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Confidence interval for a slope

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Difficulty: 1 – 5

53952012

A \(95\%\) confidence interval for the population slope relating weekly practice time to accuracy score is \((0.08, 0.42)\). An analyst claims that the population slope is positive. Is the claim supported by the interval?

Hints

- Compare both endpoints with \(0\). - A positive-slope claim requires the plausible slope values to be positive. - Check whether the interval includes any zero or negative values.

Solution

1. Every value in the interval \((0.08, 0.42)\) is greater than \(0\). 2. Therefore, the interval excludes zero and all negative slopes. 3. The interval supports the claim that the population slope is positive at the \(95\%\) confidence level.

Answer

Yes. The entire interval is above \(0\), so it supports a positive population slope.

53952212

A \(95\%\) confidence interval for the population slope relating snow depth to travel time is \((-0.25, 0.31)\). An analyst claims that the population slope is nonzero. Is the claim supported by the interval?

Hints

- Locate \(0\) relative to the two endpoints. - A nonzero-slope claim requires zero to be excluded. - Do not treat a sample estimate away from zero as sufficient by itself.

Solution

1. A nonzero-slope claim requires the confidence interval to exclude \(0\). 2. The interval \((-0.25, 0.31)\) contains \(0\). 3. Therefore, the interval does not support the claim of a nonzero population slope at the \(95\%\) confidence level.

Answer

No. The interval contains \(0\), so it does not support a nonzero population slope.

54806212

A \(95\%\) confidence interval for the population slope predicting weekly electricity use, in kilowatt-hours, from average daily operating time, in hours, is \((18, 26)\) kilowatt-hours per hour. A student says, “We are \(95\%\) confident that weekly electricity use is between \(18\) and \(26\) kilowatt-hours.” Correct the interpretation.

Hints

- Check the units attached to the interval endpoints. - Identify whether the interval estimates a response level or a rate of change. - Interpret a slope in terms of a one-unit increase in the explanatory variable.

Solution

1. The interval estimates a population regression slope, so its units are response units per explanatory-variable unit. 2. The endpoints \(18\) and \(26\) describe plausible average changes in weekly electricity use associated with a one-hour increase in average daily operating time. 3. The interval does not estimate the level of weekly electricity use itself.

Answer

The interval concerns the slope: we are \(95\%\) confident that each additional hour of average daily operating time is associated with an average increase of between \(18\) and \(26\) kilowatt-hours in weekly electricity use, under the regression conditions.

53949912

A \(95\%\) confidence interval for the population slope relating the number of pages read to reading time in minutes is \((0.42, 1.18)\). a) Interpret the interval in context. b) Does the interval provide evidence of a nonzero linear association? Explain.

Hints

- State the slope’s units as minutes per page. - Interpret the interval as a range of plausible values for the population slope. - To assess a nonzero linear association, check whether \(0\) lies in the interval.

Solution

1. a) The slope measures the change in mean reading time, in minutes, for each additional page read. We are \(95\%\) confident that the population slope is between \(0.42\) and \(1.18\) minutes per page. 2. b) The interval does not contain \(0\). 3. Therefore, the interval provides evidence that the population slope is positive and nonzero, so there is a positive linear association in the population.

Answer

a) We are \(95\%\) confident that the population slope is between \(0.42\) and \(1.18\) minutes of reading time per additional page read. b) Yes. The interval excludes \(0\), providing evidence of a positive, nonzero population slope.

53950012

A \(95\%\) confidence interval for the population slope relating distance from a heater, in feet, to temperature, in degrees Fahrenheit, is \((-2.40, -0.70)\). a) Interpret the interval in context. b) Does the interval provide evidence of a nonzero linear association? Explain.

Hints

- State the slope’s units as degrees Fahrenheit per foot. - Use the signs of both endpoints to describe the direction of the population slope. - Check whether \(0\) is included before deciding whether the interval supports a nonzero association.

Solution

1. a) The slope measures the change in mean temperature, in degrees Fahrenheit, for each additional foot from the heater. We are \(95\%\) confident that the population slope is between \(-2.40\) and \(-0.70\) degrees Fahrenheit per foot. 2. b) The interval does not contain \(0\). 3. Therefore, the interval provides evidence that the population slope is negative and nonzero, so there is a negative linear association in the population.

Answer

a) We are \(95\%\) confident that the population slope is between \(-2.40\) and \(-0.70\) degrees Fahrenheit per additional foot from the heater. b) Yes. The interval excludes \(0\), providing evidence of a negative, nonzero population slope.

53950112

A \(95\%\) confidence interval for the population slope relating vehicle age, in years, to resale price, in thousands of dollars, is \((-0.30, 0.80)\). a) Interpret the interval in context. b) Does the interval provide evidence of a nonzero linear association? Explain.

Hints

- Express the slope in thousands of dollars per year. - The interval gives plausible values for the population rate of change. - Determine whether \(0\) is one of those plausible values.

Solution

1. a) The slope measures the change in mean resale price, in thousands of dollars, for each additional year of vehicle age. We are \(95\%\) confident that the population slope is between \(-0.30\) and \(0.80\) thousand dollars per year. 2. b) The interval contains \(0\). 3. Therefore, the interval does not provide evidence that the population slope differs from \(0\). The data do not establish a nonzero linear association.

Answer

a) We are \(95\%\) confident that the population slope is between \(-0.30\) and \(0.80\) thousand dollars of resale price per additional year of vehicle age. b) No. The interval includes \(0\), so it does not provide evidence of a nonzero population slope.

53950212

Use the scatterplot and residual plot for a random sample relating water temperature to dissolving time. The sample is less than \(10\%\) of the population, and the residual distribution has no strong skew or outliers. Decide whether the usual conditions for a \(t\)-interval for the population slope are adequately met.

Figure for problem 539502

Hints

- Use the sampling information and the \(10\%\) condition to assess independence. - Check the scatterplot for an approximately straight-line form and the residual plot for a systematic pattern. - Compare the residual spread across the full range of water temperatures, then use the stated residual-distribution information.

Solution

1. The random sample and the \(10\%\) condition support independence of the observations. 2. The scatterplot shows an approximately linear relationship. 3. The residual plot shows random scatter around \(0\) with nearly constant spread, supporting linearity and equal variance. 4. The residual distribution with no strong skew or outliers supports the normality condition. 5. Therefore, the stated conditions adequately support a \(t\)-interval for the population slope.

Answer

Yes. The random-sample information and the displays support independence, linearity, approximately constant residual variance, and approximate normality, so a \(t\)-interval for the population slope is appropriate.

53950312

A researcher wants a confidence interval for the population slope relating weekly study time to exam score. Students chose whether to respond to an online survey, so the sample is voluntary response. The scatterplot and residual plot otherwise look appropriate. Decide whether the usual conditions for a \(t\)-interval for the population slope are adequately met.

Hints

- Separate conditions about the regression model from conditions about data collection. - Ask whether every member of the population had a known, chance-based path into the sample. - Appropriate graphs cannot repair bias from voluntary response.

Solution

1. The appropriate scatterplot and residual plot support the linearity and equal-variance conditions. 2. However, a voluntary-response sample is not a random sample and can be systematically biased. 3. Because the randomization condition fails, the sample does not support inference about the population slope.

Answer

No. The voluntary-response sample fails the randomization condition, so population inference about the slope is not justified even though the graphical conditions look appropriate.

53950412

Use the residual plot for a linear model relating miles traveled to fuel used. Decide whether the usual conditions for a \(t\)-interval for the population slope are adequately met.

Figure for problem 539504

Hints

- Check whether the residuals are randomly scattered around \(0\) or follow a systematic pattern. - Relate a curved residual pattern to what the fitted line failed to capture. - Identify the regression-inference condition affected by that pattern.

Solution

1. A valid linear regression model should leave residuals randomly scattered around \(0\). 2. The residual plot has a pronounced U-shaped pattern, indicating that a straight line does not adequately model the relationship. 3. Therefore, the linearity condition fails, and a \(t\)-interval for the slope of this linear model is not appropriate.

Answer

No. The U-shaped residual pattern indicates that the linearity condition fails, so a \(t\)-interval for the linear-model slope is not appropriate.

53950512

Use the residual plot for a linear model relating the number of editing passes to the number of errors remaining. Decide whether the usual conditions for a \(t\)-interval for the population slope are adequately met.

Figure for problem 539505

Hints

- Compare the vertical spread of the residuals at low and high numbers of editing passes. - A widening spread indicates that prediction variability changes with the explanatory variable. - Match the changing spread to the appropriate regression-inference condition.

Solution

1. A standard slope \(t\)-interval assumes that residuals have approximately constant spread across the explanatory-variable range. 2. The residual plot fans outward as the number of editing passes increases, so the residual spread is not approximately constant. 3. Therefore, the constant-variance condition is not met, and the usual slope \(t\)-interval is not appropriate without a different model or method.

Answer

No. The outward-fanning residual plot violates the constant-variance condition, so the usual \(t\)-interval for the population slope is not appropriate.

53950612

A researcher wants a confidence interval for the population slope relating wind speed to sailing time. The \(11\) observations form a random sample, and the relationship appears linear. Use the residual plot to decide whether the usual conditions for a \(t\)-interval are adequately met.

Figure for problem 539506

Hints

- Compare the isolated residual with the scale of the other ten residuals. - Consider how strongly one unusual observation can affect inference when \(n=11\). - A random sample and a generally linear pattern do not by themselves guarantee valid slope inference.

Solution

1. With only \(11\) observations, the regression \(t\)-procedure is sensitive to strong departures from normality and to influential observations. 2. The residual plot contains one extremely isolated residual, providing evidence of an outlier and making the residual normality condition doubtful. 3. The observation may also have a large effect on the fitted slope and its standard error. 4. Therefore, the stated information does not adequately support the usual \(t\)-interval without investigating that observation.

Answer

No. In this small sample, the extremely isolated residual makes the normality condition doubtful and may indicate an influential observation, so the usual slope \(t\)-interval is not adequately supported.

53950712

A \(95\%\) confidence interval for the population slope relating tree diameter to tree age is \((0.93, 2.57)\). Find the point estimate and the margin of error.

Hints

- The point estimate lies halfway between the two endpoints. - The margin of error is the distance from the midpoint to either endpoint. - Check that \(1.75 - 0.82 = 0.93\) and \(1.75 + 0.82 = 2.57\).

Solution

1. The point estimate is the midpoint of the interval: \(b = \frac{0.93 + 2.57}{2} = 1.75\). 2. The margin of error is half the interval’s width: \(ME = \frac{2.57 - 0.93}{2} = 0.82\).

Answer

Point estimate: \(b = 1.75\). Margin of error: \(0.82\).

53950812

For a regression of sound level on distance from a stage, a confidence interval for the population slope uses \(b = -1.20\), \(t^* = 2.12\), and \(SE_b = 0.31\). Find the two endpoints and round to three decimals.

Hints

- Use the interval form \(b \pm t^*SE_b\). - Compute the margin of error before finding the endpoints. - Keep the unrounded margin of error until the final rounding step.

Solution

1. Compute the margin of error: \(t^*SE_b = 2.12(0.31) = 0.6572\). 2. Subtract and add the margin of error: \(-1.20 - 0.6572 = -1.8572\), \(-1.20 + 0.6572 = -0.5428\). 3. Rounded to three decimals, the interval is \((-1.857, -0.543)\).

Answer

\((-1.857, -0.543)\).

53950912

A confidence interval for the population slope relating the number of volunteers to the number of boxes packed has margin of error \(0.54\) and critical value \(t^* = 2.05\). Find the standard error of the sample slope. Round to three decimals.

Hints

- Start with \(ME = t^*SE_b\). - Isolate \(SE_b\) by dividing by the critical value. - Round only after evaluating the quotient.

Solution

1. Use the relationship \(ME = t^*SE_b\). 2. Solve for the standard error: \(SE_b = \frac{ME}{t^*} = \frac{0.54}{2.05} \approx 0.263\).

Answer

\(SE_b \approx 0.263\).

53951112

Using the same sample relating trail grade to hiking speed, a researcher changes a \(90\%\) confidence interval for the population slope to a \(99\%\) confidence interval. Describe how the interval changes and explain why.

Hints

- Determine which parts of \(b \pm t^*SE_b\) change when the sample stays the same. - A higher confidence level requires capturing more of the \(t\)-distribution. - Connect the larger critical value to the margin of error.

Solution

1. The sample slope and its standard error remain unchanged because the same sample is used. 2. A higher confidence level requires a larger critical value \(t^*\). 3. Since the margin of error is \(t^*SE_b\), the larger critical value increases the margin of error. Therefore, the \(99\%\) interval is wider than the \(90\%\) interval.

Answer

The \(99\%\) confidence interval is wider because the higher confidence level uses a larger critical value, which increases the margin of error.

53951212

For a confidence interval for the population slope relating the number of exhibits visited to visit duration, the confidence level stays at \(95\%\), but the sample size increases while the data pattern remains similar. Describe how the interval generally changes and explain why.

Hints

- Start with the margin-of-error formula \(t^*SE_b\). - Consider how a larger sample affects the standard error of the slope. - Also consider the small change in \(t^*\) when the degrees of freedom increase.

Solution

1. With a larger sample and a similar data pattern, the standard error of the sample slope generally decreases. 2. The larger degrees of freedom also make the \(t\) critical value slightly smaller. 3. Both effects reduce the margin of error \(t^*SE_b\), so the confidence interval generally becomes narrower.

Answer

The interval generally becomes narrower because the slope’s standard error decreases, and the larger degrees of freedom make the critical value slightly smaller.

53951312

A confidence interval estimates the population slope relating the number of bags of road salt purchased to total cost in dollars. Every response value is converted from dollars to cents. Describe how the slope estimate and confidence interval change, and explain whether the substantive conclusion changes.

Hints

- Converting dollars to cents multiplies the response variable by \(100\). - Track how that response rescaling affects the slope and its standard error. - Check whether multiplying both endpoints by a positive number changes their signs or inclusion of \(0\).

Solution

1. Converting dollars to cents multiplies every response value by \(100\). 2. Therefore, the sample slope, its standard error, the margin of error, and both confidence-interval endpoints are multiplied by \(100\). 3. The interval represents the same relationship in cents rather than dollars. Its signs and whether it contains \(0\) do not change, so the substantive conclusion is unchanged.

Answer

The slope estimate and both confidence-interval endpoints are multiplied by \(100\). The interval is expressed in cents per bag instead of dollars per bag, but the substantive conclusion is unchanged.

53951612

A student interprets a \(95\%\) confidence interval for the population slope relating the number of paint layers to drying time by saying, “There is a \(95\%\) probability that the fixed population slope lies in this computed interval.” Correct the interpretation.

Hints

- Distinguish the fixed population parameter from the interval produced by a random sample. - Ask what would vary if the study were repeated many times. - Interpret \(95\%\) as the long-run capture rate of the interval method.

Solution

1. The population slope is a fixed parameter, while the interval is random before the sample is collected. 2. After the data are collected and the interval is computed, the interval either contains the population slope or it does not. 3. The \(95\%\) confidence level describes the method: in repeated random sampling, about \(95\%\) of intervals constructed this way would contain the true population slope.

Answer

We are \(95\%\) confident that the computed interval contains the population slope. The \(95\%\) refers to the long-run success rate of the interval procedure, not to a probability assigned to the fixed slope after the interval is computed.

53951712

A student interprets a \(95\%\) confidence interval for the population slope relating the number of musicians to setup time by saying, “\(95\%\) of individual setup times change at a rate within the interval.” Correct the interpretation.

Hints

- Identify the single parameter estimated by a slope confidence interval. - Distinguish a change in the mean response from variation among individual responses. - The confidence level describes the interval method, not a percentage of observations.

Solution

1. A confidence interval for a regression slope estimates one population parameter: the change in mean setup time associated with one additional musician. 2. It does not describe a distribution of rates for individual setup times. 3. The correct interpretation is that we are \(95\%\) confident that the population slope lies between the interval’s endpoints.

Answer

The interval estimates the population regression slope—the change in mean setup time per additional musician. It does not state that \(95\%\) of individual responses have rates within the interval.

53951812

A student interprets a confidence interval for the population slope relating distance from a lamp to light intensity by saying, “Because the interval contains \(0\), the population slope equals \(0\).” Correct the interpretation.

Hints

- Treat every value inside the confidence interval as plausible based on the data. - Check whether the interval contains values other than \(0\). - Distinguish “insufficient evidence of a nonzero slope” from “proof that the slope is zero.”

Solution

1. If the interval contains \(0\), then a zero population slope is one plausible value based on the sample. 2. The interval also contains nonzero values, so the data do not establish that the slope is exactly \(0\). 3. The correct conclusion is that the data do not provide convincing evidence of a nonzero population slope at the confidence level used.

Answer

Containing \(0\) means the interval does not provide convincing evidence of a nonzero population slope. It does not prove that the population slope is exactly \(0\).

53951912

A student interprets a narrow confidence interval for the population slope relating hours after sunrise to shadow length by saying, “The narrow interval proves that the regression model is correct.” Correct the interpretation.

Hints

- Distinguish the precision of an estimate from the validity of the model. - Identify which assumptions must be checked separately from interval width. - A precise estimate can still come from a misspecified model or biased sample.

Solution

1. A narrow confidence interval indicates that the population slope is estimated with relatively high precision, assuming the regression conditions are satisfied. 2. Interval width does not test whether the relationship is linear, whether the residual variance is constant, or whether the data were collected appropriately. 3. Therefore, a narrow interval does not prove that the regression model or study design is valid.

Answer

A narrow interval indicates a precise slope estimate under the regression assumptions. It does not prove that the model form, assumptions, or data-collection method are correct.

53952112

A \(95\%\) confidence interval for the population slope relating distance from a speaker to sound level is \((-1.80, -0.40)\). An analyst claims that the population slope is less than \(-0.20\). Is the claim supported by the interval?

Hints

- The claim must hold for every value in the interval to be fully supported. - For a “less than” claim, inspect the interval’s upper endpoint. - Compare \(-0.40\) with \(-0.20\).

Solution

1. Compare the largest value in the interval with the claimed upper bound. 2. The largest plausible slope is \(-0.40\), and \(-0.40 < -0.20\). 3. Therefore, every value in the confidence interval is less than \(-0.20\), so the interval supports the claim.

Answer

Yes. The entire interval lies below \(-0.20\), so it supports the claim that the population slope is less than \(-0.20\).

54794512

A regression report gives a sample slope of \(b=0.520\) and claims that the corresponding \(95\%\) confidence interval for the population slope is \((0.180, 0.740)\). Without recomputing the interval from raw data, determine whether the reported interval can be correct. Explain.

Hints

- Think about the symmetry of a standard two-sided confidence interval around its point estimate. - Compare the center of the reported interval with the reported estimate. - A quick structural check can reveal an inconsistency without raw data.

Solution

1. A two-sided confidence interval constructed as \(b\pm\text{margin of error}\) must be centered at the sample slope \(b\). 2. The midpoint of the reported interval is \(\frac{0.180+0.740}{2}=0.460\). 3. Since \(0.460\ne0.520\), the interval is not centered at the reported sample slope and cannot be the corresponding confidence interval.

Answer

No. The interval has midpoint \(0.460\), not \(0.520\), so it cannot be the confidence interval corresponding to the reported slope estimate.

54797512

An observational study produces a \(95\%\) confidence interval of \((0.30, 0.82)\) for the population slope relating hours of sleep to reaction-time score. A report says, “Because the entire interval is positive, getting more sleep causes reaction-time score to increase.” Evaluate the report.

Hints

- Separate what the interval says about the slope from what the study design can justify. - Ask whether the explanatory variable was assigned or merely observed. - A nonzero slope addresses association, not automatically causation.

Solution

1. The interval supports a positive population slope for the linear association between sleep and reaction-time score. 2. A confidence interval for a regression slope does not by itself establish a cause-and-effect relationship. 3. Because the study is observational, confounding or other explanations may account for the association.

Answer

The interval supports a positive linear association, but it does not establish that more sleep causes the response to increase.

54801312

A \(95\%\) confidence interval for the population slope when predicting \(y\) from \(x\) is \((0.40, 1.10)\). An analyst defines a new explanatory variable \(u=-x\) and fits the equivalent regression using \(u\). What is the corresponding \(95\%\) confidence interval for the population slope with respect to \(u\)? Explain why the endpoints must be reordered.

Hints

- Determine how the slope changes when the explanatory axis is reversed. - Apply the same transformation to every plausible slope in the original interval. - Confidence-interval endpoints are written from the smaller value to the larger value.

Solution

1. Since \(u=-x\), a one-unit increase in \(u\) corresponds to a one-unit decrease in \(x\), so the new population slope is the negative of the old slope. 2. Negating every value in \((0.40, 1.10)\) gives values from \(-0.40\) to \(-1.10\). 3. Written from smaller to larger endpoint, the interval is \((-1.10, -0.40)\).

Answer

The corresponding interval is \((-1.10, -0.40)\). Negating the slope reverses the order of the endpoints, so they must be reordered from least to greatest.

54802712

Two regression studies have the same slope estimate, \(b=2.00\), and use the same critical value, \(t^*=2.10\). Study A has slope standard error \(SE_b=0.20\), while Study B has \(SE_b=0.50\). Construct both confidence intervals and explain how greater uncertainty in the slope estimate affects interval width.

Hints

- Use the same confidence-interval structure for both studies. - Compare the quantities that multiply the standard errors. - Relate interval width directly to uncertainty in the slope estimate.

Solution

1. Study A has margin of error \(2.10\cdot0.20=0.42\), giving \((1.58, 2.42)\). 2. Study B has margin of error \(2.10\cdot0.50=1.05\), giving \((0.95, 3.05)\). 3. With the same point estimate and critical value, a larger slope standard error produces a larger margin of error and therefore a wider confidence interval.

Answer

Study A: \((1.58, 2.42)\). Study B: \((0.95, 3.05)\). The larger standard error in Study B makes its confidence interval wider.

54804312

A \(95\%\) confidence interval for the population slope predicting temperature in degrees Celsius from elevation in kilometers is \((-6.8, -5.4)\) degrees Celsius per kilometer. The response is converted to degrees Fahrenheit using \(F=32+\frac{9}{5}C\). Find the corresponding confidence interval for the slope in degrees Fahrenheit per kilometer.

Hints

- Separate the additive and multiplicative parts of the temperature conversion. - Decide which part of an affine response conversion changes a slope. - Apply that slope conversion to both endpoints of the interval.

Solution

1. Adding \(32\) to the response affects the regression intercept but not the slope. 2. Multiplying Celsius values by \(\frac{9}{5}\) multiplies every plausible slope by \(\frac{9}{5}\). 3. The transformed endpoints are \(-6.8\cdot\frac{9}{5}=-12.24\) and \(-5.4\cdot\frac{9}{5}=-9.72\). 4. The corresponding interval is \((-12.24, -9.72)\) degrees Fahrenheit per kilometer.

Answer

The confidence interval is \((-12.24, -9.72)\) degrees Fahrenheit per kilometer.

54805012

A \(95\%\) confidence interval for a population regression slope is \((0.80, 1.30)\). A student says, “Because \(1\) is inside the interval, we have shown that the population slope equals \(1\).” Evaluate the statement.

Hints

- Interpret what it means for a candidate parameter value to lie inside a confidence interval. - Compare “plausible” with “proven exactly.” - Notice that the interval contains many possible slope values, not just \(1\).

Solution

1. The interval identifies a range of population slopes that are compatible with the data at the stated confidence level. 2. The value \(1\) is one plausible value, but so are many other values between \(0.80\) and \(1.30\). 3. Including \(1\) means the data do not rule out a slope of \(1\) in the corresponding two-sided inference; it does not establish exact equality.

Answer

The statement is incorrect. A slope of \(1\) is plausible because it lies in the interval, but the interval does not prove that the population slope is exactly \(1\).

54807512

A study reports a very narrow \(95\%\) confidence interval for a population regression slope. A student says, “A narrow slope interval proves the linear association is strong.” Evaluate the statement.

Hints

- Identify what interval width measures directly. - Separate uncertainty in an estimate from descriptive strength of association. - Think about how a large sample can affect precision without changing the conceptual meaning of correlation.

Solution

1. The width of a confidence interval describes the precision of the slope estimate. 2. Precision depends on features such as sample size, explanatory-variable spread, and residual variability. 3. Strength of linear association is a different concept, commonly described by the scatterplot and correlation. 4. A slope can be estimated precisely even when the association is not especially strong, particularly with a large informative sample.

Answer

The statement is incorrect. A narrow slope confidence interval indicates a precise estimate of the population slope; it does not by itself prove that the linear association is strong.

54808312

A \(95\%\) confidence interval for a population regression slope is \((0.60, 1.10)\). Use the interval to make the decision for each corresponding two-sided test at \(\alpha=0.05\): a) \(H_0: \beta=0.75\) b) \(H_0: \beta=0.50\)

Hints

- Match the confidence level with the significance level for a two-sided test. - Check each null value against the full interval separately. - Values inside and outside the interval lead to different test decisions.

Solution

1. For a two-sided test at \(\alpha=0.05\), a null value inside the matching \(95\%\) confidence interval is not rejected. 2. The value \(0.75\) lies inside \((0.60, 1.10)\), so fail to reject \(H_0: \beta=0.75\). 3. The value \(0.50\) lies outside the interval, so reject \(H_0: \beta=0.50\).

Answer

a) Fail to reject \(H_0: \beta=0.75\). b) Reject \(H_0: \beta=0.50\).

54810012

A \(95\%\) confidence interval for the slope relating outside temperature \(x\) in degrees Fahrenheit to predicted daily heating-energy use \(y\) in kilowatt-hours is \((-4.2, -1.1)\). Interpret this interval in context. Does it imply that every warmer day uses less heating energy than every colder day?

Hints

- Identify the units of a regression slope from the response and explanatory variables. - Translate the negative endpoints into changes in the mean response. - Separate an average regression relationship from what must happen for every individual observation.

Solution

1. The interval estimates the population regression slope, the change in mean heating-energy use associated with a one-degree increase in outside temperature. 2. The analyst is \(95\%\) confident that this mean change is between a decrease of \(4.2\) and a decrease of \(1.1\) kilowatt-hours per degree Fahrenheit. 3. The interval concerns the average linear relationship, not the ordering of every pair of individual days. 4. Individual days can differ because of residual variation and other factors.

Answer

The analyst is \(95\%\) confident that each \(1\,\text{°F}\) increase in outside temperature is associated with an average decrease of between \(1.1\) and \(4.2\) kilowatt-hours in daily heating-energy use. This does not mean every warmer day must use less energy than every colder day.

54810412

An analyst constructs a standard \(95\%\) \(t\)-interval for a regression slope. The residual plot has no curvature, but the residual spread grows steadily from left to right, forming a clear fan shape. Is the usual slope confidence interval well supported by the regression conditions? Explain.

Hints

- Check the residual spread as well as the residual center. - Recall what the standard slope interval assumes about the variability of the regression errors. - Decide whether a systematic change in spread matches that assumption.

Solution

1. The usual slope \(t\)-interval assumes that the error variability is approximately constant across values of the explanatory variable. 2. A fan-shaped residual plot indicates that the error spread changes with the explanatory variable. 3. This violates the constant-variance condition used by the standard error formula for the usual interval. 4. Therefore, the reported confidence interval is not well supported without addressing the nonconstant variance.

Answer

No. The fan shape is evidence of nonconstant residual variance, so the usual slope \(t\)-interval may have an unreliable standard error and coverage.

54812212

A \(95\%\) confidence interval for a regression slope is \((-0.80, -0.20)\). The analyst then creates a new response variable by adding \(100\) to every response value and refits the regression. What happens to the \(95\%\) confidence interval for the slope?

Hints

- Determine whether adding a constant changes the rate of response change per unit of \(x\). - Consider what happens to residuals after the entire response scale is shifted. - A slope interval depends on the slope estimate and its standard error.

Solution

1. Adding the same constant to every response shifts all observations and fitted values vertically by the same amount. 2. The fitted slope and every residual remain unchanged. 3. Because the slope estimate and its standard error are unchanged, the confidence interval for the slope is unchanged.

Answer

The \(95\%\) confidence interval remains \((-0.80, -0.20)\).

54812712

A \(95\%\) confidence interval for a population slope is \((0.03, 0.09)\). A practical standard says the slope must exceed \(0.05\) to be considered large enough to matter. What does the interval support about positivity and about exceeding the practical standard?

Hints

- Check the interval against zero first. - Then compare the entire interval with the separate practical threshold. - Statistical evidence for a positive effect and evidence for a sufficiently large effect are different questions.

Solution

1. The entire interval is above \(0\), so the interval supports a positive population slope. 2. The practical threshold \(0.05\) lies inside the interval. 3. Therefore, values both below and above \(0.05\) are plausible at this confidence level, so the interval does not establish that the slope exceeds the practical standard.

Answer

The interval supports that the slope is positive, but it does not support the stronger claim that \(\beta>0.05\) because \(0.05\) lies within the interval.

53949612

A random sample of \(n = 12\) observations is used to estimate the population regression slope relating practice time to accuracy score. The sample slope is \(b = 0.800\), with standard error \(SE_b = 0.180\). Assume the conditions for inference about a regression slope are met. Construct a \(90\%\) confidence interval for the population slope \(\beta\). Use \(df = n - 2\) and round to three decimals.

Hints

- A confidence interval for a regression slope has the form \(b \pm t^*SE_b\). - Use \(df = n - 2\) to find the appropriate \(t\) critical value. - Compute the margin of error before finding the two endpoints.

Solution

1. The degrees of freedom are \(df = 12 - 2 = 10\). 2. For a \(90\%\) confidence interval with \(df = 10\), the critical value is \(t^* \approx 1.812\). 3. The margin of error is \(t^*SE_b \approx 1.812461(0.180) \approx 0.326\). 4. The confidence interval is \(b \pm t^*SE_b = 0.800 \pm 0.326\), which gives \((0.474, 1.126)\).

Answer

The \(90\%\) confidence interval for \(\beta\) is \((0.474, 1.126)\).

53949712

A random sample of \(n = 14\) observations is used to estimate the population regression slope relating distance from a heater to temperature. The sample slope is \(b = -1.050\), with standard error \(SE_b = 0.205\). Assume the conditions for inference about a regression slope are met. Construct a \(95\%\) confidence interval for the population slope \(\beta\). Use \(df = n - 2\) and round to three decimals.

Hints

- Use \(b \pm t^*SE_b\) for a confidence interval for the population slope. - Find \(df = n - 2\) before choosing the \(t\) critical value. - Because the sample slope is negative, check the order and signs of the two endpoints.

Solution

1. The degrees of freedom are \(df = 14 - 2 = 12\). 2. For a \(95\%\) confidence interval with \(df = 12\), the critical value is \(t^* \approx 2.179\). 3. The margin of error is \(t^*SE_b \approx 2.178813(0.205) \approx 0.447\). 4. The confidence interval is \(b \pm t^*SE_b = -1.050 \pm 0.447\), which gives \((-1.497, -0.603)\).

Answer

The \(95\%\) confidence interval for \(\beta\) is \((-1.497, -0.603)\).

53949812

A random sample of \(n = 16\) observations is used to estimate the population regression slope relating package weight to shipping cost. The sample slope is \(b = 1.300\), with standard error \(SE_b = 0.230\). Assume the conditions for inference about a regression slope are met. Construct a \(99\%\) confidence interval for the population slope \(\beta\). Use \(df = n - 2\) and round to three decimals.

Hints

- A confidence interval for a regression slope uses \(b \pm t^*SE_b\). - Use \(df = n - 2\) and the \(99\%\) confidence level to find \(t^*\). - Keep extra decimal places in the margin of error until computing the endpoints.

Solution

1. The degrees of freedom are \(df = 16 - 2 = 14\). 2. For a \(99\%\) confidence interval with \(df = 14\), the critical value is \(t^* \approx 2.977\). 3. The margin of error is \(t^*SE_b \approx 2.976843(0.230) \approx 0.685\). 4. The confidence interval is \(b \pm t^*SE_b = 1.300 \pm 0.685\), which gives \((0.615, 1.985)\).

Answer

The \(99\%\) confidence interval for \(\beta\) is \((0.615, 1.985)\).

53951012

A two-sided confidence interval for a population slope has sample estimate \(b = 2.40\), standard error \(SE_b = 0.28\), and lower endpoint \(1.82\). Find the critical value \(t^*\) and the upper endpoint. Round \(t^*\) to three decimals.

Hints

- Find the margin of error from the estimate and the given lower endpoint. - Use \(ME = t^*SE_b\) to solve for the critical value. - Use the symmetry of a two-sided confidence interval to find the upper endpoint.

Solution

1. The margin of error is the distance from the estimate to the lower endpoint: \(ME = 2.40 - 1.82 = 0.58\). 2. Use \(ME = t^*SE_b\): \(t^* = \frac{0.58}{0.28} \approx 2.071\). 3. A two-sided confidence interval is symmetric about \(b\), so the upper endpoint is \(2.40 + 0.58 = 2.98\).

Answer

\(t^* \approx 2.071\), and the upper endpoint is \(2.98\).

53951412

A confidence interval estimates the population slope for predicting flavor score from tea steeping time. The same data are then analyzed by predicting steeping time from flavor score. Describe what happens to the confidence interval and explain why the new interval cannot be found by simply taking reciprocals of the original endpoints.

Hints

- Identify which variable is the response before and after the interchange. - Recall the slope formulas \(r\frac{s_y}{s_x}\) and \(r\frac{s_x}{s_y}\). - A new slope and standard error require a newly calculated interval.

Solution

1. Interchanging the explanatory and response variables changes the regression question and therefore changes the population slope being estimated. 2. The original sample slope is \(b_{y\mid x} = r\frac{s_y}{s_x}\), while the reversed slope is \(b_{x\mid y} = r\frac{s_x}{s_y}\). 3. Their product is \(r^2\), not generally \(1\), so the slopes are not reciprocals unless \(|r| = 1\). 4. The standard error also changes, so the reversed regression requires a newly computed confidence interval.

Answer

The reversed analysis estimates a different population slope and requires a new confidence interval. The endpoints are not obtained by taking reciprocals because the two regression slopes are generally not reciprocals.

53951512

Regression output for predicting arrival delay, in minutes, from the number of route changes is shown below for a random sample of \(n = 18\). Assume the conditions for inference about a regression slope are met. <table> <thead><tr><th>Term</th><th>Estimate</th><th>SE Estimate</th></tr></thead> <tbody> <tr><td>Constant</td><td>5.000</td><td>0.800</td></tr> <tr><td>Route changes</td><td>1.300</td><td>0.220</td></tr> </tbody> </table> Construct a \(95\%\) confidence interval for the population slope. Use \(df = n - 2\) and round to three decimals.

Hints

- Use the explanatory-variable row, not the constant row, to identify \(b\) and \(SE_b\). - Compute \(df = n - 2\) before finding the \(t\) critical value. - Form the interval with \(b \pm t^*SE_b\).

Solution

1. From the route-changes row, \(b = 1.300\) and \(SE_b = 0.220\). 2. The degrees of freedom are \(df = 18 - 2 = 16\). For a \(95\%\) interval, \(t^* \approx 2.120\). 3. The margin of error is \(2.119905(0.220) \approx 0.466\). 4. The interval is \(1.300 \pm 0.466\), which gives \((0.834, 1.766)\).

Answer

\((0.834, 1.766)\) minutes of additional arrival delay per additional route change.

54795512

A researcher wants a confidence interval for the population slope relating typing speed to error rate. Forty people are randomly sampled, but each person completes the same task twice. The researcher treats the resulting \(80\) pairs of measurements as \(80\) independent observations in one simple linear regression. Is the usual independence condition for a slope confidence interval adequately met? Explain.

Hints

- Identify the observational unit that was randomly sampled. - Ask whether every row of the regression data comes from a different independent unit. - Repeated measurements from the same unit require special attention in inference.

Solution

1. The two observations from the same person are paired repeated measurements, so they are likely related rather than independent. 2. Treating all \(80\) measurements as independent ignores this within-person dependence. 3. Therefore, the usual simple-regression inference condition is not adequately met without a method that accounts for the repeated measurements.

Answer

No. The \(80\) observations are not independent because each sampled person contributes two related measurements.

54799312

Two independent studies estimate population regression slopes for the same response variable under two different operating conditions. Condition A has a \(95\%\) confidence interval of \((0.42, 0.88)\). Condition B has a \(95\%\) confidence interval of \((0.73, 1.19)\). A student says, “Because the intervals overlap, the two population slopes must be equal.” Evaluate the statement.

Hints

- Ask what parameter each interval estimates on its own. - Distinguish “the intervals share some plausible values” from “the parameters are equal.” - Consider what parameter would directly represent the comparison between the two conditions.

Solution

1. Each interval estimates one population slope separately, and the intervals overlap from \(0.73\) to \(0.88\). 2. Overlap of two separate confidence intervals does not establish that the population slopes are equal. 3. To make an inferential comparison between the two slopes, the analysis must directly address the difference between the population slopes using an appropriate comparison procedure.

Answer

The statement is incorrect. Overlapping \(95\%\) confidence intervals do not prove that the population slopes are equal. A direct inferential analysis of the difference between the two slopes is needed to compare them.

54799812

A random sample of \(n=10\) observations gives a sample regression slope of \(b=0.750\) with standard error \(SE_b=0.200\). Assume the conditions for inference about a regression slope are met. A student constructs a \(95\%\) confidence interval using the normal critical value \(1.96\). Explain why a \(t\) critical value should be used instead, and construct the correct interval. Use \(df=n-2\) and round the endpoints to three decimals.

Hints

- Identify the reference distribution used for inference about a regression slope when variability is estimated. - Determine the degrees of freedom from the sample size. - Combine the point estimate, critical value, and standard error symmetrically.

Solution

1. For inference about a regression slope, the standardized slope statistic follows a \(t\) distribution with \(df=n-2=8\) under the usual conditions because the error variability is estimated from the sample. 2. For \(95\%\) confidence with \(df=8\), \(t^*\approx2.306\). 3. The margin of error is \(2.306\cdot0.200\approx0.461\). 4. The confidence interval is \(0.750\pm0.461\), or approximately \((0.289, 1.211)\).

Answer

Use a \(t\) critical value with \(df=8\), not \(1.96\). The correct \(95\%\) confidence interval is approximately \((0.289, 1.211)\).

54800612

Two studies use the same sample size to estimate the slope of the same kind of linear relationship. Their residual variability is about the same. In Study A, the explanatory-variable values are tightly clustered near their mean. In Study B, the explanatory-variable values are spread across a much wider range, with no problematic outliers. Which study would generally be expected to produce the narrower confidence interval for the population slope? Explain.

Hints

- Ask which design gives more information about how the response changes across the explanatory variable. - Think about estimating a rate of change when all explanatory values are nearly the same versus widely separated. - Interval width depends on the uncertainty of the slope estimate, not only on sample size.

Solution

1. The standard error of a least-squares slope decreases as the spread of the explanatory values increases, all else being comparable. 2. Specifically, the slope standard error depends inversely on \(\sqrt{\sum (x_i-\bar x)^2}\). 3. Study B has greater explanatory-variable spread, so it provides more information about the rate of change and generally gives a smaller slope standard error. 4. With the same confidence level and comparable sample size and residual variability, the smaller standard error produces a narrower confidence interval.

Answer

Study B would generally produce the narrower slope confidence interval because its wider spread of explanatory-variable values reduces the standard error of the slope estimate, assuming the other conditions remain comparable.

54802012

Using the same regression data, an analyst reports a \(95\%\) confidence interval for the population slope of \((-0.10, 0.50)\) and an \(80\%\) confidence interval of \((0.05, 0.35)\). A student says, “These intervals contradict each other because the \(95\%\) interval includes \(0\) but the \(80\%\) interval does not.” Evaluate the statement.

Hints

- Compare the confidence levels before comparing the endpoints. - Think about how demanding more confidence affects interval width for the same data. - A wider interval can contain values that a narrower interval excludes.

Solution

1. Both intervals are centered at the same slope estimate but use different critical values. 2. The \(95\%\) interval must be wider because it is designed to capture the population slope with a higher long-run success rate. 3. A wider interval can include \(0\) even when a narrower interval from the same data does not. 4. Therefore, the two intervals are not contradictory; they reflect different confidence levels and different amounts of uncertainty included around the same estimate.

Answer

The statement is incorrect. The \(95\%\) interval is wider than the \(80\%\) interval, so it can include \(0\) even when the narrower interval excludes \(0\). The difference reflects the chosen confidence level, not inconsistent data.

54803512

An app asks users to volunteer their data for a study of daily screen time and sleep duration. More than \(10{,}000\) users opt in, and the resulting confidence interval for the regression slope is extremely narrow. A student says, “The huge sample and narrow interval make the slope estimate reliable for all app users even though participation was voluntary.” Evaluate the statement.

Hints

- Separate sampling variability from systematic selection bias. - Ask what a larger sample can reduce and what it cannot fix. - Consider whether the people who chose to participate necessarily represent the full population.

Solution

1. A large sample can reduce sampling variability and produce a narrow interval around the slope estimated from the observed volunteers. 2. Voluntary participation can create selection bias if people who opt in differ systematically from the population of all app users. 3. Increasing sample size does not remove selection bias, so a narrow interval can be precisely centered on a biased estimate. 4. The interval’s numerical precision therefore does not by itself justify generalizing the slope to all app users.

Answer

The statement is incorrect. The large sample can make the interval narrow, but voluntary-response bias can remain. Precision does not repair a nonrepresentative sampling process, so generalization to all app users is not justified from the interval alone.

54805612

A regression uses \(\ln(y)\) as the response and \(x\) as the explanatory variable. A \(95\%\) confidence interval for the population slope on the transformed scale is \((0.080, 0.120)\). A student interprets this as, “Each one-unit increase in \(x\) is associated with an increase of between \(0.080\) and \(0.120\) response units.” Correct the interpretation by converting the slope interval to multiplicative changes on the original response scale. Round factors to three decimals and percent changes to one decimal place.

Hints

- Interpret the reported interval on the scale where the regression was fitted. - Reverse the logarithm at both interval endpoints. - Convert multiplicative factors above \(1\) into percent changes only after back-transforming.

Solution

1. In a log-linear model, a slope \(\beta\) corresponds to multiplying the original response by \(e^\beta\) for a one-unit increase in \(x\). 2. Transforming the interval endpoints gives \(e^{0.080}\approx1.083\) and \(e^{0.120}\approx1.127\). 3. Thus, the plausible multiplicative change is from about \(1.083\) to \(1.127\), corresponding to increases of about \(8.3\%\) to \(12.7\%\).

Answer

The interval corresponds to multiplying the response by about \(1.083\) to \(1.127\) for each one-unit increase in \(x\), or to an increase of about \(8.3\%\) to \(12.7\%\) on the original response scale.

54806912

A software report displays a \(95\%\) confidence interval for a population regression slope as \((0.00, 0.48)\), with endpoints rounded to two decimal places. A student says, “The interval definitely contains \(0\), so a zero slope is plausible.” Evaluate the statement.

Hints

- Treat the displayed endpoint as a rounded number rather than an exact one. - Consider the range of small values that can round to \(0.00\). - Boundary conclusions require enough precision to know which side of the boundary the exact endpoint lies on.

Solution

1. The displayed lower endpoint \(0.00\) is rounded rather than exact. 2. An endpoint displayed as \(0.00\) could represent a small negative value, exactly \(0\), or a small positive value, depending on the unrounded result. 3. Therefore, the rounded interval alone does not determine whether the exact confidence interval contains \(0\). 4. More precision is needed before using the interval to make a boundary conclusion about a zero population slope.

Answer

The statement is not justified from the rounded display alone. Because the lower endpoint \(0.00\) is rounded, the exact interval may or may not contain \(0\). More precision is needed.

54810912

A data set has a visibly curved scatterplot, and its residual plot from a straight-line model shows a strong arch. Nevertheless, software reports a very narrow \(95\%\) confidence interval for the slope. Should the narrow interval be interpreted as a precise estimate of a meaningful constant linear rate of change across the observed \(x\)-range? Explain.

Hints

- Precision is meaningful only when the model being estimated is appropriate for the data. - Use the residual pattern to evaluate the straight-line form. - Separate a narrow reported interval from evidence that a constant slope is a good description.

Solution

1. The strong arch in the residual plot shows that a single straight-line form does not describe the relationship adequately across the observed range. 2. A narrow numerical interval does not repair a misspecified linear model. 3. Interpreting the slope as one meaningful constant rate of change across the range would ignore the systematic curvature. 4. The analyst should address the nonlinear form before using the usual slope interval for that interpretation.

Answer

No. The narrow interval is based on a linear model that the residual pattern contradicts, so it should not be presented as a precise constant rate of change across the range.

54813912

A slope confidence interval is \((1.4, 2.6)\), and the \(t\) critical value used to construct it was \(2.10\). Recover the slope estimate and its standard error.

Hints

- Start with the center and half-width of the interval. - The half-width combines the critical value with the standard error. - Rearrange that relationship after finding the margin of error.

Solution

1. The interval midpoint is the slope estimate: \(b=\frac{1.4+2.6}{2}=2.0\). 2. The margin of error is \(2.6-2.0=0.6\). 3. Since the margin of error equals \(t^*SE_b\), \(SE_b=\frac{0.6}{2.10}\approx0.286\).

Answer

The slope estimate is \(b=2.0\), and the standard error is approximately \(SE_b=0.286\).

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