MATH 225N Week 8 Final Exam 2024-2025: Linear Regression Exercises and Solutions, Exams of Mathematics

Download MATH 225N Week 8 Final Exam 2024-2025: Linear Regression Exercises and Solutions and more Exams Mathematics in PDF only on Docsity!

r= −0.

Correct! You nailed it. MATH 225N Week 8 Final Exam 2024-2025 (Version 2 ). Questions and Correct Verified Answers. Graded A Week 8 questions and answers Performing Linear Regressions with Technology An amateur astronomer is researching statistical properties of known stars using a variety of databases. They collect the absolute magnitude or MV and stellar mass or M⊙ for 30 stars. The absolute magnitude of a star is the intensity of light that would be observed from the star at a distance of 10 parsecs from the star. This is measured in terms of a particular band of the light spectrum, indicated by the subscript letter, which in this case is V for the visual light spectrum. The scale is logarithmic and an MV that is 1 less than another comes from a star that is 10 times more luminous than the other. The stellar mass of a star is how many times the sun's mass it has. The data is provided below. Use Excel to calculate the correlation coefficient r between the two data sets, rounding to two decimal places. Answer Explanation The correlation coefficient, rounded to two decimal places, is r≈−0.93. A market researcher looked at the quarterly sales revenue for a large e-commerce store and for a large brick-and-mortar retailer over the same period. The researcher recorded the revenue in millions of dollars for 30 quarters. The data are provided below. Use Excel to calculate the correlation coefficient r

r= −0.

Yes that's right. Keep it up! between the two data sets. Round your answer to two decimal places. Answer Explanation The correlation coefficient, rounded to two decimal places, is r≈−0.81.

Yes that's right. Keep it up! The table below contains the geographic latitudes, x, and average January temperatures, y, of 20 cities. Use Excel to find the best fit linear regression equation. Round the slope and intercept to two decimal places. 46 23 32 60 39 40 33 59 38 57 40 33 42 33 30 64 34 56 HelpCopy to ClipboardDownload CSV

y = −2.68, x147.

Thus, the equation of line of best fit with slope and intercept rounded to two decimal places is yˆ=−2.68x+147.24. An organization collects information on the life expectancy (in years) of a person in certain countries and the fertility rate per woman in those countries. The data for 21 randomly selected countries for the year 2011 is given below. Use Excel to find the best fit linear regression equation, where fertility rate is the explanatory variable. Round the slope and intercept to two decimal places. y = −4.21, x 83.68 Answer Explanation yˆ=−4.21, x+83.68. x y

r= −0.

Perfect. Your hard work is paying off

r= 0.

Great work! That's correct. That's not right - let's review the answer. Well done! You got it right. An economist is trying to understand whether there is a strong link between CEO pay ratio and corporate revenue. The economist gathered data including the CEO pay ratio and corporate revenue for 30 companies for a particular year. The pay ratio data is reported by the companies and represents the ratio of CEO compensation to the median employee salary. The data are provided below. Use Excel to calculate the correlation coefficient r between the two data sets. Round your answer to two decimal places. The correlation coefficient, rounded to two decimal places, is r≈−0.17. A researcher is interested in whether the variation in the size of human beings is proportional throughout each part of the human. To partly answer this question they looked at the correlation between the foot length (in millimeters) and height (in centimeters) of 30 randomly selected adult males. The data is provided below. Use Excel to calculate the correlation coefficient r between the two data sets. Round your answer to two decimal places. The correlation coefficient, rounded to two decimal places, is r≈0.50. The table below gives the average weight (in kilograms) of certain people ages 1 –

Use Excel to find the best fit linear regression equation, where age is the explanatory variable. Round the slope and intercept to two decimal places. Answer 1:

y = 0.35, x28.

Answer 2:

Yes that's right. Keep it up! Yes that's right. Keep it up!

y = 2.89, x 4.

Thus, the equation of line of best fit with slope and intercept rounded to two decimal places is yˆ=2.86x+4.69. In the following table, the age (in years) of the respondents is given as the x value, and the earnings (in thousands of dollars) of the respondents are given as the y value. Use Excel to find the best fit linear regression equation in thousands of dollars. Round the slope and intercept to three decimal places.

y = 0.433, x=24.

Answer Explanation Thus, the equation of line of best fit with slope and intercept rounded to three decimal places is yˆ=0.433x+24.493. PREDICITONS USING LINEAR REGRESSION Question The table shows data collected on the relationship between the time spent studying per day and the time spent reading per day. The line of best fit for the data is yˆ=0.16x+36.2. Assume the line of best fit is significant and there is a strong linear relationship between the variables. Studying (Minutes) 507090110 Reading (Minutes) 44485054 (a) According to the line of best fit, what would be the predicted number of minutes spent reading for someone who spent 67 minutes studying? Round your answer to two decimal places.

The predicted number of minutes spent reading is $$46.92.

That's incorrect - mistakes are part of learning. Keep trying! Answer Explanation

The predicted number of minutes spent reading is 1 $$.

Correct answers:

Substitute 67 for x into the line of best fit to estimate the number of minutes spent reading for someone who spent 67 minutes studying: yˆ=0.16(67)+36.2=46.92. Question The table shows data collected on the relationship between the time spent studying per day and the time spent reading per day. The line of best fit for the data is yˆ=0.16x+36.. Studying (Minutes) 507090110 Reading (Minutes) 44485054 (a) According to the line of best fit, the predicted number of minutes spent reading for someone who spent 67 minutes studying is 46.. (b) Is it reasonable to use this line of best fit to make the above prediction? The estimate, a predicted time of 46.92 minutes, is both reliable and reasonable. The estimate, a predicted time of 46.92 minutes, is both unreliable and unreasonable. The estimate, a predicted time of 46.92 minutes, is reliable but unreasonable. The estimate, a predicted time of 46.92 minutes, is unreliable but reasonable. Answer Explanation

• 46.

Not quite - review the answer explanation to help get the next one. Correct answer: The estimate, a predicted time of 46.92 minutes, is both reliable and reasonable. The data in the table only includes studying times between 50 and 110 minutes, so the line of best fit gives reliable and reasonable predictions for values of x between 50 and 110. Since 67 is between these values, the estimate is both reliable and reasonable. Your answer: The estimate, a predicted time of 46.92 minutes, is unreliable but reasonable. This estimate is both reliable and reasonable because 67 is inside the range 50 to 110 given in the table. Janet is studying the relationship between the average number of minutes spent exercising per day and math test scores and has collected the data shown in the table. The line of best fit for the data is yˆ=0.46x+66.. Minutes 15202530 Test Score 73767880 (a) According to the line of best fit, the predicted test score for someone who spent 23 minutes exercising is 76.. (b) Is it reasonable to use this line of best fit to make the above prediction? The estimate, a predicted test score of 76.98, is reliable and reasonable. The estimate, a predicted test score of 76.98, is unreliable but reasonable. The estimate, a predicted test score of 76.98, is unreliable and unreasonable. The estimate, a predicted test score of 76.98, is reliable but unreasonable. Answer Explanation

Correct answer: The estimate, a predicted test score of 76.98, is reliable and reasonable.

The data in the table only includes exercise times between 15 and 30 minutes, so the line of best fit gives reliable and reasonable predictions for values of x between 15 and 30. Since 23 is between these values, the estimate is reasonable. Your answer: The estimate, a predicted test score of 76.98, is unreliable and unreasonable. This estimate is reliable, because 23 is inside the range 15 to 30 given in the table. And, it is a realistic score, so it is reasonable. Nomenclature

• When using regression lines to make predictions, if the x-value is within the range of observed x-values, one can conclude the prediction is both reliable and reasonable. That is, the prediction is accurate and possible. For example, if a prediction were made using x=1995 in the video above, one could conclude the predicted y-value is both reliable (quite accurate) and reasonable (possible). This is an example of interpolation.
• When using regression lines to make predictions, if the x-value is outside the range of observed x-values, one cannot conclude the prediction is both reliable and reasonable. That is, the prediction is will be much less accurate and the prediction may, or may not, be possible. For example, x= is not within the range of 1950 to 2000. Therefore, the prediction is much less reliable (not as accurate) even though it is reasonable (it is possible that a person will live to be 79.72 years old). This is an example of extrapolation.

Reasonable Predictions

Note that not all predictions are reasonable using a line of best fit. Typically, it is considered reasonable to make predictions for x-values which are between the smallest and largest observed x-values. These are known as interpolated values. Typically, it is

considered unreasonable to make predictions for x-values which are not between the smallest and largest observed x-values. These are known as extrapolated values. A scatterplot has a horizontal axis labeled x from 0 to 20 in increments of 1 and a vertical axis labeled y from 0 to 28 in increments of 2. 15 plotted points strictly follow the pattern of a line that rises from left to right and passes through the points left- parenthesis 6 comma 10 right-parentheses, left-parenthesis 8 comma 13 right- parenthesis, and left-parenthesis 14 comma 2 right-parentheses. There are other plotted points at left- parenthesis 10 comma 15 right-parenthesis and left-parenthesis 13 comma 19 right-parenthesis. The regions between the horizontal axis points from 1 to 6 and 14 to 20 are shaded as unreasonable. The region between the horizontal axis points from 6 to 14 is shaded as reasonable. All coordinates are approximate In the figure above, we see that the observed values have x-values ranging from 6 to 14. So it would be reasonable to use the line of best fit to make a prediction for the x value of 9 (because it is between 6 and 14 ), but it would be unreasonable to make a prediction for the x-value of 20 (because that is outside of the range). Nomenclature

• When using regression lines to make predictions, if the x-value is within the range of observed x-values, one can conclude the prediction is both reliable and reasonable. That is, the prediction is accurate and possible. For example, if a prediction were made using x=1995 in the video above, one could conclude the predicted y-value is both reliable (quite accurate) and reasonable (possible). This is an example of interpolation.
• When using regression lines to make predictions, if the x-value is outside the range of observed x-values, one cannot conclude the prediction is both reliable and reasonable. That is, the prediction is will be much less accurate and the prediction may, or may not, be possible. For example, x= is not within the range of 1950 to 2000. Therefore, the prediction is much less reliable (not as accurate) even though it is reasonable (it is possible that a

person will live to be 79.72 years old). This is an example of extrapolation.

Reasonable Predictions

Note that not all predictions are reasonable using a line of best fit. Typically, it is considered reasonable to make predictions for x-values which are between the smallest and largest observed x-values. These are known as interpolated values. Typically, it is considered unreasonable to make predictions for x-values which are not between the smallest and largest observed x-values. These are known as extrapolated values. A scatterplot has a horizontal axis labeled x from 0 to 20 in increments of 1 and a vertical axis labeled y from 0 to 28 in increments of 2. 15 plotted points strictly follow the pattern of a line that rises from left to right and passes through the points left- parenthesis 6 comma 10 right-parentheses, left-parenthesis 8 comma 13 right- parenthesis, and left-parenthesis 14 comma 2 right-parentheses. There are other plotted points at left- parenthesis 10 comma 15 right-parenthesis and left-parenthesis 13 comma 19 right-parenthesis. The regions between the horizontal axis points from 1 to 6 and 14 to 20 are shaded as unreasonable. The region between the horizontal axis points from 6 to 14 is shaded as reasonable. All coordinates are approximate In the figure above, we see that the observed values have x-values ranging from 6 to 14. So it would be reasonable to use the line of best fit to make a prediction for the x value of 9 (because it is between 6 and 14 ), but it would be unreasonable to make a prediction for the x-value of 20 (because that is outside of the range).

Question

Erin is studying the relationship between the average number of minutes spent reading per day and math test scores and has collected the data shown in the table. The line of best fit for the data is yˆ=0.8x+51.2. According to the line of best fit, what would be the predicted test score for someone who spent 70 minutes reading? Is it reasonable to use this line of best fit to make this prediction? Minutes 3035404550 Test Score 7578858890

The predicted test score is 95.2, and the estimate is not reasonable. The predicted test score is 95.2, and the estimate is reasonable. The predicted test score is 107.2, and the estimate is not reasonable. The predicted test score is 107.2, and the estimate is reasonable. Answer Explanation Correct answer: The predicted test score is 107.2, and the estimate is not reasonable. Substitute 70 for x in the line of best fit to estimate the test score for someone who spent 70 minutes reading: yˆ=0.8(70)+51.2=107.2. The data in the table only includes reading times between 30 and 50 minutes, so the line of best fit only gives reasonable predictions for values of x between 30 and 50. Since 70 is far outside of this range of values, the estimate is not reasonable. Another thing to notice is that it predicts a test score of greater than 100 , which is typically impossible. Your answer: The predicted test score is 107.2, and the estimate is reasonable. The predicted value is not reasonable because the value of 70 minutes is not between 30 and 50 minutes. Question Data is collected on the relationship between the average number of minutes spent exercising per day and math test scores. The data is shown in the table and the line of best fit for the data is yˆ=0.42x+64.6. Assume the line of best fit is significant and there is a strong linear relationship between the variables. Minutes 25303540 Test Score 75778081

Well done! You got it right. Perfect. Your hard work is paying off (a) According to the line of best fit, what would be the predicted test score for someone who spent 38 minutes exercising? Round your answer to two decimal places.

The predicted test score is $$80.56.

Answer Explanation

The predicted test score is 1 $$.

Correct answers:

Substitute 38 for x into the line of best fit to estimate the test score for someone who spent 38 minutes exercising: yˆ=0.42(38)+64.6=80.56. Question Data is collected on the relationship between the average number of minutes spent exercising per day and math test scores. The data is shown in the table and the line of best fit for the data is yˆ=0.42x+64.. Minutes 25303540 Test Score 75778081 (a) According to the line of best fit, the predicted test score for someone who spent 38 minutes exercising is 80.. (b) Is it reasonable to use this line of best fit to make the above prediction? The estimate, a predicted test score of 80.56, is both reliable and reasonable.

• 1 $80.56$80.

That's not right - let's review the answer. The estimate, a predicted test score of 80.56, is reliable but unreasonable. The estimate, a predicted test score of 80.56, is both unreliable and unreasonable. The estimate, a predicted test score of 80.56, is unreliable but reasonable. Answer Explanation Correct ans wer: The estimate, a predicted test score of 80.56, is both reliable and reasonable. The data in the table only includes exercise times between 25 and 40 minutes, so the line of best fit gives reasonable predictions for values of x between 25 and 40. Since 38 is between these values, the estimate is both reliable and reasonable. Question Data is collected on the relationship between the average daily temperature and time spent watching television. The data is shown in the table and the line of best fit for the data is y^=−0.81x+96.7. Assume the line of best fit is significant and there is a strong linear relationship between the variables. Temperature (Degrees) 30405060 Minutes Watching Televisio n 73635748 (a) According to the line of best fit, what would be the predicted number of minutes spent watching television for an average daily temperature of 45 degrees? Round your answer to two decimal places. Ans wer 1:

The predicted number of minutes spent watching television is $$133.15.

Ans wer 2:

Keep trying - mistakes can help us grow. Correct! You nailed it.

The predicted number of minutes spent watching television is $$133.15.

Answer Explanation

The predicted number of minutes spent watching television is 1 $$.

Correct answers:

Substitute 45 for x into the line of best fit to estimate the number of minutes spent watching television for an average daily temperature of 45 degrees: y^=−0.81(45)+96.7=60.. Question Data is collected on the relationship between the average daily temperature and time spent watching television. The data is shown in the table and the line of best fit for the data is y^=−0.81x+96.. Temperature (Degrees) 30405060 Minutes Watching Televisio n 73635748 (a) According to the line of best fit, the predicted number of minutes spent watching television for an average daily temperature of 45 degrees is 60.. (b) Is it reasonable to use this line of best fit to make the above prediction? The estimate, a predicted time of 60.25 minutes, is unreliable but reasonable. The estimate, a predicted time of 60.25 minutes, is both reliable and reasonable.

• 1 $60.25$60.

Yes that's right. Keep it up! The estimate, a predicted time of 60.25 minutes, is both unreliable and unreasonable. The estimate, a predicted time of 60.25 minutes, is reliable but unreasonable. Answer Explanation Correct answer: The estimate, a predicted time of 60.25 minutes, is both reliable and reasonable. The data in the table only includes temperatures between 30 and 60 degrees, so the line of best fit only gives reliable and reasonable predictions for values of x between 30 and 60. Since 45 is between these values, the estimate is both reliable and reasonable. Question Homer is studying the relationship between the average daily temperature and time spent watching television and has collected the data shown in the table. The line of best fit for the data is yˆ=−0.6x+94.5. Assume the line of best fit is significant and there is a strong linear relationship between the variables. Temperature (Degrees) 40506070 Minutes Watching Televisio n 70655952 (a) According to the line of best fit, what would be the predicted number of minutes spent watching television for an average daily temperature of 39 degrees? Round your answer to two decimal places, as needed.

The predicted number of minutes spent watching television is $$71.1.

Answer Explanation

The predicted number of minutes spent watching television is 1 $$.

Correct answers:

Not quite - review the answer explanation to help get the next one. Substitute 39 for x into the line of best fit to estimate the number of minutes spent watching television for an average daily temperature of 39 degrees: yˆ=−0.6(39)+94.5=71.1. Question Homer is studying the relationship between the average daily temperature and time spent watching television and has collected the data shown in the table. The line of best fit for the data is yˆ=−0.6x+94.. Temperature (Degrees) 40506070 Minutes Watching Televisio n 70655952 (a) According to the line of best fit, the predicted number of minutes spent watching television for an average daily temperature of 39 degrees is 71.. (b) Is it reasonable to use this line of best fit to make the above prediction? The estimate, a predicted time of 71.1 minutes, is both unreliable and unreasonable. The estimate, a predicted time of 71.1 minutes, is both reliable and reasonable. The estimate, a predicted time of 71.1 minutes, is unreliable but reasonable. The estimate, a predicted time of 71.1 minutes, is reliable but unreasonable. Answer Explanation Correct answer: The estimate, a predicted time of 71.1 minutes, is unreliable but reasonable.

• 1 $71.1$71.

Keep trying - mistakes can help us grow. The data in the table only includes temperatures between 40 and 70 degrees, so the line of best fit gives reliable and reasonable predictions for values of x between 40 and 70. Since 39 is not between these values, the estimate is not reliable. However, 71.1 minutes is a reasonable time. Your answer: The estimate, a predicted time of 71.1 minutes, is both reliable and reasonable. This estimate is not reliable, because 39 is outside of the range 40 to 70 given in the table. Question Daniel owns a business consulting service. For each consultation, he charges $95 plus $70 per hour of work. A linear equation that expresses the total amount of money Daniel earns per consultation is y=70x+95. What are the independent and dependent variables? What is the y-intercept and the slope? The independent variable (x) is the amount, in dollars, Daniel earns for a consultation. The dependent variable (y) is the amount of time Daniel consults. Daniel charges a one-time fee of $95 (this is when x=0), so the y-intercept is 95. Daniel earns $70 for each hour he works, so the slope is 70. The independent variable (x) is the amount of time Daniel consults. The dependent variable (y) is the amount, in dollars, Daniel earns for a consultation. Daniel charges a one-time fee of $95 (this is when x=0), so the y-intercept is 95. Daniel earns $70 for each hour he works, so the slope is 70. The independent variable (x) is the amount, in dollars, Daniel earns for a consultation. The dependent variable (y) is the amount of time Daniel consults. Daniel charges a one-time fee of $70 (this is when x=0), so the y-intercept

is 70. Daniel earns $95 for each hour he works, so the slope is 95.

The independent variable (x) is the amount of time Daniel consults. The dependent variable (y) is the amount, in dollars, Daniel earns for a consultation. Daniel charges a one-time fee of $95 (this is when x=0), so the y- The independent variable (x) is the amount of time Daniel consults. The dependent variable (y) is the amount, in dollars, Daniel earns for a consultation. Daniel charges a one-time fee of $70 (this is when x=0), so the y-

$$y=−20

Well done! You got it right. The independent variable (x) is the amount of time Daniel consults. The dependent variable (y) is the amount, in dollars, Daniel earns for a consultation. Daniel charges a one-time fee of $70 (this is when x=0), so the y-intercept is 70. Daniel earns $95 for each hour he works, so the slope is 95. Answer Explanation Correct answer: The independent variable (x) is the amount of time Daniel consults because it is the value that changes. He may work different amounts per consultation, and his earnings are dependent on how many hours he works. This is why the amount, in dollars, Daniel earns for a consultation is the dependent variable (y). The y-intercept is 95 (b=95). This is his one-time fee. The slope is 70 (a=70). This is the increase for each hour he works. Your answer: Question Given the following line, find the value of y when x=2. y=−4x−12 Answer Explanation

• $y=-20$ y =−20

$$y=1

Yes that's right. Keep it up! Perfect. Your hard work is paying off

Correct answers:

Substituting x=2 in the equation, and simplifying to find y, we find y=−4x−12=−4(2)−12=−8−12=−20 Question Evaluate the linear equation, y=4x−7, at the value x=2. Answer Explanation

Correct answers:

To evaluate a linear equation at a specific value, substitute the value x=2 into the equation for the variable, x. yyyy=4x−7=4(2)−7=8−7=1 Question Evan owns a house cleaning service. For each house visit, he charges $55 plus $30 per hour of work. A linear equation that expresses the total amount of mo ne y Eva n earns per visit is y=55+30x. What are the independent and dependent variables? What is the y- intercept and the slope? The independent variable (x) the amount, in dollars, Evan earns for each session. The dependent variable (y) is the amount of time Evan works each house visit.

• $y=1$ y =1

The independent variable (x) is the amount of time Evan works each house visit. The dependent variable (y) is the amount, in dollars, Evan earns for each session. At the start of the repairs, Evan charges a one-time fee of $55 (this is when x=0), so the y-intercept is 55. Evan earns $30 for each hour he works, so the slope is 30. The independent variable (x) is the amount of time Evan works each house visit. The dependent variable (y) is the amount, in dollars, Evan earns for each session. At the start of the repairs, Evan charges a one-time fee of $55 (this is when x=0), so the y-intercept is 55. Evan earns $30 for each hour he works, so the slope is 30. The independent variable (x) the amount, in dollars, Evan earns for each session. The dependent variable (y) is the amount of time Evan works each house visit. At the start of the repairs, Evan charges a one-time fee of $30 (this is when x=0), so the y-intercept is 30. Evan earns $55 for each hour he works, so the slope is 55. The independent variable (x) is the amount of time Evan works each house visit. The dependent variable (y) is the amount, in dollars, Evan earns for each session. At the start of the repairs, Evan charges a one-time fee of $30 (this is when x=0), so the y-intercept is 30. Evan earns $55 for each hour he works, so the slope is 55. Answer Explanation Correct answer: The independent variable (x) is the amount of time Evan works each house visit because it is the value that changes. He may work different amounts per day, and his earnings are dependent on how many hours he works. This is why the amount, in dollars Evan earns for each session is the dependent variable (y). The y-intercept is 55 (b=55). This is his one-time fee. The slope is 30 (a=30). This is the increase for each hour he works Question Using a calculator or statistical software, find the linear regression line for the data in the table below.