9 12.46

10 12.7

11 12.84

12 12.58

3. A regression was run to determine if there is a relationship between hours of TV watched per day (x) and number of situps a person can do (y).

The results of the regression were:

y=ax+b

a=-1.207

b=21.568

r2=0.573049

r=-0.757

Use this to predict the number of situps a person who watches 1.5 hours of TV can do (to one decimal place)

4.

You generate a scatter plot using Excel. You then have Excel plot the trend line and report the equation and the r2 value. The regression equation is reported as

y=−98.34x+26.77

and the r2=0.0625.

What is the correlation coefficient for this data set?

r =

5.

Here is a bivariate data set.

x y

26 71

35 73.4

35.9 97.1

15.2 82.6

31.8 76.4

13.5 33

40.8 72.9

Find the correlation coefficient and report it accurate to three decimal places.

r =

What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place.

r² = %

6. The following bivariate data set contains an outlier.

x y

15.3 -362.5

28.1 66.1

20.9 -154.8

37.7 -193.3

23.9 -284.3

28.2 258.1

25.6 966.7

15.5 78.5

20.2 334.8

12.8 31.7

21.4 -128.8

12.3 555.8

22.5 -20.8

13.1 466.5

113.9 -264.4

What is the correlation coefficient with the outlier?

rw =

What is the correlation coefficient without the outlier?

rwo =

Would inclusion of the outlier change the evidence for or against a significant linear correlation?

• No. Including the outlier does not change the evidence regarding a linear correlation.

• Yes. Including the outlier changes the evidence regarding a linear correlation.

Question for thought: Would you always draw the same conclusion with the addition of an outlier?

7. Run a regression analysis on the following bivariate set of data with y as the response variable.

x y

30.3 67.8

60 70.5

65.8 81.9

57.4 88.4

31.5 61.9

33.7 65.3

22.1 66.1

16.6 64.5

23.7 60.4

34.1 73.2

29.9 63.4

22.2 57.3

22.7 68.4

-1.8 52.7

42.9 72.6

28.2 60.5

18.9 47.1

37.8 72.7

20.4 50.3

54.5 70.6

Verify that the correlation is significant at an α=0.05. If the correlation is indeed significant, predict what value (on average) for the explanatory variable will give you a value of 45.2 on the response variable.

What is the predicted explanatory value?

x =

8. Rivers in North Carolina contain small concentrations of mercury that can accumulate in fish over their lifetimes. The concentration of mercury in fish tissue can be obtained by catching fish and sending samples to a lab for analysis. A study was conducted on fish from the Waccamaw and Lumber Rivers to investigate mercury levels in tissues of largemouth bass. At several stations along each river a group of fish were caught, weighed and measured; in addition a filet from each fish was sent to a lab so that the tissue concentration of mercury could be determined. In all, 171 fish were caught at 15 different research stations along the Waccamaw and Lumber Rivers. Data from fish caught at one of these stations is shown in the following table:

length weight

1616 47.0

1862 48.7

2855 55.7

1199 45.2

1320 44.7

1225 43.8

870 38.5

1455 45.8

1220 44.0

1033 40.4

Compute the correlation between length and weight for these fish. (Assume the correlation conditions have been satisfied and round your answer to the nearest 0.001.)

9. The line of best fit through a set of data is

y=40.796−1.375x

According to this equation, what is the predicted value of the dependent variable when the independent variable has value 150?

y = Round to 1 decimal place.

10. Using your favorite statistics software package, you generate a scatter plot with a regression equation and correlation coefficient. The regression equation is reported as

y=−69.45x+97.29

and the r=−0.758.

What proportion of the variation in y can be explained by the variation in the values of x?

r² = %

Report answer as a percentage accurate to one decimal place.

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1. The correlation coefficient is 0.992.2. y = 0.10x + 11.54

3. y = -1.207*1.5+21.568 = 19.7575...

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