AP Statistics 5.4 Residuals- 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
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
RSquare 0.250401
Root Mean Square Error 0.902382
Observations 66
Most-appropriate topic codes (AP Statistics):
• Topic \(5.4\) — Residuals (Parts \( \mathrm{a} \), \( \mathrm{b} \), \( \mathrm{c} \), \( \mathrm{d} \))
• Topic \(5.5\) — Least-Squares Regression (Part \( \mathrm{a} \))
▶️ Answer/Explanation
(a)
Plug the length of 175 inches into the least squares regression equation to get the predicted FCR:
\(\widehat{\text{FCR}} = -1.595789 + 0.0372614 \times 175\)
\(\widehat{\text{FCR}} \approx 4.92 \text{ gallons per 100 miles}\)
Now compute the residual using the formula \(\text{residual} = \text{observed} – \text{predicted}\):
\(\text{residual} = 5.88 – 4.92 = 0.96\)
\(\boxed{\text{residual} \approx 0.96 \text{ gallons per 100 miles}}\)
The residual of \(0.96\) means that this car’s actual FCR is \(0.96\) gallons per 100 miles higher than what the least squares regression line would predict for a car of length 175 inches — so the model underestimates the fuel consumption for this particular car.
(b)(i)
Point A has a wheel base of 93 inches and a residual of approximately \(0.96\) gallons per 100 miles (from part a). So on Graph III, the point to circle is the one located at approximately \((93,\ 0.96)\).
\(\boxed{\text{Circle the point at wheel base} = 93 \text{ in., residual} \approx 0.96}\)
(b)(ii)
A residual very close to 0 means the observed FCR and the predicted FCR (from the regression of FCR on length) are nearly equal for that car — in other words, the length-based regression model predicts that car’s fuel consumption almost perfectly, leaving very little unexplained.
\(\boxed{\text{The car’s actual FCR} \approx \text{its FCR predicted by the length-based regression line}}\)
(c)
Graph II shows a moderate, positive, linear association between engine size and the residuals from the regression of FCR on length — as engine size increases, the residuals tend to increase as well. Graph III, on the other hand, shows little to no discernible pattern between wheel base and those same residuals; the points are scattered without any clear direction or trend. Overall, the association in Graph II is noticeably stronger than in Graph III.
(d)
Jamal should add engine size to the model along with length.
Because Graph II shows a stronger association between engine size and the residuals from the length-only regression, adding engine size will explain more of the leftover variability that length alone cannot account for. Wheel base shows almost no relationship with those residuals (Graph III), so including it would do little to improve the model’s predictions.
\(\boxed{\text{Choose engine size — it has a stronger association with the residuals, reducing unexplained variability more effectively.}}\)
Question
Most-appropriate topic codes (AP Statistics):
• Topic \(5.4\) — Residuals (Parts \(\mathrm{b}\), \(\mathrm{c}\))
• Topic \(5.5\) — Least-Squares Regression (Parts \(\mathrm{a}\), \(\mathrm{b}\), \(\mathrm{c}\), \(\mathrm{e}\))
• Topic \(4.8\) — Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Means (Part \(\mathrm{d}\))
▶️ Answer/Explanation
(a)
The slope of the least squares regression line is \(0.165\) (in thousands of dollars per square foot).
In context: for each additional square foot of house size, the predicted price of the house increases by \(0.165\) thousand dollars, or \$165, on average.
The slope tells us the rate at which the model expects price to grow with size — not a guarantee for any individual house, but the average trend across houses in this part of the city.
(b)
The residual value of 49 for this house indicates that its actual price is 49 thousand dollars higher than the model would predict for a house of its size.
(c)
We estimate the pool premium by comparing the average residuals of the two groups. If a group’s residuals average positive, the model consistently underestimates their prices; if negative, it overestimates.
Houses with a swimming pool (8 houses, residuals: \(6, 49, -18, 42, 1, 50, -23, 42\)):
\(\bar{e}_{\text{pool}} = \frac{6 + 49 + (-18) + 42 + 1 + 50 + (-23) + 42}{8} = \frac{149}{8} = 18.625 \text{ thousand dollars}\)
Houses without a swimming pool (17 houses, residuals: \(13, 26, -45, 22, 10, -46, -57, 1, -2, -69, 23, 44, -19, 26, -58, -52, 33\)):
\(\bar{e}_{\text{no pool}} = \frac{13 + 26 + (-45) + 22 + 10 + (-46) + (-57) + 1 + (-2) + (-69) + 23 + 44 + (-19) + 26 + (-58) + (-52) + 33}{17} = \frac{-150}{17} \approx -8.824 \text{ thousand dollars}\)
The estimated price premium for a swimming pool is the difference between these two averages:
\(\bar{e}_{\text{pool}} – \bar{e}_{\text{no pool}} = 18.625 – (-8.824) = \boxed{27.4 \text{ thousand dollars}}\)
This tells us that, for two houses of the same size, the one with a swimming pool is estimated to cost about \$27,400 more. The logic: pool houses have residuals that average \$18,625 above the model’s predictions, while no-pool houses sit \$8,824 below — that gap reflects the pool’s unmodeled contribution to price.
(d)
The 95% confidence interval for the true difference in slopes is \((-0.099,\ 0.110)\).
Since this interval contains zero, we cannot conclude there is a statistically significant difference between the two slopes at the 5% significance level. Zero is a plausible value for the true difference, which means it is entirely possible that the two population regression lines have the same slope.
In practical terms: the rate at which price increases with size appears to be the same for pool homes and non-pool homes — a pool shifts the price up by a roughly constant amount, but doesn’t change how sensitive the price is to square footage.
(e)
Since the two slopes are not significantly different, we pick a house size within the data range — say, \(\text{size} = 2{,}250\) sq ft (near the center of the distribution) — and compare predicted prices from both models.
Predicted price with pool:
\(\widehat{\text{Price}}_{\text{pool}} = -11.602 + 0.166 \times 2250 = -11.602 + 373.500 = 361.898 \text{ thousand dollars}\)
Predicted price without pool:
\(\widehat{\text{Price}}_{\text{no pool}} = -27.382 + 0.160 \times 2250 = -27.382 + 360.000 = 332.618 \text{ thousand dollars}\)
Estimated price premium for a pool:
\(361.898 – 332.618 = \boxed{29.280 \text{ thousand dollars} \approx \$29{,}280}\)
Comparison with part (c): The estimate from part (e), approximately \$29,280, is quite similar to the \$27,400 estimate obtained in part (c) from the residual averages. Both methods point to a pool adding roughly \$27,000–\$29,000 to the price of a house, giving us confidence that this is a reasonable estimate of the pool’s effect regardless of which approach we use.
Note — Alternative approach (difference in intercepts): Because the slopes were found not to be significantly different, we can also subtract the two fitted equations directly:
\((-11.602 + 0.166 \cdot x) – (-27.382 + 0.160 \cdot x) = 15.780 + 0.006 \cdot x\)
This gives the price difference as a function of size. For \(x = 2250\): \(15.780 + 0.006 \times 2250 = 15.780 + 13.500 = 29.280\), consistent with the calculation above.
Question
Most-appropriate topic codes (AP Statistics):
• Topic 5.3 — Linear Regression Models (Parts \(\mathrm{a}\), \(\mathrm{b}\), \(\mathrm{c}\))
• Topic 5.4 — Residuals (Part \(\mathrm{b}\))
▶️ Answer/Explanation
(a)
To draw the least squares regression line \(\hat{y} = 35.1 + 0.427x\), compute two points on the line and connect them. For example:
At \(x = 55\): \(\quad \hat{y} = 35.1 + 0.427(55) = 35.1 + 23.485 = 58.585\)
At \(x = 80\): \(\quad \hat{y} = 35.1 + 0.427(80) = 35.1 + 34.16 = 69.26\)
Plot the points \((55,\ 58.6)\) and \((80,\ 69.3)\) on the scatterplot and draw a straight line through them. The line is shown in the scatterplot above (blue line).
(b)
The point \((67,\ 61)\) is circled on the scatterplot. The predicted value at \(x = 67\) is:
\(\hat{y} = 35.1 + 0.427(67) = 35.1 + 28.609 = 63.709\)
The residual is the vertical distance from the actual point down to the regression line:
\(\text{Residual} = y – \hat{y} = 61 – 63.709\)
\(\boxed{\text{Residual} = -2.709 \text{ inches}}\)
The negative sign tells us the actual daughter’s height is about \(2.709\) inches below what the regression line predicts — the vertical dashed segment on the scatterplot drops from the line down to the actual point.
(c) — Slope:
The slope would remain about the same. The new point \((84,\ 71)\) falls very close to the existing regression line — substituting \(x = 84\) gives \(\hat{y} = 35.1 + 0.427(84) = 70.97\), which is just about \(0.03\) away from the actual \(y = 71\). Since the new point is nearly on the line, it is consistent with the existing linear pattern and will not pull the regression line in a new direction.
(c) — Correlation:
The correlation would increase. We know the relationship between slope, correlation, and the standard deviations:
\(b = r \cdot \dfrac{s_y}{s_x}\)
Adding the new point at \(x = 84\) extends the range of \(x\) values far to the right, increasing \(s_x\) considerably more than it increases \(s_y\). So the ratio \(\dfrac{s_y}{s_x}\) decreases. Since the slope \(b\) stays about the same but \(\dfrac{s_y}{s_x}\) gets smaller, \(r\) must increase to compensate. Intuitively, the new point fits the linear pattern well and sits far out in the \(x\)-direction, which strengthens the apparent linear relationship and pulls \(r\) closer to \(1\).
Question
Most-appropriate topic codes (AP Statistics):
• Topic 3.13 — Carrying Out a Test for the Difference Between Two Population Proportions (Part \(\mathrm{a}\))
• Topic 5.5 — Least-Squares Regression (Part \(\mathrm{b}\))
• Topic 5.4 — Residuals (Part \(\mathrm{c}\))
• Topic 5.5 — Least-Squares Regression (Part \(\mathrm{d}\)
▶️ Answer/Explanation
(a)
We want to test whether the proportion of species going extinct is smaller on large islands than on small islands. Let \(p_L\) be the true proportion of at-risk species that become extinct on large islands, and \(p_S\) be the true proportion on small islands.
The hypotheses are:
\(H_0: p_L – p_S = 0\)
\(H_a: p_L – p_S < 0\)
We use a two-sample \(z\)-test for the difference in proportions. The sample proportions are:
\(\hat{p}_L = \frac{19}{208} \approx 0.091, \qquad \hat{p}_S = \frac{66}{299} \approx 0.221\)
Check conditions — all expected counts must be at least 5:
\(n_L\hat{p}_L = 19,\quad n_L(1-\hat{p}_L) = 189,\quad n_S\hat{p}_S = 66,\quad n_S(1-\hat{p}_S) = 233\)
All values are well above 5, so we may proceed.
The pooled sample proportion is:
\(\hat{p} = \frac{19+66}{208+299} = \frac{85}{507} \approx 0.168\)
The test statistic is:
\(z = \frac{\hat{p}_L – \hat{p}_S}{\sqrt{\hat{p}(1-\hat{p})\left(\dfrac{1}{n_L}+\dfrac{1}{n_S}\right)}} = \frac{0.091 – 0.221}{\sqrt{(0.168)(0.832)\left(\dfrac{1}{208}+\dfrac{1}{299}\right)}} = \frac{-0.130}{0.034} \approx -3.82\)
The corresponding \(p\)-value \(\approx 0.00006\), which is essentially \(0\).
Since the \(p\)-value is far less than any reasonable significance level, we reject \(H_0\). There is very strong statistical evidence that the proportion of species going extinct is smaller for large islands than for small islands, supporting the scientist’s belief.
\(\boxed{z \approx -3.82, \quad p\text{-value} \approx 0.00006 \quad \Rightarrow \quad \text{Reject } H_0}\)
(b)
We construct a 95% confidence interval for the slope \(\beta\) of the regression of proportion extinct on \(\ln(\text{area})\).
From the regression output: \(\hat{b} = -0.05323\), \(SE_b = 0.00618\), and \(df = n – 2 = 13 – 2 = 11\).
The critical value from the \(t\)-table with \(df = 11\) at the 95% level is \(t^* = 2.201\).
The confidence interval is:
\(\hat{b} \pm t^* \cdot SE_b = -0.05323 \pm 2.201(0.00618)\)
\(-0.05323 \pm 0.01360\)
\(\boxed{(-0.0668,\ -0.0396)}\)
We are 95% confident that for every 1-unit increase in \(\ln(\text{area})\), the mean proportion of species going extinct decreases by somewhere between \(0.0396\) and \(0.0668\). In plain terms, larger islands are associated with a meaningfully lower extinction rate, and this relationship is statistically significant.
(c)
The assumption is not reasonable. The regression analysis in part (b) shows that the proportion of species going extinct decreases steadily as island area increases — it is a continuous relationship, not a step function that jumps only between “large” and “small” groups.
Within the large island group, areas ranged from 31 to 46 sq km, and within the small island group, areas ranged from 1 to 9 sq km — meaning extinction probabilities varied considerably within each group as well.
Because extinction probability depends on actual area and not just on a binary large/small classification, the assumption that all large islands share one common extinction probability and all small islands share another is not supported by the data.
(d)
We use the regression model \(\widehat{\text{prop extinct}} = 0.28996 – 0.05323\ln(\text{area})\) to estimate extinction proportions for each option.
Option 1 — One large preserve (area = 45 sq km, 70 species):
\(\widehat{\text{prop extinct}} = 0.28996 – 0.05323\ln(45) = 0.28996 – 0.05323(3.807) \approx 0.28996 – 0.20261 \approx 0.0873\)
Expected extinctions: \(70 \times 0.0873 \approx 6.1\) species
Expected survivors: \(70 – 6.1 \approx \mathbf{63.9 \approx 64}\) species
Option 2 — Five small preserves (each area = 3 sq km, 16 species each; 80 total):
\(\widehat{\text{prop extinct}} = 0.28996 – 0.05323\ln(3) = 0.28996 – 0.05323(1.099) \approx 0.28996 – 0.05850 \approx 0.2315\)
Expected extinctions per preserve: \(16 \times 0.2315 \approx 3.7\) species
Total expected extinctions: \(5 \times 3.7 \approx 18.5\) species
Expected survivors: \(80 – 18.5 \approx \mathbf{61.5 \approx 62}\) species
The one large preserve is expected to save approximately 64 species, compared to about 62 species across the five small preserves. We recommend creating one large nature preserve, as it leads to a greater expected number of surviving species, and larger areas have been shown to have substantially lower extinction rates per species.
\(\boxed{\text{Recommend: One large preserve (45 sq km)} \Rightarrow \approx 64 \text{ species saved vs. } \approx 62 \text{ for five small preserves}}\)
Question
Most-appropriate topic codes (AP Statistics):
• Topic 5.4 — Residuals (Part \(\mathrm{b}\))
• Topic 5.5 — Least-Squares Regression (Part \(\mathrm{c}\))
▶️ Answer/Explanation
(a)
Reading the coefficients directly from the computer output, the fitted regression line is:
\(\hat{y} = -2.679 + 9.5x\)
where \(\hat{y}\) represents the predicted (estimated) mean height of the soapsuds (in millimeters), and \(x\) represents the amount of detergent added to the pan (in grams).
\(\boxed{\hat{y} = -2.679 + 9.5x}\)
(b)
The value \(s = 1.99821\,\text{mm}\) is the standard deviation of the residuals.
In the context of this study, it measures a typical amount of variation in the observed heights of soapsuds from the heights predicted by the regression line — that is, for a given amount of detergent, the actual suds height typically differs from the predicted suds height by about \(1.998\,\text{mm}\).
\(\boxed{s = 1.99821\,\text{mm}}\)
(c)
The standard error of the slope is identified from the computer output as the SE Coef for the Amount row:
\(SE_b = 0.7553\,\text{mm per gram}\)
This value estimates the standard deviation of the sampling distribution of the estimated slope — in other words, it tells us how much the estimated slope \(\hat{b}_1\) would be expected to vary from experiment to experiment if the same study were repeated many times under identical conditions.
A small \(SE_b = 0.7553\) relative to the slope of \(9.5\) indicates that the estimated slope is very stable and reliable across repeated samples.
\(\boxed{SE_b = 0.7553\,\text{mm/g}}\)
Question
Most-appropriate topic codes (AP Statistics):
• Topic 5.4 — Residuals (Part \(\mathrm{a}\))
• Topic 5.2 — Correlation (Part \(\mathrm{c}\))
▶️ Answer/Explanation
(a)
Yes, a linear model is appropriate. The scatterplot of fuel consumption versus number of railcars shows a strong, positive, linear pattern with points falling close to a line. The residual plot backs this up — the residuals are scattered randomly above and below zero with no obvious curve or pattern, which means a straight line is capturing the relationship well.
\( \boxed{\text{Yes — strong linear pattern in scatterplot and no pattern in residual plot}} \)
(b)
The slope of the regression line tells us how much fuel consumption changes per railcar:
\( \text{slope} = 2.15 \text{ units/mile per railcar} \)
Each unit of fuel costs \$25, so multiply the slope by the cost per unit:
\( \text{cost change} = 2.15 \times \$25 \)
\( \text{cost change} = \$53.75 \)
\( \boxed{\$53.75 \text{ per additional railcar}} \)
(c)
The value \(r^2=96.7\%\) tells us how much of the variation in fuel consumption is accounted for by the linear relationship with the number of railcars. Putting it in plain terms:
\( \boxed{96.7\%\text{ of the variation in fuel consumption is explained by the linear relationship with number of railcars}} \)
(d)
No, it would not be reasonable. Looking at the data table, the number of railcars only ranges from 20 to 50, and 65 falls well outside that range. Using the regression line to predict fuel consumption at 65 railcars means extrapolating beyond the data we actually collected, and there’s no guarantee the same linear relationship continues to hold out there.
\( \boxed{\text{No — 65 railcars is outside the observed range (20 to 50), so this would be extrapolation}} \)
Question
Most-appropriate topic codes (AP Statistics):
• Topic 5.5 — Least-Squares Regression (Parts a, b)
• Topic 5.4 — Residuals (Regression output interpretation)
▶️ Answer/Explanation
(a)
Reading directly from the regression output, the fitted regression equation is: \[ \widehat{\text{Pulse}} = 63.457 + 16.2809 \times \text{Speed} \] where Pulse is measured in beats per minute (bpm) and Speed is measured in miles per hour (mph).
(b)
Both estimates have meaningful interpretations in this context.
Slope interpretation:
The slope \(b_1 = 16.2809\) bpm/mph means that for each additional 1 mile per hour increase in John’s walking speed, his predicted pulse rate increases by approximately 16.28 beats per minute on average.
Intercept interpretation:
The intercept \(b_0 = 63.457\) bpm means that when John’s walking speed is 0 mph — that is, when he is standing still — his predicted pulse rate is approximately 63.5 beats per minute. This is a reasonable estimate of John’s resting pulse rate, so the intercept does carry a meaningful real-world interpretation here (unlike many regression contexts where the intercept falls outside the range of observed data).
(c)
The margin of error for a confidence interval for the slope is:
\( \text{Margin of error} = t^* \times SE_{b_1} \)
From the regression output, the standard error of the slope is \(SE_{b_1} = 0.8192\).
For a 98% confidence interval, the degrees of freedom are \(df = n – 2 = 7 – 2 = 5\).
From the \(t\)-table with \(df = 5\) and a 98% confidence level (tail probability \(= 0.01\)):
\( t^* = 3.365 \)
Therefore, the margin of error is:
\( \text{Margin of error} = 3.365 \times 0.8192 \)
\( \boxed{\text{Margin of error} \approx 2.757 \text{ bpm/mph}} \)
This means John’s 98% confidence interval for the true slope would be \(16.2809 \pm 2.757\), or approximately \((13.524,\ 19.038)\) bpm per mph. The relatively narrow interval, combined with the very high \(R^2 = 98.7\%\), confirms that speed is an extremely strong linear predictor of John’s pulse rate.
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.