Chapter 5
Testing & interpreting estimates

• Population slope β₁ is unknown.
• Sample fitted slope β̂₁ = −2.28.
• n = 420 school districts.

step by step

1. Identify β₁ in population equation, with no hat.
2. Identify β̂₁ = −2.28 in fitted equation.
3. New random sample can change β̂₁, not the fixed β₁ merely because of sampling.

why it works

The population slope is beta one. It is the unknown number we want to learn about. Negative two point two eight is beta one hat, the estimate from this sample. The hat means estimated, not a different population.

Another sample could give a different estimate, even with four hundred twenty observations again. Different districts can produce a different fitted line. But choosing a new sample does not, by itself, change the underlying population relationship. Think of repeated measurements of one fixed target: the measurements move around, while the target stays put.

• Every hypothetical sample has n=420.
• Slopes: −1.92, −2.47, −2.16, −2.61.

step by step

1. Object distributed: β̂₁ across repeated samples.
2. Shape: approximately normal under the stated large-sample conditions.
3. Center: fixed β₁.
4. Spread: SD of β̂₁; estimated by SE(β̂₁).

why it works

Its shape is approximately bell-shaped in a sufficiently large sample, under the relevant least-squares assumptions. Its center is the fixed population slope, beta one. We do not automatically center it at negative two point two eight just because that was our observed estimate.

Its spread measures how much estimated slopes vary from sample to sample. A narrow spread means samples tend to produce similar estimates; a wider spread means more sampling uncertainty.

These four numbers illustrate the idea. Do not average them and announce that you have discovered the true slope. They are not four new verified regressions or the full sampling distribution.

• Intercept 698.93 has SE 9.47.
• Slope −2.28 has SE 0.48.
• The page refers to an Activity 5 qualification; printed p7 is not supplied.

step by step

1. Match coefficient with number directly below it.
2. Use 0.48 for slope tests and intervals.
3. Use 9.47 for intercept tests and intervals.
4. Do not substitute SER.

why it works

Now look directly below the fitted equation. Under six hundred ninety-eight point nine three is nine point four seven. That is the intercept's standard error. Under negative two point two eight is zero point four eight. That is the slope's standard error. Read each coefficient vertically with its own number in parentheses.

These are not the overall standard error of the regression, or S E R. S E R describes the scatter of outcomes around the fitted line. A coefficient standard error describes uncertainty in an estimated coefficient. They do different jobs.

The notes specify ordinary Excel, homoskedasticity-only standard errors here, and ask us to assume their conditions hold for these hand calculations. We use the provided errors rather than derive a new formula. Later, we will examine what happens when the standard-error method needs to change.

• β̂₁=−2.28; SE(β̂₁)=0.48.
• Null reference value 0; significance level α=0.05.

step by step

1. H₀: β₁=0.
2. H₁: β₁≠0.
3. Estimate −2.28; null value 0; matching SE 0.48.

why it works

The phrase differs from is our direction clue. A slope either above or below zero would count as a departure, so this is a two-sided test. Write the null as beta one equals zero and the alternative as beta one does not equal zero. These claims concern the population slope, without a hat.

Next, label the three numbers we will use: estimate negative two point two eight; hypothesized value zero; slope standard error zero point four eight. Do not use nine point four seven, because that belongs to the intercept.

The estimate happens to be negative, but that does not change the question into a one-sided test. The question's wording, not the sign we observed, determines the alternative.

• Estimate −2.28; null 0; SE 0.48.
• H₁: β₁≠0; large-sample standard-normal method.

step by step

1. t=(−2.28−0)/0.48=−4.75.
2. Distance: 4.75 estimated SEs below zero.
3. p=2P(Z>4.75)≈0.000002034.
4. p<0.05, so reject H₀.
5. Conclude evidence of negative population association, not automatic causation.

why it works

Subtract the claimed value from the estimate, then divide by the standard error. Negative two point two eight divided by zero point four eight is negative four point seven five. The estimate is four point seven five estimated standard errors below zero.

Because the alternative is two-sided, count unusual results in both tails of the standard normal curve. The p-value is about two millionths, which is far below point zero five. It is extremely small, not literally zero. Equivalently, the absolute statistic, four point seven five, exceeds the five-percent cutoff one point nine six.

Reject the null. The data provide evidence of a nonzero, negative population relationship between the student-to-teacher ratio and average test score, under the stated conditions. This does not by itself prove causation or say the relationship is large enough to matter in practice.

• β̂₁=−2.28; SE=0.48; 95% critical value 1.96.
• Other listed cutoffs: 99% 2.58; 98% 2.33; 90% 1.64; 80% 1.28.

step by step

1. Margin=1.96×0.48=0.9408.
2. CI=[−2.28−0.9408, −2.28+0.9408]=[−3.2208,−1.3392].
3. Units: test-score points per additional student per teacher.
4. Zero excluded: agrees with rejection at 5% two-sided.

why it works

First calculate the margin: one point nine six times zero point four eight equals zero point nine four zero eight. Subtract that margin from negative two point two eight to get negative three point two two zero eight. Add it to get negative one point three three nine two. Rounded to two decimals, the interval is negative three point two two to negative one point three four points per additional student per teacher.

This is a range for the population slope, not for individual districts' scores. The ninety-five-percent confidence level describes a method that captures the true coefficient in about ninety-five percent of repeated samples.

Zero is outside this interval. That agrees with rejecting the zero-slope null in the matching five-percent two-sided test. Keep the method and standard error the same when making that connection.

• Old STR 22; new STR 20; ΔX=−2.
• 95% slope CI [−3.2208,−1.3392].
• Slope estimate −2.28.

step by step

1. ΔX=20−22=−2.
2. Point estimate (−2.28)(−2)=4.56.
3. Multiply endpoints: 6.4416 and 2.6784.
4. Reorder CI=[2.6784,6.4416].

why it works

Start with new minus old. Twenty minus twenty-two is negative two. Multiply both endpoints of the slope interval by negative two. Negative three point two two zero eight becomes positive six point four four one six. Negative one point three three nine two becomes positive two point six seven eight four.

A negative multiplier reverses the order. Write the smaller number first: two point six seven eight four to six point four four one six, or about two point six eight to six point four four score points. The estimated mean change is positive four point five six points.

The positive signs mean an increase in the model's population mean prediction when the ratio decreases. This is not an interval for one particular district's future score. Calling it a causal effect would require additional justification beyond this regression.

• TestScore in points; SmallClass is 0/1.
• STR<20 maps to 1; STR≥20 maps to 0.

step by step

1. Y=TestScore; X=SmallClass.
2. TestScore_i=β₀+β₁SmallClass_i+u_i.
3. SmallClass=1{STR<20}; exactly 20 belongs to 0.
4. Excel shown: =IF(A2<20,1,0).

why it works

The outcome, Y, is Test Score, in points. The regressor, X, is Small Class, a zero-or-one indicator. The population model is Test Score equals beta zero plus beta one times Small Class, plus the error. Those population coefficients do not have hats.

Notice the boundary carefully: a district with a ratio of exactly twenty belongs to the zero group. In Excel, the displayed instruction uses IF: if the ratio is less than twenty, enter one; otherwise enter zero.

We are now comparing two groups, not estimating a point change for every one-unit increase in the original ratio. Also, Small Class is just the worksheet's label; the underlying measure is a district staffing ratio, not a direct count of pupils in one classroom.

• 420 observations; TestScore range C2:C421.
• Verified cross-source mean: 654.1565 in Chapter 4 guiding notes p5.

step by step

1. Use all 420 TestScore rows: =AVERAGE(C2:C421).
2. Read verified cross-source Ybar=654.1565.
3. Round to 654.16 points.

why it works

The overall mean is six hundred fifty-four point one five six five, or six hundred fifty-four point one six points to two decimals. That number is displayed in your Chapter Four summary-statistics table, which is shown alongside this question. It is not printed as a completed answer on this Chapter Five page.

The word all matters. We have not filtered to small-ratio districts or large-ratio districts yet. Also, do not simply average the two group means unless the groups have equal numbers of observations. The overall mean gives every district equal weight, so larger groups contribute more observations.

• Group: STR≥20, SmallClass=0.
• Fitted equation on p6: 649.98+7.37 SmallClass.

step by step

1. Filter SmallClass=0 with =AVERAGEIF(B2:B421,0,C2:C421).
2. Evaluate fitted equation at 0: 649.98+7.37(0)=649.98.

why it works

The result is six hundred forty-nine point nine eight points to two decimals. We can check it using the actual fitted equation printed on the next page: six hundred forty-nine point nine eight plus seven point three seven times Small Class.

Set Small Class to zero. The second term switches off, leaving six hundred forty-nine point nine eight. For an O L S regression with an intercept and just a binary regressor, that fitted value equals the zero group's sample mean.

This is not the average for every district, and it is not a population mean known with certainty. It is the observed sample average for the group selected by the condition in the question.

• Group: STR<20, SmallClass=1.
• Fitted intercept 649.98, dummy coefficient 7.37.

step by step

1. Filter SmallClass=1: =AVERAGEIF(B2:B421,1,C2:C421).
2. 649.98+7.37(1)=657.35.

why it works

Use the fitted equation to check the result. Substitute one for Small Class: six hundred forty-nine point nine eight plus seven point three seven times one. That equals six hundred fifty-seven point three five points.

Why do we add the intercept and slope? The intercept starts us at the zero-group mean. Switching the indicator from zero to one adds the difference between the two groups. The sum is therefore the one-group mean.

Do not call seven point three seven the small-ratio group's entire average score. It is only the amount added to the baseline. The complete predicted level, and hence this group's sample mean in this binary regression, is six hundred fifty-seven point three five.

• Group-1 mean 657.35; group-0 mean 649.98.
• Required order: 1 minus 0.

step by step

1. Difference=mean(D=1)−mean(D=0).
2. 657.35−649.98=7.37 points.

why it works

Subtract in that order. Six hundred fifty-seven point three five minus six hundred forty-nine point nine eight equals seven point three seven points. The positive result tells us that the below-twenty group has the higher sample average.

This subtraction also explains the coefficient in the binary regression. The slope is a group difference, not the overall mean and not the average for either group by itself.

The conclusion applies to averages. It does not say that every district in the one group scores seven point three seven points above every district in the zero group, and it does not automatically identify a causal effect.

• Fitted intercept 649.98.
• SmallClass=0 means STR≥20.

step by step

1. Read constant term: β̂₀=649.98.
2. D=0 makes slope term vanish.
3. Matches part (c), mean for STR≥20.

why it works

The intercept is six hundred forty-nine point nine eight. It is the constant term, the number left when Small Class equals zero. That is why it matches part C: the sample average for districts with ratios of twenty or higher.

Notice the hat in the question. We are reading an estimated sample coefficient, not claiming to know the population intercept exactly. The population intercept would be beta zero without a hat. A fitted intercept in a binary model has a straightforward reference-group interpretation, unlike some numerical-regressor intercepts evaluated far outside the data range.

• Coefficient multiplying SmallClass is 7.37.
• Group 1 minus group 0 mean difference is 7.37.

step by step

1. Read β̂₁=7.37.
2. β̂₁=mean(D=1)−mean(D=0).
3. Connect to part (e).

why it works

The coefficient is positive seven point three seven. The interpretation follows from how a binary variable works. At zero, the fitted score is the intercept. At one, we add seven point three seven. So the coefficient is the one-group sample mean minus the zero-group sample mean. That matches the answer to part E exactly, apart from the worksheet's rounding.

Here one means a ratio below twenty, so the below-twenty group averages seven point three seven score points higher in this sample. The coefficient's units are score points for the switch from group zero to group one. It is not seven point three seven percent and not a slope per extra student in the original ratio. We changed the regressor, so its meaning changed too.

• β̂₀=649.98; β̂₁=7.37.

step by step

1. β̂₀+β̂₁=649.98+7.37=657.35.
2. This is fitted value at SmallClass=1 and its sample group mean.

why it works

Their sum is six hundred fifty-seven point three five. Why ask for the sum instead of just another coefficient? Because it gives the fitted value when Small Class equals one. In a regression with an intercept and this single binary regressor, that fitted value equals the one group's sample mean. It therefore matches part D, the mean for districts with ratios below twenty.

Keep the three meanings separate on the screen: the intercept is the zero-group average; the slope is the difference between group averages; the intercept plus slope is the one-group average. Once those three labels are clear, a binary-regression equation is just a compact way of storing two group means.

• Difference estimate 7.37; matching SE 1.84.
• Two-sided zero-difference test at 5%.
• Source handwriting reports 6.01 incorrectly.

step by step

1. H₀: β₁=0, equivalently μ₁−μ₀=0; H₁: β₁≠0.
2. t=7.37/1.84=4.00543478≈4.01, NOT 6.01.
3. p≈0.0000619035.
4. Reject at 5%; positive population group-mean difference supported.

why it works

The null is that the population group means are equal. Because the binary slope already represents their difference, write beta one equals zero. The two-sided alternative is beta one not equal to zero. Do not subtract the regression intercept from the slope; that would test the wrong quantity.

Compute seven point three seven minus zero, divided by one point eight four. The result is about four point zero one. The handwritten answer visible on the source says six point zero one, but that division is an arithmetic error. The corrected result still rejects the null.

The two-sided normal p-value is about zero point zero zero zero zero six two, well below point zero five. There is evidence of a population group-mean difference, estimated positive for below-twenty districts. Statistical significance does not turn this group comparison into proof of causation.

• Ordinary Excel SE assumes constant conditional error variance.
• Robust SE permits changing variance under its conditions.
• Same data and OLS specification imply same coefficients across SE methods.

step by step

1. Constant conditional variance: homoskedasticity.
2. Changing conditional variance: heteroskedasticity.
3. Same fitted OLS coefficients; possibly different SEs and inference.
4. Use stated valid SE method; do not choose for significance.
5. Assignment preparation does not make the marked homework-only comparison an exam requirement.

why it works

The usual Excel coefficient standard errors assume constant conditional error variance. If that condition fails, those errors may give unreliable tests and confidence intervals. Robust standard errors allow the variance to change in large samples, while the other inference conditions still need to hold.

For the same data and fitted O L S model, switching standard-error methods does not change the estimated coefficients. It changes our uncertainty measurement, which can change test statistics, p-values, and intervals. The page tells us to use the specified standard error, not derive either formula by hand.

The preparation note also matters: work through Assignment Three, Questions One to Three, before Midterm Two. Question Four belongs with Chapter Six. The separate-regression slope comparison at the end of Question One is explicitly homework-only, not an exam topic. Keep that exception separate from the practice questions we are about to solve.

• Y monthly revenue, $1000 units.
• X monthly advertising expenditure, $1000 units.
• Target population slope exceeds 1.5.

step by step

1. Slope → β₁, not β₀.
2. Population → no hat.
3. Threshold 1.5; “exceeds” → right-tailed alternative.
4. Select B.

why it works

Identify three things before looking for an answer. We are testing the slope, so use beta one, not beta zero. We are testing a population claim, so do not put a hat on beta one. Finally, exceeds means greater than, so the alternative points to the right of one point five.

The worksheet's correct choice is B: the null slope equals one point five, and the alternative slope is greater than one point five. The null value is not zero just because zero is common in other questions.

Choice C incorrectly tests the observed estimate rather than the population parameter. Choice D tests the intercept. Choice E looks for departures in either direction, but this researcher specifically asks whether the slope exceeds the threshold. No test statistic is needed to select these hypotheses.

• Intercept estimate 54.6 thousand dollars.
• Intercept SE 1.3 thousand dollars.
• Null value $52,000 = 52 thousand dollars.

step by step

1. Convert $52,000 to 52 in model units.
2. t=(54.6−52)/1.3=2.6/1.3=2.00.
3. Select E.

why it works

Zero experience tells us to test the intercept. The units tell us to enter the null as fifty-two, not fifty-two thousand. Now use estimate minus claim over standard error: fifty-four point six minus fifty-two equals two point six; divide by one point three to get two point zero zero. Choice E is correct.

The numerator is a difference in thousand-dollar units, and the denominator uses the same units, so the statistic has no dollar units. Dividing fifty-four point six by one point three would test a zero intercept instead. Subtracting fifty-two thousand from fifty-four point six would mix dollars and thousands of dollars. Both mistakes answer a different, incorrectly scaled question.

• β̂₁=−0.42; SE=0.20; null 0.
• Right-tailed alternative β₁>0; α=0.05.

step by step

1. t=(−0.42−0)/0.20=−2.10.
2. Right-tail p=P(Z>−2.10)=0.9821356≈0.9821.
3. p>0.05; fail to reject.
4. Select C.

why it works

Calculate the statistic first. Negative zero point four two divided by zero point two zero equals negative two point one zero. Now stop and look at the alternative again: greater than means the right tail. We need the normal area to the right of negative two point one zero, not the small area to its left.

That right-tail area is about zero point nine eight two one. It is much larger than point zero five, so fail to reject. Choice C is correct.

The estimate lies strongly in the opposite direction from the positive alternative. A small left-tail probability is not evidence for the positive claim. Do not take an absolute value automatically in a one-sided test, and remember that a p-value can never be negative.

• Intercept $1,250; SE $100.
• 99% critical value 2.58.
• Null intercept $900; matching two-sided α=0.01.

step by step

1. Margin=2.58×100=$258.
2. 99% CI=[1250−258,1250+258]=[$992,$1508].
3. 900 outside interval → reject at 1% two-sided.
4. Select D.

why it works

For ninety-nine percent, the worksheet's critical value is two point five eight. Multiply by the standard error: two point five eight times one hundred gives a margin of two hundred fifty-eight dollars. Subtract and add that margin to twelve hundred fifty. The interval runs from nine hundred ninety-two to one thousand five hundred eight dollars.

Now compare the specific null value, nine hundred, with this interval. Nine hundred is below its lower endpoint, so it is outside. Reject the null in the matching one-percent two-sided test. Choice D has both pieces correct.

Choice B has the right interval but the wrong decision. Choice E confuses a coefficient interval with a range of individual rents. This interval concerns the population intercept, meaning the model's mean at zero regressor value, not ninety-nine percent of people's actual rent payments.

• β̂₁=2.40; SE=0.60; confidence level 98%.
• Course critical value 2.33.

step by step

1. 98% → critical value 2.33.
2. Margin=2.33×0.60=1.398.
3. CI=[2.40−1.398,2.40+1.398]=[1.002,3.798].
4. Select A.

why it works

The worksheet lists two point three three for ninety-eight percent. Multiply two point three three by zero point six zero to get the margin, one point three nine eight. Then subtract that margin from two point four zero and add it to two point four zero.

The lower endpoint is one point zero zero two, and the upper endpoint is three point seven nine eight. That is choice A.

The interval must be centered on the sample estimate, two point four, not on zero. Using one point nine six would produce the ninety-five-percent interval in a distractor, while two point five eight would produce a ninety-nine-percent interval. The method stays the same; the requested confidence level changes the multiplier.

• 95% CI [−0.90,0.30].
• Null −0.50; two-sided 5%; same method and SE.

step by step

1. Check −0.90≤−0.50≤0.30.
2. Null value lies inside matching 95% CI.
3. Fail to reject at 5%; do not conclude exact equality.
4. Select C.

why it works

Put negative zero point five on a number line between the interval endpoints. It lies inside. Therefore, fail to reject that null at five percent. Choice C is correct.

Notice that the question is not asking whether the interval includes zero. Zero happens to be inside too, but the proposed value we must evaluate is negative zero point five. The lower endpoint being negative is not itself a reason to reject anything.

We also do not need the sample size separately, because the supplied interval already gives the matching test decision. Finally, failure to reject is not proof that the slope equals negative zero point five. It means this sample does not rule out that value at the stated significance level.

• Express=1 mean31 minutes; Express=0 mean43 minutes.
• Only regressor is Express, with an intercept.

step by step

1. β̂₀=mean(Express=0)=43.
2. β̂₁=31−43=−12.
3. Time_hat=43−12Express.
4. Check D=0→43, D=1→31.
5. Select A; group-average interpretation only.

why it works

Start with the zero group. Standard orders have Express equal to zero, so the intercept is their mean: forty-three. Then calculate the slope as one-group mean minus zero-group mean: thirty-one minus forty-three equals negative twelve.

The fitted equation is Time hat equals forty-three minus twelve times Express. Check both groups. Zero gives forty-three; one gives thirty-one. Choice A is correct: express orders have a sample average delivery time twelve minutes lower.

Choice E uses the same equation but makes a stronger, incorrect claim about every individual order. The regression reproduces group means, not a fixed twelve-minute gap between every possible pair. The intercept is also not automatically the unweighted average of the two group means.

• Slope 0.48; ordinary SE0.20; robust SE0.30.
• Heteroskedasticity present; other robust conditions hold.
• Two-sided test H₀: β₁=0, α=0.05.

step by step

1. Use robust SE=0.30 under given variance conditions.
2. t=0.48/0.30=1.60.
3. p=2P(Z>1.60)=0.1095986≈0.1096.
4. Fail to reject at 5%.
5. Select B.

why it works

Use the robust standard error, zero point three zero. The statistic is zero point four eight minus zero, divided by zero point three zero, which equals one point six zero. The one-tail normal area above one point six is about zero point zero five four eight. Since the test is two-sided, double it to get zero point one zero nine six.

That exceeds point zero five, so fail to reject. Choice B is correct. Choice C forgot the second tail.

Do not choose the ordinary error simply to obtain significance. Do not replace the slope with the standard error. Robust standard errors change uncertainty measurement, not the fitted coefficient, and they are not guaranteed to be larger in every dataset.

• Original fitted equation 52−8Auto.
• Auto=1 automated; Auto=0 manual.
• Manual=1−Auto, so Auto=1−Manual.

step by step

1. Auto=1−Manual.
2. 52−8(1−Manual)=44+8Manual.
3. New intercept44: automated mean, Manual=0.
4. New slope+8: manual minus automated mean.

why it works

Since Manual equals one minus Auto, Auto also equals one minus Manual. Substitute that into the original equation: fifty-two minus eight times the quantity one minus Manual. Distribute the negative eight. We get fifty-two minus eight plus eight times Manual, which simplifies to forty-four plus eight times Manual.

The new zero group is automated processes, so the new intercept, forty-four minutes, is their sample mean. The new slope, positive eight minutes, is manual minus automated mean time. Manual processes average eight minutes longer.

Check both codings: automated is forty-four, manual is fifty-two. We changed the labels and the equation's appearance, not the actual group predictions. Do not assume the new intercept has the old intercept's standard error; that was not provided.

• Equivalent equations 52−8Auto and 44+8Manual.
• Residual=actual Minutes−fitted Minutes.

step by step

1. Automated: 52−8(1)=44+8(0)=44.
2. Manual: 52−8(0)=44+8(1)=52.
3. Actual unchanged and fitted unchanged → residual unchanged.

why it works

Take an automated observation first. Its original Auto value is one, giving fifty-two minus eight, or forty-four minutes. Its new Manual value is zero, giving forty-four plus zero, again forty-four. A manual observation similarly receives fifty-two minutes in either coding.

Thus no observation's fitted value changes. A residual equals actual outcome minus fitted outcome. The actual minutes have not changed, and we just showed that the fitted minutes have not changed, so the residual cannot change either.

This is a relabeling of the same two groups, not the introduction of new information or a better-fitting model. A sign change in the displayed slope does not necessarily mean predictions changed; here it reflects which group we chose as the baseline.

• Original slope −8; recoded slope +8.
• Both slope SEs3.20.
• Both null slopes0; two-sided tests at5%.

step by step

1. Auto t=(−8−0)/3.20=−2.50.
2. Manual t=(8−0)/3.20=+2.50.
3. Both p=2P(Z>2.5)=0.0124193≈0.0124.
4. Both reject at5%.

why it works

For Auto, calculate negative eight divided by three point two. The statistic is negative two point five. For Manual, positive eight divided by three point two gives positive two point five.

The signs differ, but both statistics are two and a half standard errors away from zero. A two-sided test counts extremeness in either direction, so both p-values are identical: approximately zero point zero one two four. That is below point zero five. Reject the zero-slope null under both codings.

Changing which group is called one reverses the estimated difference, but not the evidence that the means differ. Be careful that this argument concerns two-sided tests. For a one-sided test, reversing labels would require matching the direction of the substantive claim too.

• Slope95%CI[−16,−4] dollars per kilometer.
• Distance decreases3km, so ΔX=−3.

step by step

1. ΔX=−3km.
2. (−16)(−3)=48; (−4)(−3)=12.
3. Reorder to [12,48] dollars.
4. Interpret positive change in model’s population mean monthly fee.

why it works

A decrease of three means delta X is negative three. Multiply both slope endpoints by negative three. Negative sixteen times negative three is positive forty-eight. Negative four times negative three is positive twelve.

Because the multiplier is negative, the endpoint order reverses. Write the final interval as twelve to forty-eight dollars, not forty-eight to twelve. Kilometers cancel, so the answer is a change in the monthly fee measured in dollars.

The model associates a three-kilometer decrease with a population mean fee increase in that range. The positive result is not a mistake: both the slope and distance change are negative. Follow the stated model rather than changing the sign to match an intuition about what moving closer ought to cost.

• Interval[12,48] dollars concerns β₁ΔX, not a fee level.
• Outcome monthly fee; regressor distance.

step by step

1. β₁ΔX is a mean change, not final outcome level.
2. Individual future outcomes include unexplained variation.
3. Coefficient CI is not an individual prediction interval.
4. Causal moving interpretation requires justification beyond association.

why it works

There are two distinctions. First, our interval describes a change, not the final fee level. Twelve to forty-eight dollars is the estimated range of increase in the population regression mean, not the entire future bill.

Second, the interval concerns that mean relationship, not one facility's actual outcome. An individual facility can differ from the line because of other influences. Predicting its future bill would require an appropriate prediction calculation, including individual uncertainty, and enough information about the level.

There is also a causality caution. A regression association between distance and fees does not, by itself, prove what moving one facility will cause. Keep the model's average association separate from an individual forecast or a causal promise.

• Estimate −2.28; SE0.48; null −3.
• “Differs from” gives two-sided alternative; α=.05.

step by step

1. H₀:β₁=−3; H₁:β₁≠−3.
2. t=[−2.28−(−3)]/.48=.72/.48=1.50.
3. p=2P(Z>1.50)=.1336144≈.1336.
4. Fail to reject at5%.

why it works

The null is beta one equals negative three. The alternative is beta one not equal to negative three. The numerator is estimate minus null: negative two point two eight minus negative three. Subtracting a negative gives positive zero point seven two.

Divide zero point seven two by zero point four eight to get one point five zero. For a two-sided standard-normal test, the p-value is approximately zero point one three three six. Since that exceeds point zero five, fail to reject.

The same slope estimate strongly rejected zero earlier, but does not reject negative three. That is sensible: negative three is much closer to the observed estimate than zero is, relative to the same uncertainty. A test result depends on the proposed value, not only on whether the estimated coefficient is negative.

• 95% slopeCI[−3.2208,−1.3392].
• Null−3; test in part(a) failed to reject.

step by step

1. −3 lies in [−3.2208,−1.3392].
2. Matching test therefore fails to reject.
3. Nonrejection does not identify −3 as uniquely true.

why it works

Place negative three between the endpoints. It is inside the interval, so the matching five-percent two-sided test fails to reject. Yes, the interval agrees with the calculation in part A.

But no, that does not prove exact equality. The data leave a range of population slopes compatible with this method. Negative three is one of those values, not the only possible value.

Say, we do not have sufficient evidence to reject negative three at five percent. Do not say, we proved the slope is negative three. A failure to detect a difference and a demonstration of exact equality are different claims.

• Repeated large same-size samples; same population and model.
• Relevant least-squares assumptions hold.

step by step

1. Estimator β̂₁ varies; population β₁ fixed.
2. Large-sample distribution centered at β₁ under assumptions.
3. SE estimates estimator-distribution SD.
4. Select D; normal raw X/Y not required.

why it works

The estimated slopes vary from sample to sample. Their large-sample distribution is centered at the fixed population slope, and a standard error estimates that distribution's standard deviation. In plain language, different samples give slightly different measurements of one underlying relationship, and the standard error estimates how much those measurements tend to vary.

Choice A incorrectly moves the population parameter around. Choice B incorrectly assumes equal sample size means identical data or estimates. Choice C incorrectly centers the distribution at our first sample's estimate instead of the population coefficient. Choice E adds a condition we do not need: the observed X and Y do not themselves have to be normally distributed for the estimator's stated large-sample approximation.

This brings us back to Activity One: keep the fixed population target, the changing sample estimate, and the estimate's uncertainty separate.

The correct choice is D.