Free math worksheets

- 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.

- 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.

- Think about whether changing units changes the relative positions of the observations. - Positive shifts and rescalings leave standardized values unchanged. - Use the relationship between \(r\) and \(r^2\).

1. A positive linear conversion shifts and/or rescales a variable without changing its standardized values. 2. Therefore, the correlation \(r\) is unchanged. 3. Because \(r^2\) is the square of the correlation, \(r^2\) remains \(0.79\).

\(r^2\) remains \(0.79\) because positive linear unit conversions do not change the correlation.

- 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 exact quantity minimized by the least-squares method. - Consider what happens when one residual is positive and another is negative. - Think about why squaring each residual before adding makes the criterion useful.

1. The least-squares regression line minimizes the sum of the squared residuals, not the sum of the residuals. 2. Positive and negative residuals can cancel. In fact, any fitted line that passes through \((\bar x, \bar y)\) has residuals that sum to \(0\). 3. Because many lines can have a residual sum of \(0\), that criterion does not identify a unique best-fitting line. Squaring the residuals prevents cancellation and penalizes larger errors.

The least-squares regression line minimizes the sum of squared residuals. Unsquared residuals can cancel, and many lines can have a residual sum of \(0\), so minimizing their sum does not identify a unique best-fitting line.

- 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 how one residual enters the least-squares objective. - Compare the squared contribution before and after the change. - Think about how squaring changes the impact of a residual as its magnitude grows.

1. With residual \(2.0\), the observation contributes \((2.0)^2=4.0\) to the sum of squared residuals. 2. With residual \(4.0\), it contributes \((4.0)^2=16.0\). 3. The contribution increases by \(16.0-4.0=12.0\), even though the residual itself only doubled. 4. Squaring makes large residuals contribute disproportionately more to the least-squares criterion.

The contribution increases by \(12.0\), from \(4.0\) to \(16.0\). This illustrates that least squares penalizes larger residuals more heavily because residuals are squared.

- 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.

- 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.

- 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 \(\bar y = 32.0\).

- Rearrange the regression-slope formula to isolate \(r\). - The sign of \(r\) must match the positive slope. - Check that the result lies within the allowable interval \([-1, 1]\).

1. Start with \(b = r\frac{s_y}{s_x}\). 2. Solve for correlation: \(r = b\frac{s_x}{s_y}\). 3. Substitute: \(r = 1.875\left(\frac{2.0}{5.0}\right) = 0.750\).

\(r = 0.750\).

- 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 \(r^2 = 0.74\).

- 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 student’s 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.

- 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 how duplicating every point affects the objective value for any one candidate line. - Ask whether multiplying every candidate objective by the same factor changes the minimizer. - Then count how many times each original squared residual appears in the duplicated data set.

1. For any candidate line, duplicating every observation duplicates every residual and therefore doubles that line's SSE. 2. Multiplying every candidate SSE by the same positive factor does not change which line minimizes SSE. 3. Therefore, the least-squares slope and intercept are unchanged. 4. At the unchanged fitted line, each original squared residual appears twice, so the fitted SSE doubles.

The least-squares slope and intercept stay the same, while 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 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 \(\text{SSR}_A = 0.00 + 0.04 + 0.01 + 0.04 + 0.09 = 0.18\). 2. For line B, the residuals are \(-0.8, -0.4, -0.3, -0.4, 0.3\), so \(\text{SSR}_B = 0.64 + 0.16 + 0.09 + 0.16 + 0.09 = 1.14\). 3. For line C, the residuals are \(0.3, 0.3, 0.0, -0.5, -0.2\), so \(\text{SSR}_C = 0.09 + 0.09 + 0.00 + 0.25 + 0.04 = 0.47\). 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 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 \(0\).

- 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.