- Interpret \(r^2\) using variation in the response variable, boxes packed. - Convert the decimal \(0.64\) to a percentage. - Subtract the explained percentage from \(100\%\) to find the unexplained portion.

1. Convert \(r^2 = 0.64\) to \(64\%\). 2. About \(64\%\) of the observed variation in boxes packed is explained by the least-squares linear relationship with number of volunteers. 3. The unexplained percentage is \(100\% - 64\% = 36\%\).

About \(64\%\) of the observed variation in boxes packed is explained by its linear relationship with number of volunteers; about \(36\%\) is not explained by that linear relationship.

- Identify which residual measure the least-squares method minimizes. - The equal sums of absolute residuals do not decide the least-squares comparison. - Compare \(19\) and \(28\).

1. The least-squares criterion selects the line with the smaller sum of squared residuals. 2. Since \(19 < 28\), Line B is preferred.

Line B, because its sum of squared residuals is smaller.

- Add the claimed signed residuals. - Recall the balancing property of least-squares residuals. - Compare the two totals.

1. Six residuals of \(2\) would have signed sum \(6\cdot 2=12\). 2. Residuals from an unrounded least-squares line with an intercept must sum to \(0\). 3. Therefore, the claimed residual list cannot come from that least-squares fit.

It is impossible because the residuals would sum to \(12\), not \(0\).

- Residuals are vertical differences from an observed point to the fitted line at the same \(x\). - Compare the lengths of the two vertical residual segments. - Squaring preserves the ordering of positive residual magnitudes.

1. At \(x=2\), the fitted line has \(\hat y=3\), while \(A\) is at \(y=4\), so its residual has magnitude \(1\). 2. At \(x=5\), the fitted line has \(\hat y=6\), while \(B\) is at \(y=9\), so its residual has magnitude \(3\). 3. Because least squares uses squared residuals, \(B\)'s contribution is larger: \(3^2\) is greater than \(1^2\).

Observation \(B\). Its vertical residual is larger in magnitude, so its squared residual contributes more to the sum of squared residuals.

- First add the errors without changing their signs. - Next square each error before adding. - Why do the two operations give different results?

1. The signed errors cancel: \(2+(-2)=0\). 2. Squaring gives \(2^2+(-2)^2=4+4=8\). 3. Squared errors do not cancel when one prediction is too high and the other is too low.

Signed-error sum: \(0\). Squared-error sum: \(8\).

- What operation is performed on each residual? - How do units change when a measured quantity is squared?

1. Squaring a residual squares both its numerical value and its unit. 2. Adding those squares does not change the unit: the SSE has units of square pounds.

Square pounds, \(\text{lb}^2\).

- What does a sum of squared residuals tell you about a proposed line's errors? - What would the sum of squared residuals be if every prediction were exact?

1. Least squares chooses the line with the smaller sum of squared residuals. 2. Line B has the smaller total, \(9\), compared with \(14\) for line A.

Line B, because its sum of squared residuals is smaller.

- Look at vertical distance from P to the fitted line. - Is P unusually far left or right? - Is the vertical size of a residual the same as the change a point causes in the fitted line?

1. P is far above the fitted line, giving a much larger positive residual than any other point. 2. Its input lies near the middle of the observed values, not at an extreme input. 3. Residual size alone does not measure the change in the line when P is removed. That requires comparing the two fits.

Yes, P has a large positive residual. No, the effect of removing P must be checked by comparing fits.

- Enter ramp heights and speeds in corresponding technology lists. - Use linear regression output for both the intercept and slope rather than estimating from two points. - Square the reported correlation only if the technology does not display \(r^2\) directly.

1. Enter the paired ramp-height and speed data and run linear regression. 2. The unrounded slope is \(1.482857\ldots\), and the unrounded intercept is \(4.126667\ldots\). 3. The least-squares regression line is \(\hat y \approx 4.127 + 1.483x\). 4. The coefficient of determination is \(r^2 = 0.964213\ldots \approx 0.964\).

\(\hat y \approx 4.127 + 1.483x\), with \(r^2 \approx 0.964\).

- Use the fact that the least-squares line contains the point of sample means. - Substitute \(\bar x\), \(\bar y\), and the given slope into \(\bar y = a + b\bar x\). - Isolate \(a\) and check that the line predicts \(22.0\) when \(x = 8.0\).

1. A least-squares regression line passes through \((\bar x, \bar y) = (8.0, 22.0)\). 2. Use \(\bar y = a + b\bar x\): \(22.0 = a + 1.40(8.0)\). 3. Solve: \(a = 22.0 - 11.20 = 10.80\).

The intercept is \(a = 10.80\).

- Start with the point \((\bar x, \bar y)\) on the fitted line. - Keep the negative sign on the slope when substituting. - Rearranging a subtraction term requires adding \(1.90(11.0)\) to both sides.

1. A least-squares regression line passes through \((\bar x, \bar y) = (11.0, 27.0)\). 2. Use \(\bar y = a + b\bar x\): \(27.0 = a - 1.90(11.0)\). 3. Solve: \(a = 27.0 + 20.90 = 47.90\).

The intercept is \(a = 47.90\).

- Use the defining point \((\bar x, \bar y)\) on the least-squares line. - Substitute the mean x-value directly into the fitted equation. - The arithmetic is exact with the given coefficients, so use \(=\), not \(\approx\).

1. A least-squares regression line passes through \((\bar x, \bar y)\). 2. Substitute \(\bar x = 14.0\): \(\bar y = 49.848 - 1.632(14.0)\). 3. \(\bar y = 49.848 - 22.848 = 27.000\).

\(\bar y = 27.000\) minutes.

- Compare the two \(r^2\) values: larger indicates more explained variation. - Compare the two sums of squared residuals: smaller indicates less total squared prediction error. - These comparisons are valid because both models use the same response data.

1. For models fitted to the same response data, a larger \(r^2\) means the model explains more of the variation in the response. 2. A smaller sum of squared residuals means the observed values are closer to the fitted values overall. 3. Model B has both the larger \(r^2\), \(0.81 > 0.72\), and the smaller sum of squared residuals, \(12.5 < 18.4\). Therefore, Model B gives the better in-sample fit.

Model B gives the better in-sample fit because it has the larger \(r^2\) and the smaller sum of squared residuals.

- Recall the sum-of-residuals property for a least-squares line with an intercept. - Add the four known residuals, keeping their signs. - Choose the missing value that makes the total \(0\).

1. For a least-squares line with an intercept, the residuals sum to \(0\). 2. Let the missing residual be \(e\): \(-2.4+1.1+0.8-0.5+e=0\). 3. The known residuals sum to \(-1.0\), so \(-1.0+e=0\). 4. Therefore, \(e=1.0\).

The missing residual is \(1.0\).

- Add the residuals with their signs before dividing. - The mean equals the residual sum divided by \(5\). - Compare the result with the residual-mean property of a least-squares fit with an intercept.

1. Add the residuals: \(-1.2+0.7+0.4-0.3+0.4=0.0\). 2. Divide by the number of residuals: \(\frac{0.0}{5}=0.0\). 3. A least-squares line with an intercept has residuals with mean \(0\), so this result is consistent with that property.

The mean residual is \(0.0\), which is consistent with a least-squares regression line that includes an intercept.

- Generate one predicted value for each input. - Use the same subtraction order for every residual. - Check whether positive and negative errors balance.

1. The predicted values are \(14, 18, 22, 26, 30\). 2. The residuals \(y-\hat y\) are \(-1, 1, 0, 1, -1\). 3. Their sum is \(-1+1+0+1-1=0\). 4. Least-squares residuals from a model with an intercept balance to a sum of \(0\), apart from possible rounding.

Residuals: \(-1, 1, 0, 1, -1\) Sum: \(0\)

- Use the displayed equation consistently for all four predictions. - Keep the residual subtraction order unchanged. - Compare the small total with what exact least-squares coefficients would produce.

1. The rounded predictions are \(1.59, 3.97, 7.54, 9.92\). 2. The residuals are \(-0.59, 1.03, -0.54, 0.08\). 3. Their sum is \(-0.02\). 4. The exact least-squares coefficients make the signed residuals sum to zero; rounding the coefficients changes the predictions slightly.

Residuals: \(-0.59, 1.03, -0.54, 0.08\) Sum: \(-0.02\) The small nonzero sum is caused by rounded regression coefficients.

- What must the total signed residual be for this fitted line? - Add the values you know, keeping their signs. - What missing value would balance that sum?

1. Residuals from an unrounded least-squares line with an intercept add to \(0\). 2. The four known values total \(1.2-0.5+0.8-1.1=0.4\). 3. Thus \(0.4+e=0\), giving \(e=-0.4\).

\(e=-0.4\).

- Apply each error rule separately. - Notice how squaring changes the influence of one large residual. - Use the criterion named in part c).

1. Model A's absolute-error sum is \(2+0+0+2=4\), and Model B's is \(1+1+1+1=4\). 2. Model A's squared-error sum is \(4+0+0+4=8\), while Model B's is \(1+1+1+1=4\). 3. Least squares prefers Model B because it has the smaller sum of squared residuals. 4. Both residual lists sum to \(0\), so the comparison does not conflict with the balancing property of least-squares residuals.

a) Both absolute-error sums are \(4\). b) Model A: \(8\); Model B: \(4\) c) Model B is preferred by least squares.

- Use the mean point as a point on the fitted line. - Substitute the known slope and both mean coordinates. - Solve the resulting equation for the constant term.

1. The least-squares line passes through \((12, 31)\). 2. Substitute into \(\hat y=1.8x+a\): \(31=1.8\cdot 12+a\). 3. Solving gives \(a=9.4\). 4. The equation is \(\hat y=1.8x+9.4\).

Y-intercept: \(9.4\) Regression equation: \(\hat y=1.8x+9.4\)

- Substitute the mean \(x\)-value into each candidate equation. - Compare each prediction with the mean \(y\)-value. - Decide whether passing through one point uniquely determines a line.

1. Line A gives \(2\cdot 5+2=12\), so it passes through the mean point. 2. Line B gives \(5+7=12\), so it also passes through the mean point. 3. Line C gives \(2\cdot 5+3=13\), so it cannot be the least-squares line. 4. The mean-point property eliminates C but cannot choose between A and B without more information.

Line C is impossible. The mean-point property alone cannot determine whether A or B is the actual regression line.

- Identify the special point through which the fitted line passes. - Compare the actual and predicted \(y\)-values at the mean \(x\)-value. - Apply the residual definition.

1. The fitted least-squares line passes through \((\bar x, \bar y)\). 2. At \(x=\bar x\), the predicted response is \(\hat y=\bar y\). 3. The residual is \(y-\hat y=\bar y-\bar y=0\).

The residual is \(0\).

- Translate zero residuals into a geometric statement about all points. - Recall how many points with different \(x\)-values determine a line. - Compare zero squared error with any positive squared error.

1. Zero residual for every observation means every point lies on the line. 2. Two points with different \(x\)-values determine exactly one nonvertical line. 3. Any different line would miss at least one of those points and have a positive sum of squared residuals. 4. The zero-error line is therefore the unique least-squares fit.

The line is unique because two observations with different \(x\)-values determine one line, and no other line can retain zero residual for every point.

- Check that each residual list balances around zero. - Compute the squared-error totals and the largest absolute residuals separately. - Do not assume one model must win under both criteria.

1. Both residual lists sum to \(0\), so each could be consistent with a least-squares line that includes an intercept. 2. Model A's squared-error sum is \(9+1+1+1=12\). 3. Model B's squared-error sum is \(4\cdot1.8^2=12.96\). 4. Least squares therefore prefers Model A. 5. Model A's maximum absolute residual is \(3\), while Model B's is \(1.8\), so Model B has the smaller worst individual error.

Least squares prefers Model A. Model B has the smaller maximum absolute residual.

- Compare the direction represented by the correlation with the direction represented by the fitted line. - Think about whether changing scale factors can reverse the sign in the relationship between these two statistics.

1. In simple linear regression, the fitted slope and the correlation coefficient must have the same sign because the two standard deviations are positive. 2. The reported correlation is positive, while the reported slope is negative. 3. Therefore, the two values cannot both come from the same least-squares regression.

No. A positive correlation requires a positive least-squares slope, so \(r=0.84\) and \(b=-2.10\) are inconsistent.

- Ask whether \(r^2\) summarizes point counts or variation. - Connect the statistic to explained and unexplained response variability. - A fitted line can explain much variation even when no point lies exactly on it.

1. The coefficient of determination does not count how many observations lie exactly on the fitted line. 2. The value \(r^2=0.81\) means that \(81\%\) of the variation in the response variable is explained by its linear relationship with the explanatory variable in this model. 3. The remaining \(19\%\) of response variation is not explained by that linear relationship.

\(r^2=0.81\) means that \(81\%\) of the variation in the response is explained by the fitted linear relationship, not that \(81\%\) of the points lie on the line.

- Think about the form of a least-squares line when both variables are measured in standard-deviation units. - Connect the standardized fitted slope to the strength and direction of the linear association. - Square the correlation only after identifying its sign.

1. For standardized variables, the least-squares regression line has intercept \(0\) and slope equal to the correlation. 2. Therefore, \(r=-0.65\). 3. The coefficient of determination is \(r^2=(-0.65)^2=0.4225\).

\(r=-0.65\) and \(r^2=0.4225\).

- Focus on this observation's contribution while the model remains fixed. - How does one residual enter a sum of squared errors? - Compare the two squared values, not just the residuals themselves.

1. The original squared residual is \((2.0)^2=4.0\). 2. The new squared residual is \((4.0)^2=16.0\). 3. Its contribution rises by \(16.0-4.0=12.0\). This happens without refitting the line. 4. Squaring gives large residuals a much greater effect on the total error than small residuals.

The contribution increases by \(12.0\), from \(4.0\) to \(16.0\). Squaring makes larger errors count more heavily.

- Use a property of the residuals from a least-squares fit with an intercept. - Write the residual sum as the difference between two totals. - You do not need any individual prediction to determine the total.

1. For a least-squares regression line with an intercept, the residuals sum to \(0\): \(\sum(y_i-\hat y_i)=0\). 2. Therefore, \(\sum y_i=\sum\hat y_i\). 3. Since the observed responses total \(250\), the predicted responses also total \(250\).

The predicted responses total \(250\). For a least-squares line with an intercept, the residuals sum to \(0\), so the sum of the fitted responses equals the sum of the observed responses.

- Add the example residuals and count their signs separately. - Distinguish a balance of signed values from a balance of counts. - A few larger residuals can offset several smaller residuals of the opposite sign.

1. The residuals \(-8, 2, 2, 2, 2\) sum to \(0\). 2. This set has four positive residuals and only one negative residual, so a zero residual sum does not require equal counts on the two sides of \(0\). 3. The least-squares property balances signed residual magnitudes in total, not the number of positive and negative residuals.

The statement is incorrect. The residuals \(-8, 2, 2, 2, 2\) sum to \(0\) even though four are positive and one is negative. A zero residual sum does not imply equal numbers above and below the fitted line.

- Identify which data set the mean-point property refers to. - Ask whether deleting an observation can change the sample means. - The regression is refitted to the remaining observations, not constrained by summaries from the old data set.

1. The mean-point property applies to the data set used for the particular least-squares fit. 2. Removing an observation can change both the sample mean of \(x\) and the sample mean of \(y\). 3. The refitted line must pass through the new means of the remaining observations, not necessarily through the old means from the full data set.

The statement is incorrect. After an observation is removed, the refitted least-squares line passes through the new sample means of the remaining data, which need not equal the original \((\bar x,\bar y)\).

- Consider how a vertical shift of every observed point affects the best-fitting line. - Compare an old residual with the corresponding residual after both the observation and its fitted value shift. - The sum of squared errors depends only on the residuals.

1. Adding \(10\) to every response shifts every observed point upward by \(10\) without changing the linear association with \(x\). 2. The fitted value at each \(x\) also increases by \(10\), so the new line is \(\hat y^*=15+2x\). 3. Each new residual is \(y^*-\hat y^*=(y+10)-(\hat y+10)=y-\hat y\), so every residual is unchanged. 4. Because the residuals are unchanged, the sum of squared errors is unchanged.

The new least-squares line is \(\hat y^*=15+2x\). All residuals stay the same, so the sum of squared errors also stays the same.

- Recall the point that every least-squares line with an intercept must pass through. - Substitute the given means into the reported equation. - If the check fails, use the known slope and mean point to determine the intercept.

1. A least-squares regression line with an intercept passes through \((\bar x, \bar y)=(8, 25)\). 2. The reported line gives \(\hat y=40-1.5\cdot8=28\), not \(25\), so it cannot be the stated least-squares line. 3. The correct intercept satisfies \(25=a-1.5\cdot8\). 4. Thus \(a=25+12=37\), giving \(\hat y=37-1.5x\).

The reported line is incorrect because it does not pass through \((8, 25)\). The correct line is \(\hat y=37-1.5x\).

- Identify the quantity that least-squares regression minimizes. - Apply that criterion separately to the two residual lists. - Do not let the equal sums of absolute residuals decide the comparison.

1. For line A, the sum of squared residuals is \((-3)^2+0^2+3^2=18\). 2. For line B, the sum of squared residuals is \((-2)^2+(-1)^2+3^2=14\). 3. Least squares compares squared residuals, not absolute residuals. 4. Therefore, line B has the smaller least-squares criterion even though both absolute-residual sums equal \(6\).

Line B. Its sum of squared residuals is \(14\), compared with \(18\) for line A.

- Express the original explanatory variable in terms of the rescaled one. - Substitute that expression into the original fitted equation. - Decide whether changing only the horizontal units changes any vertical prediction errors.

1. Because \(x=\frac{x^*}{3}\), substitute into the original equation: \(\hat y=4+6\left(\frac{x^*}{3}\right)=4+2x^*\). 2. Each observation represents the same underlying explanatory value, only on a rescaled axis, so its fitted response is unchanged. 3. Since observed and fitted responses are unchanged, every residual is unchanged. 4. Therefore, the SSE is also unchanged.

The new equation is \(\hat y=4+2x^*\). The fitted values, residuals, and SSE all remain unchanged.

- Use the residual degrees of freedom for a simple linear regression. - Divide the SSE by those degrees of freedom before taking the square root. - Interpret the result as a typical vertical prediction error in the response variable's units.

1. The residual degrees of freedom are \(12-2=10\). 2. The residual mean square is \(\frac{50}{10}=5\). 3. Therefore, \(s=\sqrt{5}\approx2.236\) minutes. 4. The observed delivery times typically differ from their fitted values by about \(2.24\) minutes, as measured by the residual standard deviation.

\(s\approx2.24\,\text{minutes}\). Observed delivery times typically fall about \(2.24\) minutes from the values predicted by the fitted line.

- For each observation, read the line's predicted height at the same \(x\)-value. - Compute each residual as observed minus predicted before squaring. - Compare the totals of the squared residuals, not the signed residual sums.

1. For line \(m\) in panel a), the fitted values at \(x=1,2,3,4\) are \(2,3,4,5\). The residuals are \(0,0,1,1\), so \(SSE_m=0^2+0^2+1^2+1^2=2\). 2. For line \(n\) in panel b), the fitted values are \(2,3.5,5,6.5\). The residuals are \(0,-0.5,0,-0.5\), so \(SSE_n=0^2+(-0.5)^2+0^2+(-0.5)^2=0.5\). 3. The least-squares criterion prefers the line with the smaller sum of squared residuals, so line \(n\) is preferred.

\(SSE_m=2\) and \(SSE_n=0.5\). The least-squares criterion prefers line \(n\).

- Each colored segment measures how far the observed value is from the line at the same input. - Include the sign of each residual when reading the segments. - How does least squares combine errors of different sizes and signs?

1. At \(x=1\), the line predicts 3 goals while 4 were observed, so the residual is \(1\). 2. At \(x=3\), the line predicts 5 goals while 2 were observed, so the residual is \(-3\). 3. The sum of squared residuals is \(1^2+(-3)^2=10\) goals squared.

\(10\) goals squared.

- Read each plotted input and response as a pair. - Find the mean on each axis separately. - Does a least-squares line have to pass through one of the original dots?

1. The points are \((1,1)\), \((2,5)\), and \((3,6)\). 2. The means are \(\bar x=(1+2+3)\div3=2\) and \(\bar y=(1+5+6)\div3=4\). 3. A least-squares line with an intercept passes through \((\bar x,\bar y)=(2,4)\), which is not one of the plotted observations.

Mean water \(2\) cups, mean leaves \(4\); the line passes through \((2,4)\).

- For each point, compare its height with each line's height at the same input. - Square the three vertical differences for each candidate. - Which total is smaller?

1. The points are \((1,3)\), \((2,5)\), and \((3,5)\). 2. Line a, \(\hat y=x+2\), predicts \(3,4,5\). Its squared residuals sum to \(0^2+1^2+0^2=1\). 3. Line b, \(\hat y=2x\), predicts \(2,4,6\). Its squared residuals sum to \(1^2+1^2+(-1)^2=3\). 4. Line a fits these data better by the least-squares rule.

Line a; its sum of squared residuals is \(1\), compared with \(3\) for line b.

- Compare the points' horizontal positions with the rest of the data. - Compare each point's vertical distance from the line. - How is a point's effect on the fitted line different from either distance?

1. Most practice times lie from \(1\) to \(5\) hours, but P is at \(x=9\), so P has the more extreme input and higher leverage. 2. Q is at \((3,10)\), whereas the line predicts \(7\). Q's residual is \(3\); P is only \(0.5\) above the line, so Q has the larger residual. 3. Influence is the change in the fitted line when a point is removed. Neither an extreme input nor a large residual alone determines that change.

P has the more extreme input value. Q has the larger residual. To check influence, compare fits with and without each point.

- Read the residual height of each dot, including its sign. - Square the residuals before adding them within each panel. - Negative residuals still add positive amounts to the squared-error total.

1. Panel a) has residuals \(1,-1,1\), so its sum of squares is \(1+1+1=3\). 2. Panel b) has residuals \(0,-1,2\), so its sum of squares is \(0+1+4=5\). 3. Model a) has the smaller sum of squared residuals.

Model a), with \(3\) versus \(5\) for model b).

- Find a squared-error total for each line. - Keep the squaring step separate from adding the signed residuals. - Can two different sets of errors have the same signed sum but different squared-error totals?

1. Line A has sum of squared residuals \(1^2+(-2)^2+1^2=6\). 2. Line B has sum of squared residuals \(2^2+(-2)^2+0^2=8\). 3. Least squares chooses A. Both signed sums equal \(0\), but the squared-error sums are different.

Line A: its squared-error sum is \(6\), less than B's \(8\). Both signed sums are \(0\), so they cannot distinguish the lines.

- Use \(b = r\frac{s_y}{s_x}\) to find the regression slope from summary statistics. - The least-squares line passes through \((\bar x, \bar y)\), which determines the intercept. - Square the correlation and round only after computing the exact decimal.

1. Compute the slope: \(b = r\frac{s_y}{s_x} = 0.82\left(\frac{5.5}{2.5}\right) = 1.804\). 2. Compute the intercept: \(a = \bar y - b\bar x = 30.0 - 1.804(10.0) = 11.960\). 3. The least-squares regression line is \(\hat y = 11.960 + 1.804x\). 4. The coefficient of determination is \(r^2 = (0.82)^2 = 0.6724 \approx 0.672\).

\(\hat y = 11.960 + 1.804x\), with \(r^2 \approx 0.672\).

- Use the negative correlation in \(b = r\frac{s_y}{s_x}\), so the fitted slope must be negative. - Find the intercept by forcing the line through \((\bar x, \bar y)\). - Squaring \(r\) removes the sign; round \(r^2\) only at the end.

1. Compute the slope: \(b = r\frac{s_y}{s_x} = -0.71\left(\frac{6.2}{2.9}\right) = -1.517931\ldots \approx -1.518\). 2. Compute the intercept: \(a = \bar y - b\bar x = 33.0 - (-1.517931\ldots)(12.0) = 51.215172\ldots \approx 51.215\). 3. The least-squares regression line is \(\hat y = 51.215 - 1.518x\). 4. The coefficient of determination is \(r^2 = (-0.71)^2 = 0.5041 \approx 0.504\).

\(\hat y = 51.215 - 1.518x\), with \(r^2 \approx 0.504\).

- Recall the point that every least-squares regression line must contain. - Evaluate the proposed equation at the given mean x-value. - Compare that prediction with the given mean y-value.

1. A least-squares regression line must pass through \((\bar x, \bar y) = (14.0, 32.0)\). 2. At \(x = 14.0\), the proposed line gives \(\hat y = 1.40 + 2.40(14.0) = 35.00\). 3. Because \(35.00 \ne 32.0\), the proposed line does not pass through the sample-mean point and cannot be the least-squares line.

The claim is false. At \(x = \bar x = 14.0\), the proposed line predicts \(35.00\), but a least-squares line must predict \(\bar y = 32.0\).

- Multiplying every response value changes both fitted-response coefficients. - Apply the factor \(10\) to the intercept and slope. - Positive rescaling does not change standardized association or \(r^2\).

1. Let the transformed response be \(y^* = 10y\), so \(\hat y^* = 10\hat y\). 2. Multiply the entire fitted equation by \(10\): \(\hat y^* = 10(20.00 + 1.50x) = 200.00 + 15.00x\). 3. Positive rescaling of the response does not change correlation, so \(r^2\) remains \(0.64\).

\(\hat y^* = 200.00 + 15.00x\), with \(r^2 = 0.64\).

- Adding a constant to every response shifts all fitted values vertically. - A vertical shift changes the intercept but not the slope. - Correlation and \(r^2\) are unaffected by adding a constant.

1. Let the transformed response be \(y^* = y + 50\), so \(\hat y^* = \hat y + 50\). 2. Add \(50\) to the intercept: \(\hat y^* = 74.00 - 1.90x\); the slope is unchanged. 3. Shifting the response by a constant does not change correlation, so \(r^2\) remains \(0.69\).

\(\hat y^* = 74.00 - 1.90x\), with \(r^2 = 0.69\).

- Express the old explanatory variable \(x\) in terms of the new variable \(x^*\). - Substitute that expression into the original regression equation. - A positive rescaling of \(x\) changes the slope’s units but not the fitted values or \(r^2\).

1. Since \(x^* = 100x\), write \(x = \frac{x^*}{100}\). 2. Substitute into the original line: \(\hat y = 28.00 + 2.30\left(\frac{x^*}{100}\right) = 28.00 + 0.0230x^*\). 3. Rescaling the explanatory variable by a positive constant does not change the fitted values or \(r^2\). Therefore, the new slope is \(0.0230\), the intercept is \(28.00\), and \(r^2 = 0.74\).

\(\hat y = 28.00 + 0.0230x^*\); slope \(0.0230\), intercept \(28.00\), and \(r^2 = 0.74\).

- Use regression technology on the four plotted observations to obtain one of the lines. - For each line, compare every observed \(y\)-value with the predicted value at the same \(x\)-value. - The least-squares criterion compares sums of squared residuals, not just the signed residual totals.

1. The least-squares line is \(\hat y=1.4x+0.9\). 2. Its residuals are \(0.1, 0.7, -1.7, 0.9\), giving a squared-residual sum of \(4.20\). 3. The endpoint line has residuals \(0, \frac{1}{3}, -\frac{7}{3}, 0\), giving a squared-residual sum of \(\frac{50}{9}\approx 5.56\). 4. The least-squares line fits better because \(4.20<5.56\).

Least-squares line: \(\hat y=1.4x+0.9\) Least-squares SSE: \(4.20\) Endpoint-line SSE: approximately \(5.56\) The least-squares line fits better.

- Substitute the new \(x\)-value into the current line. - Determine the new point's residual. - Consider what a zero-error observation contributes to the fitting criterion.

1. The new point lies exactly on the current line because \(3\cdot 100+2=302\). 2. Its residual from the current line is \(0\), so it adds no squared error and no imbalance to the least-squares conditions. 3. The original line therefore remains the least-squares line after the point is added.

The least-squares line remains \(\hat y=3x+2\).

- Find \(\bar x\) and \(\bar y\) before forming the deviations in the slope formula. - Keep the numerator and denominator of the slope formula separate until the final division. - After finding the line, compute each residual as observed minus predicted before squaring.

1. The means are \(\bar x=0\) and \(\bar y=3\). 2. The slope numerator is \(9+1+0+12=22\), and the denominator is \(9+1+1+9=20\). Thus \(b=\frac{22}{20}=1.1\). 3. Since \(\bar x=0\), \(a=\bar y-b\bar x=3\). The least-squares line is \(\hat y=1.1x+3\). 4. The predicted values are \(-0.3, 1.9, 4.1, 6.3\), so the residuals are \(0.3, 0.1, -1.1, 0.7\). 5. The sum of squared residuals is \(0.3^2+0.1^2+(-1.1)^2+0.7^2=1.80\).

Least-squares line: \(\hat y=1.1x+3\) Residuals: \(0.3, 0.1, -1.1, 0.7\) Sum of squared residuals: \(1.80\)

- Compare what the summary statistics imply about the direction and scale of the fitted line. - A least-squares line has a required relationship to the two sample means. - Check more than one feature of each candidate before deciding.

1. The least-squares slope is \(b=r\frac{s_y}{s_x}=0.75\cdot\frac{3.2}{4.0}=0.600\). 2. A least-squares regression line passes through \((\bar x,\bar y)=(16.0,11.0)\), so \(a=11.0-0.600\cdot16.0=1.4\). 3. Therefore, only line A has both the required slope and intercept.

Line A, \(\hat y=1.4+0.6x\).

- Compare explained variation with the part left in the residuals. - The two models use the same response data, so the total response variation is the same. - Convert the unexplained fraction into the explained fraction before comparing the models.

1. For Model A, the unexplained proportion is \(\frac{72}{320}=0.225\). 2. Therefore, \(r^2=1-0.225=0.775\). 3. Since \(0.775>0.740\), Model A explains the larger proportion of the response variation.

Model A has \(r^2=0.775\), so Model A explains more of the variation than Model B.

- Separate a property that a fitted line may have from the criterion that defines “least squares.” - Compare the two measures of prediction error that are given. - Ask which quantity the fitting procedure actually minimizes.

1. A zero residual sum is a property of an ordinary least-squares line with an intercept, but that property alone does not identify the least-squares line. 2. The least-squares criterion selects the line with the smallest sum of squared residuals. 3. Since \(9.7<14.2\), Line B is preferred.

The claim is incorrect. A residual sum of \(0\) is not enough by itself. Line B is preferred because its sum of squared residuals, \(9.7\), is smaller.

- Use a defining location property of the least-squares regression line to recover the missing mean. - Express the original variables in terms of their centered versions. - Think about whether subtracting constants changes rates of change.

1. A least-squares regression line passes through \((\bar x, \bar y)\), so \(\bar y=18.0+1.4\cdot10.0=32.0\). 2. Since \(x=X+10.0\) and \(\hat y=\hat Y+32.0\), substitute into the fitted line to get \(\hat Y+32.0=18.0+1.4(X+10.0)\). 3. Simplifying gives \(\hat Y=1.4X\). 4. Centering subtracts constants from the variables, so it changes the intercept but not the least-squares slope.

\(\bar y=32.0\), and the centered regression line is \(\hat Y=1.4X\). The slope remains \(1.4\).

- Apply the response transformation to the fitted values as well as the observations. - Track how the transformation changes each residual before squaring it. - Compare explained and total variation after the same response scaling.

1. Multiplying every response by \(-2\) multiplies every fitted response by \(-2\), so the new line is \(\hat y^*=-8.0-3.0x\). 2. Each new residual is \(-2\) times the corresponding original residual. 3. Squaring the residuals multiplies each squared residual by \(4\), so the new sum of squared residuals is \(4\cdot20.0=80.0\). 4. A nonzero linear rescaling of the response does not change the proportion of response variation explained, so \(r^2\) is unchanged.

\(\hat y^*=-8.0-3.0x\), the new sum of squared residuals is \(80.0\), and \(r^2\) is unchanged.

- First check whether the fitted line reproduces every response exactly. - Then recall what correlation requires about the variability of both variables. - A perfect prediction in a degenerate data set is not the same as a perfect increasing linear association.

1. The horizontal line \(\hat y=7\) predicts every observed response exactly, so every residual is \(0\). 2. Therefore, \(\hat y=7\) is the least-squares line and the sum of squared residuals is \(0\). 3. However, the response standard deviation is \(0\) because every \(y\)-value is identical. 4. Correlation is undefined when either variable has zero standard deviation, so the correlation is not \(1\).

The least-squares line is \(\hat y=7\) and its sum of squared residuals is \(0\). The correlation is undefined because the response variable has no variation.

- Identify the units carried by a squared residual. - Ask whether the two sums of squares are measured on the same response scale. - A numerical comparison is meaningful only when the quantities being compared have a common scale and role.

1. A sum of squared residuals is measured in squared response units, so its numerical size depends on the scale of the response variable. 2. Model A and Model B use different responses and different units, so their raw \(SSE\) values are not on a common scale. 3. A smaller numerical \(SSE\) can support a better fit when models are compared on the same observations and response scale, but it cannot be used directly to rank unrelated fits with different response scales.

The comparison is not valid. Raw \(SSE\) values depend on the response scale and have squared response units, so \(420\) dollars squared and \(18{,}000\) cents squared are not directly comparable measures of fit across different responses.

- Distinguish the exact fitted coefficients from their rounded display values. - Compare the size of the discrepancy with the precision of the reported coefficients. - A defining property applies to the exact fitted line, not necessarily to every rounded representation of it.

1. An exact least-squares regression line with an intercept passes through \((\bar x,\bar y)\). 2. The displayed coefficients have been rounded, so the displayed line need not reproduce \(\bar y\) exactly when evaluated at \(\bar x\). 3. Here, the discrepancy is only \(6.014-6.01=0.004\), which is consistent with coefficient rounding. 4. The rounded display therefore does not show that the underlying fitted line violates the mean-point property.

The claim is not justified. The exact least-squares line passes through \((\bar x,\bar y)\), but rounded coefficients can produce a small discrepancy. A difference of \(0.004\) is consistent with rounding.

- Use the relationship among least-squares slope, correlation, and the two standard deviations. - Apply the allowable range of the correlation coefficient. - Consider what equality in that bound would require.

1. The least-squares slope is \(b=r\frac{s_y}{s_x}\). 2. Because \(|r|\le1\), \(|b|\le\frac{12.0}{3.0}=4.0\). 3. If \(|b|=4.0\), then \(|r|=1\), so the observations have a perfect linear association. 4. The sign of the slope would determine whether the perfect association is increasing or decreasing.

The largest possible slope magnitude is \(4.0\). Reaching that magnitude would require \(|r|=1\), a perfect linear association.

- Correlation has no units, while a regression slope depends on the scales of both variables. - Relate slope to correlation and the two standard deviations. - Compare the scale ratios in the two data sets.

1. Correlation is standardized, but the least-squares slope also depends on the relative scales of the two variables. 2. For data set A, \(b=0.60\cdot\frac{6}{2}=1.8\). 3. For data set B, \(b=0.60\cdot\frac{2}{6}=0.2\). 4. Therefore, equal correlations do not imply equal regression slopes.

No. Data set A has slope \(1.8\), while data set B has slope \(0.2\). The same correlation can correspond to different slopes when the variable spreads differ.

- Use the centered cross-product information to determine the slope. - Then use the fact that the least-squares line passes through the point of sample means. - Check the final equation by substituting \(x=\bar x\).

1. The least-squares slope is \(b=\frac{\sum (x_i-\bar x)(y_i-\bar y)}{\sum (x_i-\bar x)^2}=\frac{60}{24}=2.5\). 2. The intercept satisfies \(a=\bar y-b\bar x\). 3. Thus \(a=13-2.5\cdot4=3\).

The least-squares regression line is \(\hat y=3+2.5x\).

- Think about what happens to the squared errors when every observation is counted twice. - Would doubling the SSE of every possible line change which line fits best? - Consider the coefficients separately from the total error.

1. For any candidate line, doubling every observation doubles its SSE. 2. Multiplying every candidate line's SSE by \(2\) does not change which line gives the smallest SSE. 3. Because at least two \(x\)-values differ, the fitted slope and intercept are uniquely determined and stay the same. 4. At that line, every squared residual occurs twice, so the SSE doubles.

The least-squares slope and intercept stay the same, and the SSE doubles.

- Identify which fitted coefficient allows a line to shift vertically. - The usual residual properties depend on which parameters are free in the optimization. - Compare estimating an intercept with fixing it in advance.

1. The zero-sum residual property comes from estimating a free intercept in ordinary least-squares regression. 2. When the intercept is fixed at \(0\), the optimization chooses only the slope and does not impose the same condition on the residual sum. 3. Therefore, a through-origin least-squares fit can have \(\sum e_i\ne0\).

No. Residuals are guaranteed to sum to zero for an ordinary least-squares fit with a fitted intercept, but not when the intercept is constrained to \(0\).

- For each observation, compare its height with each line at the same input. - Consider the size of every prediction error, not just the number of exact matches. - What happens to a particularly large error under the least-squares rule?

1. The observed points are \((1,2)\), \((2,3)\), \((3,4)\), and \((4,8)\). 2. Line a is \(\hat y=x+1\), with residuals \(0,0,0,3\) and squared-error sum \(9\). 3. Line b is \(\hat y=x+2\), with residuals \(-1,-1,-1,2\) and squared-error sum \(1+1+1+4=7\). 4. Line b has the smaller squared-error sum even though a passes through more observations.

No. Line b has the smaller sum of squared residuals: \(7\), compared with \(9\) for line a.

- For each candidate, calculate \(\text{residual} = y - \hat y\) at all five x-values. - Square every residual before adding so positive and negative errors do not cancel. - The least-squares criterion selects the candidate with the smallest total squared residual.

1. For line A, the residuals are \(0.0, 0.2, 0.1, -0.2, 0.3\), so \(\text{SSE}_A = 0.00 + 0.04 + 0.01 + 0.04 + 0.09 = 0.18\). 2. For line B, the residuals are \(-0.8, -0.4, -0.3, -0.4, 0.3\), so \(\text{SSE}_B = 0.64 + 0.16 + 0.09 + 0.16 + 0.09 = 1.14\). 3. For line C, the residuals are \(0.3, 0.3, 0.0, -0.5, -0.2\), so \(\text{SSE}_C = 0.09 + 0.09 + 0.00 + 0.25 + 0.04 = 0.47\). 4. Line A has the smallest sum of squared residuals, so it is best among the three candidate lines.

A: \(0.18\); B: \(1.14\); C: \(0.47\). Line A is best among the three because it has the smallest sum of squared residuals.

- Write the residual for a general line \(\hat y=a+bx\). - Imagine moving the entire proposed line upward while keeping its slope fixed. What happens to each signed error? - Ask whether the signed sum has a smallest possible value, then compare it with a sum of squares.

1. For a proposed line \(\hat y=a+bx\), the sum of signed residuals is \(S=\sum_{i=1}^{n}(y_i-a-bx_i)=\sum_{i=1}^{n}y_i-na-b\sum_{i=1}^{n}x_i\). 2. For any fixed slope \(b\), increasing the intercept \(a\) makes \(S\) smaller without limit. Therefore, the signed sum has no minimum. 3. Positive and negative residuals can also cancel, so a sum of zero does not imply a good fit—and zero is not the minimum. 4. Least squares instead minimizes \(\sum_{i=1}^{n}(y_i-\hat y_i)^2\), which cannot decrease without limit.

The sum of signed residuals has no minimum: raising the intercept makes it arbitrarily negative. Least squares minimizes the sum of squared residuals, \(\sum_{i=1}^{n}(y_i-\hat y_i)^2\), instead.

- Write the slope relationship once for each choice of response variable. - Compare the product of the two fitted slopes with the correlation. - Think about what would have to be true for the slopes to be exact reciprocals.

1. The forward slope is \(b_{y\text{ on }x}=r\frac{s_y}{s_x}\), and the reverse slope is \(b_{x\text{ on }y}=r\frac{s_x}{s_y}\). 2. Their product is \(r^2=(0.60)^2=0.36\). 3. Thus \(b_{x\text{ on }y}=\frac{0.36}{1.50}=0.24\). 4. The two slopes are reciprocals only when \(r^2=1\), meaning a perfect linear association.

The reverse-regression slope is \(0.24\), not \(\frac{1}{1.50}\). Regression is not symmetric unless the linear association is perfect.

- Distinguish removing one residual contribution from refitting the entire model. - Ask whether the fitted line must stay the same after an observation is deleted. - Remember that all residuals depend on the fitted line, not only on their observed points.

1. Under the current fitted line, the observation contributes \((-4.0)^2=16.0\) to the sum of squared residuals. 2. Removing that observation would leave a sum of squared residuals of \(36.0\) only if the original fitted line were kept unchanged for the remaining observations. 3. A least-squares regression is refitted after an observation is removed, so the slope and intercept can change and the residuals of the remaining observations can also change. 4. The refitted line cannot have a larger sum of squared residuals than the unchanged original line on the remaining data, so the new sum is at most \(36.0\), but it need not equal \(36.0\).

The statement is incorrect. The removed observation contributes \(16.0\) to the old sum of squared residuals, but deleting it requires refitting the regression line. The new least-squares sum of squared residuals is at most \(36.0\), but it is not necessarily equal to \(36.0\).

- Compare the set of lines allowed by the two fitting procedures. - Ask whether the first fitted line is still an available candidate in the second procedure. - A minimum over a larger set of choices cannot be worse than the value from one specific available choice.

1. The ordinary least-squares fit considers all lines of the form \(a+bx\), including the special case \(a=0\). 2. Therefore, the through-origin line with sum of squared residuals \(50.0\) is one candidate available to the ordinary least-squares fit. 3. The minimizing line cannot have a larger sum of squared residuals than an available candidate, so its sum of squared residuals must be at most \(50.0\). 4. A reported value of \(58.0\) is therefore impossible for the ordinary least-squares fit to the same data.

No. Allowing an intercept gives the least-squares procedure at least all the choices available to the through-origin fit, so the new sum of squared residuals must be \(50.0\) or smaller, not \(58.0\).

- Ask what information about change in \(y\) is available when \(x\) never changes. - Think about how many different lines can pass through the same vertical location at \(x=4\). - Check what happens to the explanatory-variable spread in the usual slope formula.

1. Because every \(x_i=4\), the data contain no horizontal variation with which to estimate how the response changes as \(x\) changes. 2. Any line whose fitted value at \(x=4\) equals the mean response gives the same fitted value for every observation. 3. Infinitely many intercept-slope pairs can produce that same value at \(x=4\), so there is no unique least-squares slope. 4. Equivalently, the usual slope calculation is undefined because the explanatory-variable standard deviation is \(0\).

A unique slope cannot be estimated because all explanatory values are identical. With no variation in \(x\), many different lines give the same fitted values at \(x=4\), so the regression slope is not identifiable.

- Identify a simple candidate line that is always available when an intercept is allowed. - Compare the candidate line’s squared error with the minimum that least squares is supposed to achieve. - Check the usual relationship between unexplained and total response variation.

1. The horizontal line \(\hat y=\bar y\) is an available candidate line and has a sum of squared residuals equal to \(SST=200\). 2. The ordinary least-squares line minimizes the sum of squared residuals over all intercept-slope lines, so its \(SSE\) cannot exceed \(200\). 3. The reported \(SSE=240\) is therefore impossible for the least-squares fit with an intercept. 4. The reported ratio would give \(1-\frac{240}{200}=-0.20\), another sign that the values are inconsistent with an ordinary least-squares fit with an intercept.

No. The least-squares fit must have \(SSE\le200\) because the horizontal mean-response line already achieves \(SSE=200\). The reported values give \(1-\frac{240}{200}=-0.20\), so they are inconsistent with the stated fit.

- Express each new residual in terms of the original residual and the vertical shift. - Expand the squared expression before summing across observations. - Use a standard property of residuals from a least-squares fit with an intercept.

1. Let the least-squares residuals be \(e_i\). Because the fitted model has an intercept, \(\sum e_i=0\). 2. Shifting the fitted line upward by \(1.5\) changes each residual to \(e_i-1.5\). 3. The new squared-error sum is \(\sum(e_i-1.5)^2=\sum e_i^2-3\sum e_i+20\cdot(1.5)^2\). 4. Using \(\sum e_i=0\) and \(\sum e_i^2=80\), the new value is \(80+20\cdot2.25=125\).

The shifted line has \(SSE=125\).

- Check the ordinary residual sum for each set first. - Then weight each residual by its corresponding explanatory value and add those products. - A set must satisfy both stated conditions, not just one of them.

1. For Set A, \(1-2+1=0\), and \(1\cdot1+2\cdot(-2)+3\cdot1=0\). 2. For Set B, \(-1+0+1=0\), but \(1\cdot(-1)+2\cdot0+3\cdot1=2\ne0\). 3. Therefore, Set A satisfies both least-squares residual conditions, while Set B satisfies only the zero-sum condition.

Set A is consistent with both conditions. Set B is not because \(\sum x_ie_i=2\), not \(0\).

- Express the second regression slope using a centered cross-product in the numerator. - Relate fitted responses to the original least-squares line and its residuals. - Use the condition that the fitted values are not all identical when considering the denominator.

1. In an ordinary least-squares fit with an intercept, residuals are orthogonal to the fitted values, so \(\sum (\hat{y}_i-\bar{\hat{y}})e_i=0\). 2. The slope from regressing residuals on fitted values is this centered cross-product sum divided by \(\sum(\hat{y}_i-\bar{\hat{y}})^2\). 3. The numerator is \(0\), and the denominator is positive because the fitted values are not all identical. 4. Therefore, the second least-squares line has slope \(0\).

The second fitted line has slope \(0\).

- Duplicating one point changes its weight relative to the others. - Recompute the sample means before finding the slope. - Use the new mean point to determine the intercept after finding the slope.

1. In the duplicated data, \(\bar x=1.25\) and \(\bar y=0.25\). 2. The centered cross-product sum is \(-0.25\), and the centered square sum for \(x\) is \(2.75\). 3. Thus the new slope is \(b=\frac{-0.25}{2.75}=-\frac{1}{11}\). 4. The intercept is \(a=0.25-\left(-\frac{1}{11}\right)\cdot1.25=\frac{4}{11}\). 5. Therefore, duplicating only one point changes the fit to \(\hat y=\frac{4}{11}-\frac{1}{11}x\).

No. The new least-squares line is \(\hat y=\frac{4}{11}-\frac{1}{11}x\), approximately \(\hat y=0.364-0.091x\).

- Use the two residual conditions satisfied by an ordinary least-squares fit with an intercept. - Determine what the removed point contributes to those conditions when its residual is zero. - Use the condition on the remaining \(x\)-values to establish uniqueness of the refitted line.

1. For the original least-squares line, the residuals satisfy \(\sum e_i=0\) and \(\sum x_ie_i=0\). 2. The removed observation contributes \(0\) to both sums because its residual is exactly \(0\). 3. Therefore, the same slope and intercept still satisfy both least-squares normal conditions for the remaining observations. 4. Because the remaining \(x\)-values are not all identical, the least-squares line is unique, so refitting gives the same coefficients.

No. Under the stated conditions, removing an observation with residual exactly \(0\) leaves the least-squares slope and intercept unchanged.