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Math Bellevue

In: Science

Submitted By bubbalmt
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1. | What is the slope of the regression line? | | Answer |
The slope is the coefficient before the x variable (D in this case). Thus the answer is 0.0138. | Points Earned: | 1/1 | Correct Answer: | 0.0138 | Your Response: | 0.0138 | 2. | Explain in specific language what this slope says about this penguin's dives. | | A. | If the depth of the dive is increased by one meter, it adds 0.0138 minutes to the time spent under water. | B. | If the depth of the dive is decreased by one meter, it adds 0.0138 minutes to the time spent under water. | C. | If the depth of the dive is increased by 0.0138 meter, it adds one minute to the time spent under water. | |
In the equation of a line, ŷ = a + bx, b is the slope. The slope is the amount by which y changes when x increases by one unit. | Points Earned: | 1/1 | Correct Answer: | A | Your Response: | A | 3. | According to the regression line, how long does a typical dive to a depth of 180 meters last? Answer to 3 decimal places. | | Answer |
The predicted value of the dive duration (DD) to a depth D = 180 is given by the regression equation DD = 2.69 + 0.0138D = 2.69 + 0.0138 × 180 = 5.174 | Points Earned: | 1/1 | Correct Answer: | 5.174 | Your Response: | 5.174 | 4. | The dives varied from 40 meters to 300 meters in depth. Plot the regression line from x = 40 to x = 300. Which of the lines in the figure below is the correct regression line? | | A. | Blue | B. | Yellow | C. | Red | |
To plot the line, compute DD = 3.242 minutes when D = 40 meters, and DD = 6.83 minutes when D = 300 meters. | Points Earned: | 0/1 | Correct Answer: | A | Your Response: | C | | Biochemical oxygen demand (BOD) measures organic pollutants in water by measuring the amount of oxygen consumed by microorganisms that break down these compounds. BOD is hard to measure accurately. Total organic carbon (TOC) is easy to measure, so it is common to measure TOC and use regression to predict BOD. A typical regression equation for water entering a municipal treatment plant isBOD = −55.43 + 1.507 TOCBoth BOD and TOC are measured in milligrams per liter of water. |

5. | What does the slope of this line say about the relationship between BOD and TOC? | | A. | For every unit increase in BOD, the TOC increases by 1.507 units. | B. | For every unit increase in TOC, the BOD increases by 1.507 units. | C. | For every unit increase in TOC, the BOD decreases by 1.507 units. | D. | For every unit increase in TOC, the BOD decreases by -55.34 units. | |
The slope of the line measures the rate of increase/decrease of the y variable for a change of one unit in the x variable. | Points Earned: | 1/1 | Correct Answer: | B | Your Response: | B | 6. | What is the predicted BOD when TOC = 1? Give your answer up to three decimal places. | | Answer |
The predicted value of the BOD at TOC = 1 is given by the regression equation ŷ = −55.43 + 1.507 x = −55.43 + 1.507 × 1 = −53.923 | Points Earned: | 1/1 | Correct Answer: | -53.923 | Your Response: | -53.923 | | Keeping water supplies clean requires regular measurement of levels of pollutants. The measurements are indirect—a typical analysis involves forming a dye by a chemical reaction with the dissolved pollutant, then passing light through the solution and measuring its “absorbence.” To calibrate such measurements, the laboratory measures known standard solutions and uses regression to relate absorbence and pollutant concentration. This is usually done every day. Here is one series of data on the absorbence for different levels of nitrates. Nitrates are measured in milligrams per liter of water.

Data Set |

7. | Chemical theory says that these data should lie on a straight line. If the correlation is not at least 0.997, something went wrong and the calibration procedure is repeated. Plot the data and find the correlation. Must the calibration be done again? | | A. | Yes | B. | No | |
Using a calculator or spreadsheet we find the correlation to be 0.999939232, thus there is no need to repeat the experiment. | Points Earned: | 1/1 | Correct Answer: | B | Your Response: | B | 8. | The calibration process sets nitrate level and measures absorbence. Once established, the linear relationship will be used to estimate the nitrate level in water from a measurement of absorbence. What is the equation of the line used for estimation? | | A. | ŷ = −8.8250 − 14.5271x | B. | ŷ = −14.5217 + 8.8250x | C. | ŷ = 1.333 − .1657x | D. | ŷ = 14.522 + 8.825x | |
Using the spreadsheet or calculator we find that the slope is given by b = 8.8250, while the intercept is a = −-14.5217 Thus the equation that matches ŷ = a + bx. | Points Earned: | 1/1 | Correct Answer: | B | Your Response: | B | 9. | Based on the equation exactly as it appears in your choice in the previous question (to avoid roundoff errors), what is the estimated nitrate level in a water specimen with absorbence 224? Give your answer to one decimal place. | | Answer |
Evaluating the regression equation with x = 224 gives (result rounded to one decimal place) ŷ = a + bx = −14.522 + 8.825 × 224 = 1962.3 | Points Earned: | 1/1 | Correct Answer: | 1962.3 | Your Response: | 1962.3 | 10. | Do you expect estimates of nitrate level from absorbence to be quite accurate? | | A. | Yes | B. | No | |
The results should be very accurate since r2 is very close to unity. | Points Earned: | 1/1 | Correct Answer: | A | Your Response: | A | 11. | Make a scatterplot of cerebellum weight (y) against body weight (x) and another scatterplot using the transformed values.
Compare your scatterplots to the ones below and choose the correct ones. | | A. | Scatterplot I | B. | Scatterplot II | C. | Scatterplot III | D. | Scatterplot IV | |
Scatterplots II and III are correct.
Scatterplots I and IV are incorrect because they do not match the data. | Points Earned: | 2/2 | Correct Answer: | B, C | Your Response: | B, C | 12. | Describe and compare both scatterplots.
What did you learn about the relationship between cerebellum and body weight from these graphs? | | A. | Small mammal species are much more variable and rarer than the larger species. Creating a graph from the transformed data helps to reduce that variability and uncover a linear relationship between the two transformed variables. | B. | Large mammal species are much more variable and rarer than the smaller species. Creating a graph from the transformed data helps to reduce that variability and uncover a linear relationship between the two transformed variables. | C. | Large mammal species are much less variable and rarer than the smaller species. Creating a graph from the transformed data helps to reduce that variability and uncover a non-linear relationship between the two transformed variables. | |

| Points Earned: | 1/1 | Correct Answer: | B | Your Response: | B | | Measuring tree height is not an easy task. How well does trunk diameter predict tree height?
A survey of 958 live trees in an old-growth forest in Canada provides us with the following information: the mean tree height is 15.6 m with standard deviation 13.4 m; the mean diameter, measured at "breast height" (1.3 m above ground), is 23.4 cm with standard deviation 23.5 cm.
The correlation between the height and diameter is very high: r = 0.96. |

13. | What is the slope of the regression line to predict tree height (in m) from trunk diameter (in cm)?
Give your answer in m/cm to four decimal places.
Fill in the blank: | | Answer |
The diameters are the x values and the heights are the y values.
The slope and intercept are b = r · sy/sx = 0.96 · 13.4/23.5= 0.5474 and a = y − bx = 15.6 − (0.5474)(23.4) = 2.791.
The regression equation is yˆ = 2.791 + 0.5474x. | Points Earned: | 0/1 | Correct Answer: | 0.5474 | Your Response: | .5679 | 14. | What is the intercept of the regression line to predict tree height?
Give your answer to three decimal places.
Fill in the blank: | | Answer |

| Points Earned: | 0/1 | Correct Answer: | 2.791 | Your Response: | 2.310 | 15. | Draw a graph of this regression line.
Compare your graph to the one below and choose the correct line. | | A. | Black | B. | Green | C. | Blue | |
The correct regression line is black. | Points Earned: | 0/1 | Correct Answer: | A | Your Response: | B | 16. | The tree diameters ranged from 1 to 101 cm.
Predict the height of a tree in this area that is 50 cm wide in diameter at breast height.
Give your answer in meters to 2 decimal places.
Fill in the blank: | | Answer |
For x = 50 cm we predict yˆ = 2.791 + 0.5474 × 50 = 30.16 m. | Points Earned: | 0/1 | Correct Answer: | 30.16 | Your Response: | 54 | 17. | Use r2 to argue the reliability of this prediction. | | A. | r2 is high implying that the prediction is very precise, provided that the relationship is linear. | B. | r2 is low implying that the prediction is very precise, provided that the relationship is linear. | C. | r2 is high implying that the prediction is very precise, provided that the relationship is non-linear. | D. | r2 is low implying that the prediction is very precise, provided that the relationship is non-linear. | | r2 = 0.962 = 0.9216.
A high r2 implies that the prediction is very precise, provided that the relationship is linear. | Points Earned: | 0/1 | Correct Answer: | A | Your Response: | C | | The EESEE data set on breakfast cereals examined the nutritional content of 77 brands. |

18. | As expected, sugar content is positively correlated with calories per serving and explains 32% of the variations in calories per serving.
What is the numerical value of the correlation between sugar content and calories per serving?
Give your answer to 3 decimal places.
Fill in the blank: | | Answer |
Because the association between sugar content and calories is positive, r must be positive.
Therefore, r = √r2 = √0.32 = 0.566.
A moderate correlation. | Points Earned: | 0/1 | Correct Answer: | 0.566 | Your Response: | 1.897 | 19. | Fiber content explains 9% of the calories per serving in this data set.
What is the numerical value of the correlation between fiber content and calories per serving?
What information are you missing to be certain of the value of the correlation? | | A. | Since we do not know what the relationship between sugar content and fiber content is we can only calculate a predicted value of the correlation, which is 0.3. | B. | Since we do not have information on the other ingredients of the cereal, we can only calculate a predicted value of the correlation, which is 0.3. | C. | Since we do not know whether the association between fiber content and calories is positive or negative we can only calculate the absolute value of the correlation, which is 0.3. | |
This time, we do not know whether the association between fiber content and calories is positive or negative.
Therefore, we can only calculate the absolute value of the correlation:
√0.09 = 0.3.
A weak correlation. | Points Earned: | 1/1 | Correct Answer: | C | Your Response: | C | | Researchers performed a study that shows that social exclusion causes “real pain.” That is, activity in an area of the brain that responds to physical pain goes up as distress from social exclusion goes up. A scatterplot shows a moderately strong linear relationship. The following figure shows regression output from software for these data.

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20. | What is the equation of the least-squares regression line for predicting brain activity from social distress score? | | A. | ŷ = 0.127 + 0.061x | B. | ŷ = 0.127x + 0.061 | C. | ŷ = -0.126 + 0.061x | D. | ŷ = -0.127x + 0.061 | |
From the table we see that the intercept is a = −0.127, and the slope is in the "pairs" row, giving b = 0.061. | Points Earned: | 0/1 | Correct Answer: | C | Your Response: | B | 21. | Use the equation exactly as it appears in your choice in question 1 (to avoid roundoff errors) to predict brain activity for a social distress score of 1.8. Your answer should be accurate to 3 decimal places. | | Answer |
The regression equation is ŷ = a + bx = −0.126 + 0.061x
Now we evaluate it for x = 1.8 to obtain (result rounded to 3 decimal places) ŷ = −0.127 + 0.061 × 1.8 = −0.017 | Points Earned: | 0/1 | Correct Answer: | -0.016 | Your Response: | 2.908 | 22. | What percent of the variation in brain activity among these subjects is explained by the straight-line relationship with social distress score? (give your answer to one decimal points) | | Answer |
The fraction of variation explained by the linear relationship is given by r2. In this case the required percentage is calculated as r2 × 100% = (0.8782)2 × 100% = 77.1% | Points Earned: | 0/1 | Correct Answer: | 77.1% | Your Response: | 87.8% | | Researchers collected data on the number of breeding pairs of merlins in an isolated area in each of nine years and the percent of males who returned the next year. The data show that the percent returning is lower after successful breeding seasons and that the relationship is roughly linear. The following shows software regression output for these data.

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23. | What is the equation of the least-squares regression line for predicting the percent of males that return from the number of breeding pairs? | | A. | ŷ = 157.682 − 2.993x | B. | ŷ = 157.682x − 2.993 | C. | ŷ = 157.682 + 2.993x | D. | ŷ = 157.682x + 2.993 | |
From the table we see that the intercept is a = 157.682, and the slope is in the "pairs" row, giving b = −2.993. | Points Earned: | 0/1 | Correct Answer: | A | Your Response: | C | 24. | Use the equation exactly as chosen in the first question (to avoid roundoff errors) to predict the percent of returning males after a season with 29 breeding pairs. Give your answer to two decimal places. | | Answer |
The regression equation is ŷ = a + bx = 157.682 − 2.993x
Now we evaluate it for x = 29 to obtain (result rounded to 2 decimal places) ŷ = 157.682 − 2.993 × 29 = 70.89 | Points Earned: | 0/1 | Correct Answer: | 70.89 | Your Response: | 23.89 | 25. | What percent of the year-to-year variation in percent of returning males is explained by the straight-line relationship with number of breeding pairs the previous year? Give your answer to one decimal point. | | Answer |
The fraction of variation explained by the linear relationship is given by r2. In this case the required percentage is calculated as r2 × 100% = (−0.7943)2 × 100% = 63.1% | Points Earned: | 0/1 | Correct Answer: | 63.1 | Your Response: | 1.2 | 26. | Make a scatterplot of mice survival rate (response) against aspartame concentration (explanatory) and add the regression line.
Compare your scatterplot to the one below and choose the correct regression line. | | A. | Red | B. | Green | C. | Blue | D. | Gray | E. | Black | |
The correct regression line is blue. | Points Earned: | 0/1 | Correct Answer: | C | Your Response: | B | 27. | Describe the direction, form, and strength of the relationship. | | A. | The plot shows a fairly strong, positive linear relationship, with no outliers. | B. | The plot shows a mild, negative linear relationship, with possibly an outlier and an influential point. | C. | The plot shows a fairly strong, negative linear relationship with one outlier. | D. | The plot shows a fairly strong, non-linear relationship. | |
The correct answer is B. The plot shows a very mild, negative linear relationship, with possibly an outlier and an influential point. | Points Earned: | 0/1 | Correct Answer: | B | Your Response: | C | 28. | What percent of the variation in mice survival rate can be explained by variations in aspartame concentrations?
Give your answer in percents rounded off to a whole number.
Fill in the blank: | | Answer |
Concentration explains only about 7% of the variation in survival rate. | Points Earned: | 0/1 | Correct Answer: | 7 | Your Response: | 34 | 29. | Make another scatterplot, this time with all the data points except the one for an aspartame concentration of 6250 ppm. Add the regression line.
Compare your new regression line to the ones in the scatterplot above and choose the correct one. | | A. | Red | B. | Green | C. | Blue | D. | Gray | E. | Black | |
Now the correct regression line is black. | Points Earned: | 0/1 | Correct Answer: | E | Your Response: | D | 30. | Compare the regression line with the one obtained from the data with all data points. What happened to the regression line?
What does this say about the data point you just removed? | | A. | The plot now shows a less steep negative linear relationship with an outlier. The removed point was an influential point for calculating the regression line. | B. | The plot now shows a stronger positive linear relationship with one outlier. The removed point was an outlier with little influence on the regression line. | C. | The plot now shows a mild, but positive linear relationship with an outlier. The removed point was an influential point for calculating the regression line. | D. | The plot now shows a steeper, negative relationship. The removed point was an outlier with little influence on the regression line. | |
Concentration now explains about 40% of the variation in survival rate.
The experiment is very challenging to interpret.
An association is not at all evident. | Points Earned: | 1/1 | Correct Answer: | D | Your Response: | D | 31. | Make another scatterplot, this time ignoring the last data point (an aspartame concentration of 50,000 ppm).
Add the regression line.
Compare your new regression line to the ones in the scatterplot above and choose the correct one. | | A. | Red | B. | Green | C. | Blue | D. | Gray | E. | Black | |
Now the correct regression line is red. | Points Earned: | 0/1 | Correct Answer: | A | Your Response: | B | 32. | What happened to the regression line?
What does this say about the data point corresponding to the aspartame concentration of 50,000 ppm? | | A. | The plot now shows a mild, but positive linear relationship with an outlier. The removed point was an influential point for calculating the regression line. | B. | The plot now shows a stronger negative linear relationship. The removed point was an influential point for calculating the regression line. | C. | The plot now shows a positive strong linear relationship with two outliers. The removed point was an outlier, with little influence on the regression line. | D. | The plot now shows weaker negative linear relationship with an outlier. The removed point was an influential point for calculating the regression line. | |
Now concentration explains only about 15% of the variation in survival rate. | Points Earned: | 0/1 | Correct Answer: | A | Your Response: | C | | Drilling down beneath a lake in Alaska yields chemical evidence of past changes in climate. Biological silicon, left by the skeletons of single-celled creatures called diatoms, is a measure of the abundance of life in the lake. A rather complex variable based on the ratio of certain isotopes relative to ocean water gives an indirect measure of moisture, mostly from snow. As we drill down, we look further into the past. Here are data from 2300 to 12,000 years ago:

Data Set |

33. | Make a scatter plot of silicon (response) against isotope (explanatory). Ignoring the outlier, choose the correct description for the direction, form and strength of the relationship. The researchers say that this and relationships among other variables they measured are evidence for cyclic changes in climate that are linked to changes in the sun's activity. Pick all of the descriptions below that apply to your plot. | | A. | Linear relationship | B. | Non-linear relationship | C. | Strong correlation | D. | Moderately strong correlation | E. | Positive correlation | F. | Negative correlation | |
Ignoring the outlier we see that we have a linear relationship, with weak correlation due to the large scattering of points, and this correlation is negative, since the slope of the linear relationship is negative. | Points Earned: | 1/3 | Correct Answer: | A, D, F | Your Response: | A | 34. | The researchers single out one point: “The open circle in the plot is an outlier that was excluded in the correlation analysis.” Of the four points circled on the graph below, which one is the correct outlier that the researchers circled? | | A. | A | B. | B | C. | C | D. | D | |
It is clear that point B lies outside the general negative slope linear trend. | Points Earned: | 1/1 | Correct Answer: | B | Your Response: | B | 35. | Calculate the correlation with the outlier point. Your answers should be accurate to 3 decimal places. | | Answer |
Use your spreadsheet application or calculator to calculate the correlation with the outlier taken into account. | Points Earned: | 0/1 | Correct Answer: | -0.339 | Your Response: | .876 | 36. | Calculate the correlation without the outlier point. Your answers should be accurate to 3 decimal places. | | Answer |
Use your spreadsheet application or calculator to calculate the correlation without the outlier taken into account. | Points Earned: | 0/1 | Correct Answer: | -0.787 | Your Response: | .987 | 37. | Based on your previous results, determine if the following statement is true."Removal of the outlier point weakens the correlation." | | Answer |
From the previous calculations we saw that removal of the outlier moves the correlation closer to -1, indicating a stronger correlation with the outlier removed. | Points Earned: | 0/1 | Correct Answer: | False | Your Response: | True | 38. | Now we would like to examine if the outlier you found in the previous question is strongly influential for the regression line. Calculate and draw on your graph two regression lines, with and without the outlier taken into account. Examine the plot below and choose the correct regression line corresponding to all of the points taken into account. | | A. | A | B. | B | C. | C | D. | D | |
Your plot of the regression line for all the points should look similar to choice C. | Points Earned: | 1/1 | Correct Answer: | C | Your Response: | C | 39. | Choose the correct regression line corresponding to ignoring the outlier point. | | A. | A | B. | B | C. | C | D. | D | |
Your plot of the regression line ignoring the outlier should look similar to choice D. | Points Earned: | 0/1 | Correct Answer: | D | Your Response: | C | 40. | Choose the correct explanation for the movement of the regression line when the outlier is taken into account. | | A. | The regression line gets a more negative slope since the outlier pushes the regression line away. | B. | The regression line gets a more negative slope since the outlier pulls the regression line towards it. | C. | The regression line gets a more positive slope since the outlier pushes the regression line away. | D. | The regression line gets a more positive slope since the outlier pulls the regression line towards it. | |
The regression line always get pulled towards outliers when they are taken into account. Since the outlier in this case is in the upper right corner, when the regression line get pulls towards it, the slope increases, i.e. the slope becomes more positive. | Points Earned: | 0/1 | Correct Answer: | D | Your Response: | B | 41. | Make a scatterplot of the relationship between latitude of anglerfish distribution (y) and temperature (x).
Describe the shape and strength of that relationship. | | A. | There is a reasonably mild, negative linear relationship between temperature and latitude. | B. | There is a reasonably mild, positive non-linear relationship between temperature and latitude. | C. | There is no apparent relationship between temperature and latitude. | D. | There is a reasonably strong, positive linear relationship between temperature and latitude. | |

| Points Earned: | 0/1 | Correct Answer: | D | Your Response: | B | 42. | Give the equation for the regression line to predict latitude from temperature and the value of the correlation.
Add the regression line to your scatterplot. | | A. | yˆ = 51.261 + 0.984x, r=0.7767. | B. | yˆ = 52.452 + 0.818x, r=0.6033. | C. | yˆ = 54.253 + 0.748x, r=0.3640. | |

The correlation is r = 0.6033. | Points Earned: | 0/1 | Correct Answer: | B | Your Response: | A | 43. | The scatterplot shows one outlier of the relationship.
Take that point out of the data set and calculate the new regression line and correlation.
Add the new regression line to the original scatterplot, making sure to use a different color or line thickness.
Compare your scatterplot to the one below and choose the correct new regression line. | | A. | Black | B. | Green | C. | Blue | |
The correct new regression line is green.
The regression line with the outlier is blue.
The outlier is the 6th observation.
The new regression line has for equation yˆ = 51.26 + 0.9843x and the correlation is now r = 0.7643. | Points Earned: | 1/1 | Correct Answer: | B | Your Response: | B | 44. | How do the regression line and correlation omitting the outlier compare with the original ones in (2)?
Use your findings to discuss how influential the outlier is. | | A. | The outlier strengthens the correlation and makes the slope less steep. | B. | The outlier strengthens the correlation and makes the slope steeper. | C. | The outlier weakens the correlation and makes the slope steeper. | D. | The outlier weakens the correlation and makes the slope less steep. | |

| Points Earned: | 0/1 | Correct Answer: | D | Your Response: | B | | The Correlation and Regression applet allows you to animate Figure 4.6. Click to create a group of 10 points in the lower-left corner of the scatterplot with a strong straight-line pattern (correlation about 0.9). Click the “Show least-squares line” box to display the regression line. |

45. | Add one point at the upper right that is far from the other 10 points but exactly on the regression line. Why does this outlier have no effect on the line even though it changes the correlation? | | A. | Placing a new point on the existing line doesn't change the averages x and y, thus the parameters of the regression line remain the same. | B. | The new point doesn't change the correlation, and this in turn, means that the regression line doesn't change either. | C. | The new point on the existing line doesn't change the standard deviations of x and y, meaning that the slope of the regression line cannot change. | D. | The new point on the existing line exactly matches the prediction, thus it has no residual. Since the regression line results in the least sum of residual squares, adding a point with zero residual doesn't change the line. | | | Points Earned: | 0/1 | Correct Answer: | D | Your Response: | C | 46. | Now use the mouse to drag this last point straight down. You see that one end of the least-squares line chases this single point, while the other end remains near the middle of the original group of 10. What makes the last point so influential? Select the correct factors from the list below. | | A. | The outlier is far from the group, both in the x and y directions, and therefore has a strong influence on the means x and y. The change in the means has a strong influence on the line. | B. | The line always passes through (x,y), and the group of 10 points dominates the average, thus the line has to pass through the group in the lower left corner, i.e. the group acts as a "pivot point" for the line. | C. | The outlier is far from the group, both in the x and y directions, and therefore has a strong influence on the standard deviations of both variables x and y. The change in the standard deviations strongly influences the parameters of the regression line. | D. | The outlier is far from the group, both in the x and y directions, and therefore has a strong influence on the correlation, which in turn, affects the slope of the line. | E. | The correlation is dominated by the group of points in the lower left corner, while the additional point strongly affects the averages x and y. This causes large changes in the regression line parameters. | | | Points Earned: | 1/2 | Correct Answer: | B, D | Your Response: | B |

47. | People who do well tend to feel good about themselves.
Perhaps helping people feel good about themselves will help them do better in school and life.
Raising self-esteem became for a time a goal in many schools.
California even created a state commission to advance the cause, hoping that it would help reduce crime, drug use, and teen pregnancy (the commission was ended in 1995).
Can you think of explanations for the association between high self-esteem and good school performance other than "Self-esteem causes better work in school"?
Choose one or more plausible explanations. | | A. | Good work and scholarly success may result in higher self-esteem. | B. | Good school work helps reduce teen pregnancy, which in turn leads to higher self-esteem. | C. | Scholarly success and strong self-esteem can both be explained by reduced crime. | D. | Both strong self-esteem and good school work can be explained by parental attitudes. | | | Points Earned: | 0/2 | Correct Answer: | A, D | Your Response: | B |

| One way to assess the link between smoking and lung cancer is to ask whether groups of individuals who tend to smoke more than the population average also experience higher rates of death from lung cancer than the general population.
The data set below was obtained for men in 25 occupational groups in England (for example, farmers, sales workers).
For each group, smoking extent and mortality from lung cancer are represented in the form of an index.
An index value of 100 corresponds exactly to the national average, while index values above 100 represent group values above the national average.

Data Set

Analyze these data to uncover the nature and strength of the effect of smoking on mortality from lung cancer.
Follow the four-step process (page 55) in reporting your work. | 48. | STATE: Choose one of the following statements that best describes the question we are trying to answer: | | A. | What is the relationship between various occupations and mortality caused by smoking among men? | B. | What is the relationship between mortality from smoking and mortality from lung cancer among men in various occupational groups? | C. | What is the relationship between smoking and mortality from lung cancer among men in various occupational groups? | D. | What is the relationship between various occupations and smoking among men with lung cancer? | |

| Points Earned: | 1/1 | Correct Answer: | C | Your Response: | C | 49. | FORMULATE: What kind of graph would you use to study the data? | | A. | A scatterplot with smoking as the explanatory variable and mortality from lung cancer as the response variable.
If appropriate, the regression line and correlation would be added. | B. | A scatterplot with mortality from smoking as the explanatory variable and mortality from lung cancer as the response variable.
In addition, the regression line and correlation would be added. | C. | A scatterplot with mortality from smoking as the explanatory variable and mortality from lung cancer as the response variable.
If the relationship is non-linear, the regression line and correlation would be added. | D. | A scatterplot with mortality from lung cancer as the explanatory variable and smoking as the response variable.
If appropriate, the regression line and correlation would be added. | |

| Points Earned: | 0/1 | Correct Answer: | A | Your Response: | B | 50. | SOLVE: Describe the shape and strength of the relationship between the two variables.
If appropriate find the regression line and calculate r2. | | A. | The scatterplot shows a positive non-linear association.
Regression appears to be inappropriate. | B. | The scatterplot shows a negative linear association.
Regression appears to be appropriate; the regression line is yˆ = −1.0875 − 2.8853x and r2 = 0.513. | C. | The scatterplot shows a positive linear association.
Regression appears to be appropriate; the regression line is yˆ = −2.8853 + 1.0875x and r2 = 0.513. | D. | The scatterplot shows a positive linear association.
Regression appears to be appropriate; the regression line is yˆ = −1.8853 + 1.0875x and r2 = 0.713. | |
The straight-line relationship explains r2 = 51.3% of the variation in lung cancer mortality.

| Points Earned: | 0/1 | Correct Answer: | C | Your Response: | B | 51. | CONCLUDE: Choose one of the following statements that best describes your conclusions. | | A. | The weak positive association supports the idea that certain occupations may increase the risk of dying from smoking. | B. | The moderately weak negative association supports the idea that smoking may increase the risk of dying from lung cancer. | C. | The non-linear positive association supports the idea that lung cancer may increase the risk of dying from smoking. | D. | The moderately strong positive association supports the idea that smoking may increase the risk of dying from lung cancer. | | | Points Earned: | 1/1 | Correct Answer: | D | Your Response: | D |

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