AP Statistics 5.1 Graphical Representations Between Two Quantitative Variables- Exam Style Questions - FRQs - New Syllabus
Question
(ii) Calculate the residual for this male tule elk. Show your work.
$H_a: \beta \ne 4.5$
(ii) At a significance level of $\alpha = 0.05$, what conclusion should the wildlife biologist make regarding the slope of the population regression line for male tule elk? Justify your response.
Most-appropriate topic codes (AP Statistics):
• Topic \(5.3\) — Linear Regression Models (Parts \( \mathrm{b} \), \( \mathrm{c} \))
• Topic \(5.4\) — Residuals (Part \( \mathrm{d} \))
• Topic \(5.5\) — Least-Squares Regression (Part \( \mathrm{d} \))
▶️ Answer/Explanation
(a)
The relationship between chest circumference and weight of male tule elk is strong, positive, and linear. As the chest circumference of male tule elk increases, their weight tends to increase.
When describing a scatterplot, always cover the Direction, Form, and Strength (DFS), and make sure to include context by explicitly naming the variables (chest circumference and weight).
(b) (i)
$\text{predicted weight} = -350.3 + 3.7455(145.9)$
$\text{predicted weight} = 196.168 \text{ kg}$
(b) (ii)
$\text{residual} = \text{actual weight} – \text{predicted weight}$
$\text{residual} = 204.3 – 196.168$
$\text{residual} = 8.132 \text{ kg}$
(c)
For each additional centimeter increase in chest circumference, the predicted weight of a male tule elk increases by $3.7455$ kilograms.
(d) (i)
The degrees of freedom for regression slope inference is $df = n – 2$. With a sample size of $n = 30$, $df = 30 – 2 = 28$.
Using the t-distribution table with $df = 28$ and a test statistic of $t = 3.408$, the one-tail area is exactly $0.001$.
Since the alternative hypothesis ($H_a: \beta \ne 4.5$) is two-sided, the p-value is $2 \times 0.001 = 0.002$.
(d) (ii)
Because the p-value of $0.002$ is less than the significance level of $\alpha = 0.05$, we reject the null hypothesis.
There is convincing statistical evidence to conclude that the slope of the population regression line relating chest circumference to weight for male tule elk is different than $4.5$ kg/cm.
Question
(ii) Does the least-squares regression line overestimate or underestimate the mass of the bullfrog identified in part (d-i)? Explain your answer.
Most-appropriate topic codes (AP Statistics):
• Topic \(5.3\) — Linear Regression Models (Parts \( \mathrm{b} \), \( \mathrm{c} \), \( \mathrm{d} \))
▶️ Answer/Explanation
(a)
The scatterplot reveals a strong, positive, roughly linear association between the mass and length of bullfrogs. There are no points that seriously deviate from the straight-line pattern of the points in the plot.
(b)
The value of the slope of the least-squares regression line is 6.086. This value indicates that the predicted mass of a bullfrog increases by 6.086 grams for each additional millimeter of length.
(c)
The coefficient of determination is $r^2 \approx 0.819$. This value indicates that 81.9% of the variation in bullfrog mass can be explained by variation in bullfrog length as described by the least-squares line.
(d)(i)
The largest residual in absolute value belongs to the bullfrog with length 162 mm and mass 356 grams.
(d)(ii)
The least-squares regression line overestimates the mass of the bullfrog with length 162 mm. Plot 2 shows that the point for the bullfrog with length 162 mm is below the least-squares regression line.
Question
ii. Graph II shows the same information as Graph I, but also indicates the old and new stadiums. Does Graph II suggest that the rate at which attendance changes as number of games won increases is different in the new stadium compared to the old stadium? Explain your reasoning.
Most-appropriate topic codes (AP Statistics):
• Topic \(5.1\) — Graphical Representations Between Two Quantitative Variables (Parts \( \mathrm{b} \), \( \mathrm{c} \))
• Topic \(5.3\) — Linear Regression Models (Parts \( \mathrm{c} \), \( \mathrm{d} \))
▶️ Answer/Explanation
(a)
The median average attendance is noticeably higher in the new stadium (around \(25,000\)) compared to the old stadium (around \(16,000\)).
While the interquartile ranges are fairly similar, indicating comparable variability in the middle \(50\%\) of attendance figures, the overall range is slightly wider for the new stadium.
Neither stadium’s distribution shows any clear outliers.
(b)
Average attendance in the old stadium remained relatively flat over time with no clear upward or downward trend.
In contrast, the new stadium exhibits a strong, positive, upward trend in average attendance over the years it was actively used.
(c)(i)
There is a strong, positive, linear relationship between the number of games won and the average attendance for each year.
(c)(ii)
No, the rate of change appears to be roughly the same for both stadiums.
If you drew separate lines of best fit through the points representing the old stadium and the new stadium, the slopes of both lines would be nearly identical, meaning attendance increases at a similar rate per win regardless of the venue.
(d)
The number of games won could serve as a confounding variable when assessing the relationship between the stadium type and average attendance.
Since the team won significantly more games playing in the new stadium, and winning is strongly associated with higher attendance, it’s impossible to tell if the attendance spike was caused by the appeal of the new stadium or simply because the team was performing better on the field.
Question
Most-appropriate topic codes (AP Statistics):
• Topic \(5.3\) — Linear Regression Models (Parts \( \mathrm{a} \), \( \mathrm{b} \))
▶️ Answer/Explanation
(a)
The estimate of the intercept is $72.95$.
Detailed Solution:
To find the intercept, we look at the computer output table under the “Coef” column right next to “Constant”, which is $72.95$.
In the context of the problem, the $y$-intercept represents the predicted $y$-value when $x=0$.
Therefore, if a customer gets into a line with $0$ people ahead of them, we predict their checkout process will take an average of $72.95$ seconds.
(b)
The coefficient of determination is $r^{2}=73.33\%$.
Detailed Solution:
The coefficient of determination is clearly labeled as “$R\text{-}Sq$” in the given computer output, which reads $73.33\%$.
This metric tells us the proportion of variance in the response variable that is predictable from the explanatory variable.
So, we can comfortably say that about $73.33\%$ of the changes in the total checkout time are directly accounted for by the linear relationship with the number of customers waiting in line.
(c)
The outlier is the point with $x=3$ and $y$ close to $100$ (this point should be circled on the scatterplot).
This point is considered an outlier because the combination of $x$ and $y$ values differs from the pattern of the rest of the data. Specifically, the value of $y$ (time to finish checkout) is much lower than would be expected when there are $x=3$ customers in line in front of the selected customer, given the remaining data.
Detailed Solution:
When observing the scatterplot, most data points cluster relatively close to the least-squares regression line, showing a clear positive trend.
However, there is one distinct point at $(3, \sim 100)$ that sits far below the regression line.
This point is considered an outlier because the checkout time is unusually fast for having $3$ customers in front, severely breaking the linear pattern established by the rest of the sample.
Question
Most-appropriate topic codes (AP Statistics):
• Topic \(5.3\) — Linear Regression Models (Part \( \mathrm{b} \))
• Topic \(5.4\) — Residuals (Part \( \mathrm{c} \))
▶️ Answer/Explanation
(a)(i) Positive:
A positive relationship means that wolves with greater length also tend to have greater weight — so as one variable goes up, so does the other. Think of it visually: the data points on the scatterplot trend upward from left to right when length \(x\) is plotted on the horizontal axis and weight \(y\) on the vertical axis.
(a)(ii) Linear:
A linear relationship means that as the length of a wolf increases by one meter, the weight tends to change by a roughly constant amount on average. In other words, the pattern of the data follows a straight-line shape rather than a curve.
(a)(iii) Strong:
A strong relationship means that the data points fall close to the regression line, so there is little scatter around it. Equivalently, the observed weights are close to the predicted weights — the residuals are generally small.
(b)
The least-squares regression equation is
\(\hat{y} = -16.46 + 35.02x\)
The slope \(35.02\) means that for every one-meter increase in the length of a wolf, the predicted weight increases by \(35.02\) kilograms, on average. In other words, two wolves that differ by one meter in length are predicted to differ by \(35.02\) kg in weight, with the longer wolf expected to be heavier.
\(\boxed{\text{For each 1-meter increase in length, predicted weight increases by } 35.02 \text{ kg}}\)
(c)
Recall the residual formula:
\(\text{Residual} = \text{Actual weight} – \text{Predicted weight}\)
First, find the predicted weight for \(x = 1.4\) meters:
\(\hat{y} = -16.46 + 35.02(1.4)\)
\(\hat{y} = -16.46 + 49.028 = 32.568 \text{ kg}\)
Now use the residual to find the actual weight:
\(\text{Actual weight} = \hat{y} + \text{Residual}\)
\(\text{Actual weight} = 32.568 + (-9.67)\)
\(\boxed{\text{Actual weight} = 22.898 \approx 22.9 \text{ kg}}\)
Question
Most-appropriate topic codes (AP Statistics):
• Topic \(5.2\) — Correlation (Part \( \mathrm{e} \))
• Topic \(5.3\) — Linear Regression Models (Part \( \mathrm{b} \))
• Topic \(1.9\) — Comparisons of the Distributions for One Quantitative Variable (Part \( \mathrm{d} \))
▶️ Answer/Explanation
(a)
Yes, the scatterplot supports the newspaper report. The scatterplot shows a positive association between the number of semesters needed to complete an academic program and starting salary — as the number of semesters increases, starting salary tends to increase as well. This is consistent with the newspaper’s claim that more semesters are associated with a greater starting salary.
(b)
The slope of the least-squares regression line is \(b_1 = 1.1594\).
The least-squares regression equation is:
\( \hat{y} = 34.018 + 1.1594x \)
Interpretation: For each additional semester needed to complete an academic program, the predicted starting salary in the first year of a job increases by approximately €1,159.40 (i.e., 1.1594 thousand euros).
(c)
For the business majors alone, there is a strong, negative, linear association between the number of semesters and starting salary. Business majors who need more semesters to complete their academic program tend to have lower starting salaries — which is the opposite direction from the overall trend seen in the combined scatterplot.
(d)
Business majors have the lowest median starting salary, at approximately €38,000. Physics majors have the next highest median starting salary, at approximately €48,000. Chemistry majors have the highest median starting salary, at approximately €55,000.
So in order from lowest to highest median starting salary: Business \(\approx\) €38,000 < Physics \(\approx\) €48,000 < Chemistry \(\approx\) €55,000.
(e)
The newspaper report should be modified to account for the major of each person. The overall positive association in the original report is largely explained by the fact that different majors — chemistry, physics, and business — tend to require different numbers of semesters and also have very different starting salary levels. Chemistry majors take more semesters on average and also earn the highest salaries; business majors take fewer semesters and earn the lowest salaries. This creates an apparent positive association when all three majors are pooled together.
However, within each individual major, students who take a greater number of semesters to complete their program tend to have lower starting salaries, not higher. The newspaper report should therefore be revised to state that, while majors requiring more semesters overall tend to have higher starting salaries (chemistry highest, physics next, business lowest), within any given major, taking more semesters to complete the program is associated with a lower starting salary.
Question
Most-appropriate topic codes (AP Statistics):
• Topic \(5.5\) — Least-Squares Regression (Part \( \mathrm{c} \))
▶️ Answer/Explanation
(a)
There is a moderately strong, positive, linear relationship between height and arm span so that taller students tend to have longer arm spans.
(b)(i)
The line in Graph \(2\) is the one that is helpful. For each student, the graph illustrates whether arm span is equal to height (points on the line), arm span is greater than height (points above the line), or arm span is less than height (points below the line).
(b)(ii)
The frequencies of the \(12\) seniors classified by body shape are:
Square: \(3\)
Tall Rectangle: \(4\)
Short Rectangle: \(5\)
(c)
The predicted arm span is calculated using the given least squares regression line equation:
\( \hat{y} = 11.74 + 0.8247x \)
\( \hat{y} = 11.74 + 0.8247(61) \)
\( \hat{y} = 62.05 \text{ inches} \)
Question
Most-appropriate topic codes (AP Statistics):
• Topic 5.3 — Linear Regression Models (Part b)
• Topic 5.2 — Correlation (Part c)
▶️ Answer/Explanation
(a)
The data show a weak to moderate, positive association between price and quality rating for these sewing machines.
The overall form of the relationship is curved or nonlinear rather than a straight line.
Among the machines costing under \$500, there appears to be little to no visible association between price and quality rating.
However, machines costing above \$500 generally tend to achieve much higher quality ratings than the cheaper group, which creates the overall positive direction.
(b)
The machine that most heavily influences and reduces the appropriateness of a linear regression model is the one located at an approximate price of \(\$2,200\) with a quality rating of approximately \(65\).
The general trend among the other four machines priced over \$500 suggests that quality starts to level off or approach a maximum potential limit rather than increasing continuously with price.
This particular \(\$2,200\) sewing machine is the absolute most expensive model in the entire dataset, yet its quality rating drops significantly compared to the models around \$1,500.
Including this point in a linear regression model would heavily drag the least-squares line down toward it, creating a poor overall fit for the rest of the data points.
(c)
According to Chris’s rule, he wants a machine only if there isn’t another choice available that is both cheaper and higher in quality. Following this strategy, only two models fit his standard and should be circled on the scatterplot:
1. The model positioned at a price slightly above \(\$100\) with a quality rating of \(65\).
2. The model positioned at a price slightly below \(\$500\) with a quality rating of \(81\) (or \(82\)).
The data points corresponding to these two machines have been circled on the scatterplot below.
Question
ii. the experimental units
iii. the response that will be measured
ii. Based on your graph, do you think a linear regression model is appropriate? Explain.
Most-appropriate topic codes (AP Statistics):
• Topic 5.1 — Graphical Representations Between Two Quantitative Variables (Part b.i)
• Topic 5.3 — Linear Regression Models (Part b.ii)
▶️ Answer/Explanation
(a)
i. The treatments: The five different concentrations of garlic oil infused into the food granules, which are 0%, 2%, 10%, 25%, and 50%.
ii. The experimental units: The 40 individual European starlings (or the individual cages housing each bird).
iii. The response that will be measured: The number of food granules consumed by an individual bird during the two-hour period.
(b)
i. Graph construction:
To investigate linearity, we construct a scatterplot with Garlic Oil Concentration (%) on the horizontal $x$-axis and Mean Number of Food Granules Consumed on the vertical $y$-axis.
The points to plot are $(0, 58)$, $(2, 48)$, $(10, 29)$, $(25, 24)$, and $(50, 20)$.
ii. Linearity Assessment:
No, a linear regression model is not appropriate.
The scatterplot reveals a clear, distinct curved pattern rather than a straight line trend.
As the garlic oil concentration increases, the mean number of food granules consumed drops sharply at first and then begins to level off, indicating a non-linear relationship.
Question
Most-appropriate topic codes (AP Statistics):
• Topic 4.4 — Setting Up a Test for a Population Mean or Population Mean Difference (Part \(\mathrm{c}\))
• Topic 4.5 — Carrying Out a Test for a Population Mean or Population Mean Difference (Part \(\mathrm{c}\))
• Topic 1.5 — Graphical Representations for One Quantitative Variable (Part \(\mathrm{d}\))
• Topic 1.9 — Comparisons of the Distributions for One Quantitative Variable (Part \(\mathrm{d}\))
▶️ Answer/Explanation
(a)
The trait that distinguishes the two groups in the scatterplot is the dominant foot (left or right). All the points in the upper-left cluster represent patients whose dominant foot is the right foot, while all the points in the lower-right cluster represent patients whose dominant foot is the left foot. The dominant foot type is the common trait, and it differs between the two groups.
(b)
Two conclusions become clear from this scatterplot that were not visible before:
First, there is a positive linear relationship between swelling in the dominant foot and swelling in the nondominant foot — as swelling in the dominant foot increases, swelling in the nondominant foot tends to increase as well.
Second, and importantly, every single point lies below the line \(y = x\), which means swelling in the dominant foot is consistently greater than swelling in the nondominant foot for all patients in the sample. This pattern across both groups combined is something you simply could not see in the left-foot vs. right-foot scatterplot from part (a).
(c)
We perform a matched-pairs \(t\)-test on the differences \(d_i = \text{(dominant swelling)} – \text{(nondominant swelling)}\).
The 12 differences are:
\(0.30,\ 0.30,\ 0.45,\ 0.15,\ 0.30,\ 0.35,\ 0.25,\ 0.35,\ 0.20,\ 0.25,\ 0.40,\ 0.15\)
State hypotheses (where \(\mu_d\) is the mean difference, dominant minus nondominant):
\(H_0: \mu_d = 0\)
\(H_a: \mu_d \neq 0\)
Check conditions:
1. We are told a random sample was selected from the population of adult females with Morton’s neuroma.
2. A dotplot of the differences shows a roughly symmetric, unimodal distribution with no outliers — it is reasonable to treat the population of differences as approximately normal.
Compute the test statistic:
\(\bar{x}_d = 0.2875, \quad s_d = 0.0932, \quad n = 12, \quad df = 11\)
\(t = \dfrac{\bar{x}_d – 0}{\dfrac{s_d}{\sqrt{n}}} = \dfrac{0.2875 – 0}{\dfrac{0.0932}{\sqrt{12}}} = 10.68\)
\(p\text{-value} \approx 0.0000004 \approx 0\)
Since the \(p\)-value is essentially \(0\), which is far less than any reasonable significance level \(\alpha\), we reject \(H_0\). There is very convincing statistical evidence that the mean swelling in the dominant foot is different from (and specifically greater than) the mean swelling in the nondominant foot for adult females who have Morton’s neuroma in at least one foot.
(d)
To suggest a diagnostic criterion, we separate all 24 swelling measurements into two groups: the 17 foot measurements from feet that have Morton’s neuroma and the 7 foot measurements from feet that do not have Morton’s neuroma. A stacked dotplot of the two groups is shown below:
The dotplot makes it visually clear that all 7 feet without Morton’s neuroma have swelling measurements of \(1.40\) or below, while the feet with Morton’s neuroma have swelling values of \(1.40\) and above (with the measurements extending up to \(1.85\)). Based on this graphical display, a reasonable diagnostic criterion is:
\(\boxed{\text{Swelling measurement} \geq 1.4 \Rightarrow \text{diagnose Morton’s neuroma}}\)
A cutoff of approximately \(1.4\) or higher serves as a sensible threshold for diagnosing Morton’s neuroma, since it cleanly separates the feet with and without the condition in this dataset.
Question
• Connect these two points with a line segment.
• Plot the two means (suburban and urban) for the children who played outside at the two types of day-care centers.
• Connect these two points with a second line segment.
Most-appropriate topic codes (AP Statistics):
• Topic 4.8 — Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Means (Part a, Part c)
• Topic 5.1 — Graphical Representations Between Two Quantitative Variables (Part b)
▶️ Answer/Explanation
(a)
First check conditions for a two-sample \(t\)-interval: the two groups of urban children were assigned at random and independently to play inside or outside, and dotplots of each group’s data show no strong skew or outliers, so it’s reasonable to treat the underlying populations as approximately normal.
Summary statistics for the urban sample:
\( n_{\text{in}}=9,\quad \bar{x}_{\text{in}}=4.56,\quad s_{\text{in}}=0.846 \)
\( n_{\text{out}}=9,\quad \bar{x}_{\text{out}}=17.56,\quad s_{\text{out}}=4.61 \)
The two-sample \(t\)-confidence interval formula is
\( (\bar{x}_{\text{in}}-\bar{x}_{\text{out}})\pm t^*\sqrt{\dfrac{s_{\text{in}}^2}{n_{\text{in}}}+\dfrac{s_{\text{out}}^2}{n_{\text{out}}}} \)
Using the conservative degrees of freedom \(df=\min(n_{\text{in}}-1,\,n_{\text{out}}-1)=8\), so \(t^*=2.306\):
\( (4.56-17.56)\pm 2.306\sqrt{\dfrac{(0.846)^2}{9}+\dfrac{(4.61)^2}{9}} \)
\( -13.00\pm 2.306(1.564) \)
\( -13.00\pm 3.61 \)
\( \boxed{(-16.60,\ -9.40)\text{ mcg}} \)
Interpretation: we are 95% confident that, for the population of urban day-care children, the mean amount of lead on the dominant hand after an hour of play inside is between 9.40 and 16.60 mcg lower than after an hour of play outside. Since this interval doesn’t contain zero, the difference is meaningful — urban children who play outside pick up noticeably more lead on their hands.
(b)
Plot the points \((\text{Suburban},3.75)\) and \((\text{Urban},4.56)\), connect them with a line labeled “inside.” Then plot \((\text{Suburban},5.65)\) and \((\text{Urban},17.56)\), connect them with a line labeled “outside.” The “inside” line should be nearly flat and low on the graph, while the “outside” line should rise sharply from suburban to urban.
\( \boxed{\text{Inside line: nearly flat, low values; Outside line: steep increase from suburban to urban}} \)
(c)
Setting (inside vs. outside): In both suburban and urban environments, children who played outside ended up with more lead on their hands than children who played inside. This is supported by the fact that all four endpoints of the two confidence intervals (inside minus outside) are negative, and the graph shows the “outside” line sitting above the “inside” line everywhere.
Environment (suburban vs. urban): For both inside and outside play, urban children had more lead on their hands on average than suburban children. The graph shows both lines sloping upward from suburban to urban.
Relationship between the two: The effect of going inside versus outside depends heavily on the environment. In the suburban setting, the inside and outside means are fairly close together (3.75 vs. 5.65), but in the urban setting the gap is much larger (4.56 vs. 17.56). In other words, playing outside makes a much bigger difference in lead exposure in the urban environment than in the suburban environment.
\( \boxed{\text{Outside > Inside in both settings; Urban > Suburban in both settings; the inside/outside gap is much larger for urban than suburban}} \)
Question
Most-appropriate topic codes (AP Statistics):
• Topic 5.2 — Correlation (Part b)
• Topic 5.3 — Linear Regression Models (Parts b, c)
• Topic 5.4 — Residuals (Part c)
▶️ Answer/Explanation
(a)
The relationship between impact rate and age is negative and nonlinear (curved). As age increases, impact rate decreases, but not at a constant rate — the decrease is very steep for craters with ages less than about 0.7 billion years, and then the impact rate levels off and remains close to zero for older craters. The spread of the data also decreases as age increases, suggesting a fan-shaped, non-constant variance pattern.
(b)
The value of \(r^2 = 0.894\), or \(89.4\%\).
This means that approximately \(89.4\%\) of the variability in \(\ln(\text{rate})\) is explained by the linear relationship with \(\ln(\text{age})\). In other words, after taking logarithms of both variables, the linear model accounts for nearly \(89.4\%\) of the variation seen in the log-transformed impact rate values among the 11 craters studied.
(c)
The linear regression model is appropriate for the log-transformed variables. Here’s the reasoning:
First, the residual plot shows no obvious curved pattern — the residuals appear to be scattered roughly randomly around zero, with no systematic curvature, which supports linearity.
Second, the residuals do not show a clear fan shape (no dramatic increase in spread), so the equal variance condition appears reasonably satisfied.
Third, the \(R^2\) value of \(89.4\%\) is quite high, indicating the linear model fits the transformed data well.
However, one concern is that the residual plot shows a slight tendency for residuals to be negative in the middle range of fitted values and positive at the extremes, hinting at a mild pattern. Given the small sample size of only 11 observations, this could simply be due to natural sampling variability. Overall, the linear model on the log-transformed data is a reasonable fit.
Question
Most-appropriate topic codes (AP Statistics):
• Topic 5.4 — Residuals (Part a)
• Topic 5.5 — Least-Squares Regression (Parts a, b)
• Topic 5.1 — Graphical Representations Between Two Quantitative Variables (Part b)
▶️ Answer/Explanation
(a)
Yes, point \(P\) does exercise a large influence on the regression line.
When point \(P\) is included, the slope of the regression line is \(b_1 = 0.4919\), and it is statistically significant (\(p = 0.040\)).
When point \(P\) is removed, the slope drops dramatically to \(b_1 = 0.1500\), which is no longer statistically significant (\(p = 0.709\)).
The intercept also changes from \(8.107\) to \(11.123\), and \(R^2\) falls sharply from \(47.6\%\) to only \(2.5\%\).
Because removing a single point caused such substantial changes in the slope, intercept, statistical significance, and \(R^2\), point \(P\) is clearly an influential point — it is extreme in the \(x\)-direction (a work value of about 30, far beyond the rest of the data), which gives it high leverage and strong pull on the regression line.
(b)
In the original data, point \(P\) is at approximately \((\text{Work} = 30,\ \text{Study} = 25)\), which is high on both variables and pulls the regression line upward to the right, producing a positive slope.
The corrected data point \(Q\) is at approximately \((\text{Work} = 30,\ \text{Study} = 6)\), which is far below the trend of the other data points at that work value.
With \(Q\) replacing \(P\), the regression line for the corrected data will have a negative slope rather than a positive slope — the corrected point now pulls the line downward on the right side.
The intercept would also be considerably larger for the corrected data, since the line must start higher on the \(y\)-axis to accommodate the negative slope while passing near the rest of the data.