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MATH225 Week 8 Final Exam
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. Correct! You nailed it.
r= −0.
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 between the two data sets. Round your answer to two decimal places.
Yes that's right. Keep it up!
r= −0.
Answer Explanation
The correlation coefficient, rounded to two decimal places, is r≈−0.81.
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. x y 46 23 32 60 39 40 33 59 38 57 40 33 42 33 30 64 34 56 HelpCopy to ClipboardDownload CSV Yes that's right. Keep it up!
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.
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. Yes that's right. Keep it up!
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.
Yes that's right. Keep it up!
The predicted number of minutes spent reading is $$46.92.
Answer Explanation
The predicted number of minutes spent reading is 1 $$.
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? That's incorrect - mistakes are part of learning. Keep trying! 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 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.
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=2020 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=2020 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.
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
(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.
Well done! You got it right.
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? Perfect. Your hard work is paying off 😀 The estimate, a predicted test score of 80.56, is both reliable and reasonable. 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 answer:
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..
Temperature (Degrees) 30405060 Minutes Watching Television 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? Correct! You nailed it. 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. 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 Television 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. Yes that's right. Keep it up!
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:
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 Television 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 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 $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. 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 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:
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.
Question
Given the following line, find the value of y when x=2.
y=−4x−
Well done! You got it right.
$$ y =−
Answer Explanation
Correct answers:
$y=-20$ y =−
Substituting x=2 in the equation, and simplifying to find y, we find
y=−4x−12=−4(2)−12=−8−12=−
Question
Evaluate the linear equation, y=4x−7, at the value x=2.
Yes that's right. Keep it up!
$$ y =
Answer Explanation
Correct answers:
$y=1$ y =
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=
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.
Enter your answer in the form y=mx+b, with m and b both rounded to two decimal places.
x y 0
1
2
3
4
5
HelpCopy to ClipboardDownload CSV Answer 1: Keep trying - mistakes can help us grow.
$$ y =0.53 x +1.
Answer 2: Keep trying - mistakes can help us grow.
$$ y =0.53 x +1.
Answer Explanation
Correct answers:
$y=0.54x+1.59$ y =0.54 x +1. If you use a TI-83 or TI-84 calculator, you press STAT, and then ENTER, which brings you to
the edit menu where you can enter values. In the L1 list, you enter the values of x from the table
above, 0,1,2,3,4,5. Then, in the L2 list, you enter the values of y from the table
above, 2.12,2.19,1.92,2.79,3.81,4.72.
Now, press STAT again, and arrow to the right, to CALC. Arrow down to the LinReg option and
press ENTER. The resulting a and b are the slope m and y-intercept b of the linear regression
line. You should find that m≈0.54 and b≈1.59. So the final answer is
y=0.54x+1.
Using spreadsheet software or other statistical software should give you the same result. Question Using a calculator or statistical software, find the linear regression line for the data in the table below.
Enter your answer in the form y=mx+b, with m and b both rounded to two decimal places.
x y 0
1
2
3
4
5
HelpCopy to ClipboardDownload CSV
