Test for a slope
A sample of plots shows a fitted line that rises. Is the rise real across the whole coast, or could a flat population line have produced a sample slope this steep by chance?
Where you are 4 in Inference for slopes
When you reach for it
Somchai restores mangroves on the coast of Thailand. The project has 320 planted plots, and he takes a random sample of 16 (made-up data). For each plot he records x, the planted seedling density in seedlings per square meter, and y, the number of crab burrows per square meter, a sign that the mud is coming back to life. His program prints:
In the sample, denser plots have more burrows: 0.844 more per square meter for each extra seedling per square meter (Figure 1). But sixteen plots scatter widely, and even if burrows had nothing to do with density across all 320 plots, a sample would rarely give a slope of exactly 0. The question is the one One sample test for a mean asked of a mean, now asked of a slope: is the sample result too far from the claimed value to blame on sampling alone? Interval for a slope gave the plausible slopes as b \pm t^*\,SE_b; the test weighs one claimed slope, almost always 0.
The procedure
Day 1: the test from computer output
The parameter and the hypotheses. As in Sampling distribution of the slope, the model is a population line \mu_y = \alpha + \beta x giving the mean of y at each x. Here \beta is the change in mean burrows per square meter for each additional seedling per square meter, among all 320 plots. A slope of 0 means the mean number of burrows is the same at every density: no linear relationship. So the usual test is
or \beta > 0, or \beta < 0, if a direction was named before the data were seen.
The conditions. The same five as for the interval, from Conditions for regression inference: a linear trend, independent observations (a sample under a tenth of the population when sampling without replacement), normal residuals, equal spread of residuals, and a random sample or random assignment. They are checked with the scatter plot, the residual plot, and a plot of the residuals.
The statistic. If H_0 is true, b varies around 0 with estimated spread SE_b, and
follows a t distribution. It counts how many standard errors the fitted slope lies from 0. The degrees of freedom are those of s, the residual standard deviation, which loses two for the two fitted numbers a and b.
The p-value, decision and conclusion. The p-value (p-values and the significance level) is the chance, if \beta = 0, of a t at least as far out as the observed one in the direction H_a names. If it is at most \alpha, reject H_0; otherwise fail to reject (Deciding and concluding from a p-value). The conclusion names \beta in context.
Reading the printout for inference. Reading regression output read the line from the Coef column and the fit from S and R-Sq. For a test, the row that matters is the explanatory variable’s row:
| Slope row | Is | In the test |
|---|---|---|
| Coef | b | numerator |
| SE Coef | SE_b | denominator |
| T | b/SE_b | t for \beta_0 = 0 |
| P | two-sided p | halve if one-sided |
The Constant row holds a and a test that the population intercept is 0; the slope test does not use it. S is s, the typical size of a residual, which enters only through SE_b. R-Sq and R-Sq(adj) describe the fit and are not used.
The printed P assumes H_a\colon \beta \neq 0, because the program cannot know which direction you named in advance. For a one-sided alternative, the p-value is half the printed P when b falls on the side H_a names, and 1 minus half of it when b falls on the other side.
Day 2: the standard error, other null values, and the interval
Where SE_b comes from. With raw data instead of a printout, fit the line, compute the residuals, and find \displaystyle s = \sqrt{\text{SSE}/(n - 2)} and s_x. Then
The two forms agree because s_x^2 (n - 1) = \sum (x - \bar{x})^2. More scatter about the line makes the slope less certain; a wider spread of x makes it more certain. A graphing calculator’s LinRegTTest prints t and the p-value for the alternative you select, with \text{df}, a, b, s, r^2 and r, but not SE_b; since t = b/SE_b, the standard error is b/t.
A null slope other than 0. To test whether the slope equals some claimed rate \beta_0,
The printout’s T and P are for \beta_0 = 0 only, so here they do not apply: compute t from Coef and SE Coef.
Agreement with the interval. The test and the interval use the same b, SE_b and t curve. A two-sided test at level \alpha rejects \beta = \beta_0 exactly when \beta_0 lies outside the 1 - \alpha confidence interval.
What rejecting says. Rejecting \beta = 0 is convincing evidence of a linear association between x and y in the population. It does not say the association is strong (look at R-Sq and S), or large (look at the interval), or that x causes y. Cause needs the x values to have been assigned at random to the units; in an observational study a lurking variable may drive both (Correlation is not causation: lurking and confounding variables). Failing to reject does not show \beta = 0.
Worked examples
Day 1: Somchai’s test, start to finish
Use Somchai’s printout to test, at \alpha = 0.05, whether there is a linear relationship between seedling density and crab burrows across the 320 plots. His scatter plot shows a straight band, the residual plot has no curve and no fan, and a dot plot of the residuals is roughly symmetric with no outliers.
SolutionParameter and hypotheses. \beta is the change in mean crab burrows per square meter for each additional seedling per square meter, among all 320 plots. “Any linear relationship” has no direction, so H_0\colon \beta = 0 and H_a\colon \beta \neq 0.
Conditions. Linear: the scatter and residual plots show no curve. Independent: 16 plots sampled without replacement is under a tenth of 320. Normal: the residuals’ dot plot has no strong skew or outliers. Equal spread: no fan. Random: a random sample.
Statistic. From the Density row, b = 0.844 and SE_b = 0.311, so
This is the T entry: the fitted slope lies about 2.7 standard errors above 0.
P-value. Both tails beyond \pm 2.71 under the t curve with 14 degrees of freedom (Figure 2): `2*tcdf(2.71, 10^99, 14)` gives about 0.017, the printed P.
Decision and conclusion. Since 0.017 \le 0.05, reject H_0. If burrow numbers had no linear relationship with density across the 320 plots, a sample slope this far from 0 would turn up in only about 1.7\% of samples. There is convincing evidence of a linear relationship between planted seedling density and crab burrows per square meter in the project’s plots; the sample slope says it is positive.
Day 1: every number in the printout, and a one-sided test
Before sampling, Somchai’s colleague Anong predicted that denser planting goes with more burrows. (a) Say what each entry of the printout tells her and which ones her test needs. (b) Carry out her test of H_a\colon \beta > 0 at \alpha = 0.05.
Solution(a) The Density row is the one that matters. Coef 0.844 is b; SE Coef 0.311 is SE_b; T 2.71 is 0.844/0.311; P 0.017 is the two-sided p-value for \beta = 0. The Constant row gives a = 2.777 burrows per square meter, the predicted count at density 0, and its T and P test whether the population intercept is 0. A plot with no seedlings lies outside the densities sampled, 0.6 to 4.0, so that test answers no question anyone asked. \text{S} = 1.225 says plots typically sit about 1.2 burrows per square meter off the line; it enters SE_b but not the test directly. R-Sq = 34.5\% says the line accounts for about a third of the variation in burrows. It is tied to the test, not separate from it: \dfrac{t^2}{t^2 + \text{df}} = \dfrac{2.713^2}{2.713^2 + 14} \approx 0.345.
(b) H_0\colon \beta = 0, H_a\colon \beta > 0, conditions as in Example 1. The statistic is the same, t \approx 2.71 with \text{df} = 14, but the p-value is now one tail only. The sample slope is positive, on the side H_a names, so the p-value is half the printed one: about 0.017/2 \approx 0.0084. Since 0.0084 \le 0.05, reject H_0: convincing evidence that mean burrows rise with seedling density across the plots. Halving is honest only because Anong named the direction before seeing the data.
Day 1: a test that does not reject
Anong suspects saltier soil holds fewer crabs. Before sampling she chooses H_a\colon \beta < 0 and \alpha = 0.05, then measures a random sample of 12 of the 320 plots, with x the soil salinity in parts per thousand (18 to 34) and y the crab burrows per square meter (made-up data). The conditions are met.
\beta is the change in mean burrows per square meter for each additional part per thousand of salinity, among all 320 plots. H_0\colon \beta = 0, H_a\colon \beta < 0.
The sample slope is negative, the side H_a names, so the p-value is the lower tail alone, half the printed P: about 0.148/2 = 0.074. Since 0.074 > 0.05, fail to reject H_0. The data do not give convincing evidence that mean burrows fall as salinity rises. They do not show salinity is unrelated either: twelve plots with this much scatter leave both a zero slope and a sizable negative one plausible.
Had Anong’s alternative been \beta > 0, the sample slope would lie on the wrong side, and the p-value would be the upper tail, 1 - 0.074 = 0.926.
Day 2: the standard error from raw data
Niran runs a planting trial. Eight new plots are given densities from 0.5 to 4.0 seedlings per square meter, each density assigned to a plot at random, and two years later he counts crab burrows (made-up data). He decided beforehand to test H_a\colon \beta > 0 at \alpha = 0.05. The product column is (x - \bar{x})(y - \bar{y}); the conditions hold.
| x | y | product | residual |
|---|---|---|---|
| 0.5 | 3.9 | 2.49375 | 0.892 |
| 1.0 | 3.1 | 2.78125 | -0.570 |
| 1.5 | 5.0 | 0.24375 | 0.668 |
| 2.0 | 4.4 | 0.23125 | -0.594 |
| 2.5 | 5.1 | -0.05625 | -0.556 |
| 3.0 | 5.4 | 0.05625 | -0.918 |
| 3.5 | 6.7 | 1.71875 | -0.280 |
| 4.0 | 9.0 | 6.43125 | 1.358 |
Means. \bar{x} = 18/8 = 2.25 and \bar{y} = 42.6/8 = 5.325.
Slope and intercept. The products add to 13.9. The deviations x - \bar{x} run from -1.75 to 1.75 in steps of 0.5, and their squares add to 2(3.0625 + 1.5625 + 0.5625 + 0.0625) = 10.5. So
s. The residuals in the last column square and add to \text{SSE} \approx 4.994, so \displaystyle s = \sqrt{4.994/6} \approx 0.9123 burrows per square meter, dividing by n - 2 = 6.
Standard error. \displaystyle s_x = \sqrt{10.5/7} \approx 1.2247, and
Test. t = \dfrac{1.3238}{0.2816} \approx 4.70 with \text{df} = 6. The slope is positive, so the p-value is the upper tail, `tcdf(4.70, 10^99, 6)`, about 0.0017. Reject H_0: there is convincing evidence that mean burrows rise with planted density.
Day 2: the calculator’s screen
Niran enters the same eight plots and runs LinRegTTest with the alternative \beta > 0. The screen shows
Match it with the hand computation and recover SE_b.
SolutionThe second line repeats the chosen alternative; the calculator states it for the population correlation \rho too, since a zero slope and a zero correlation are the same hypothesis. Here p is the upper tail, the one-sided p-value asked for, not a two-sided P as on a printout: 0.00166, matching 0.0017. The values t, \text{df}, a, b and s match the hand results. The standard error is
as found by hand. The trial’s line accounts for about 78.7\% of the variation in burrows among the eight plots.
Day 2: a slope other than zero, and the interval
A restoration report from another coast says each extra seedling per square meter adds 1.5 burrows per square meter. Somchai asks whether his plots differ from that rate. Test H_0\colon \beta = 1.5 against H_a\colon \beta \neq 1.5 at \alpha = 0.05, then compare with a 95\% interval for \beta.
SolutionThe printed T and P are for \beta_0 = 0 and do not apply. From Coef and SE Coef,
Both tails beyond \pm 2.11 give a p-value of about 0.053. Since 0.053 > 0.05, fail to reject: the data do not give convincing evidence that the slope in Somchai’s plots differs from 1.5 burrows per seedling.
At \text{df} = 14, t^* = 2.145, and 2.145 \times 0.311 = 0.667095, so the 95\% interval is
The claimed 1.5 lies inside, barely, which matches a p-value barely above 0.05. And 0 lies outside, which matches Example 1, where the two-sided test rejected \beta = 0 with p \approx 0.017.
Day 2: what each rejection licenses
Somchai’s test (Example 1) and Niran’s trial (Example 4) both rejected \beta = 0. In Somchai’s project, planting crews chose each plot’s density, putting more seedlings where the mud was firm enough to hold them. What may each conclude, and what does Somchai’s interval say about an increase of 2 seedlings per square meter?
SolutionSomchai has a random sample, so his conclusion reaches all 320 plots, but the densities were not assigned. Firmer mud may itself suit crabs, so density and burrows could rise together with neither causing the other. He may say denser plots have more burrows on average; not that planting more seedlings brings more crabs. Niran assigned densities at random, which spreads mud type and every other plot difference evenly across densities in the long run, so his test is evidence that density itself raises burrow counts, though only for plots like his eight and the densities he tried.
Neither rejection says the relationship is strong: Somchai’s R-Sq is 34.5\%, and plots scatter about 1.2 burrows from his line. Nor does it say the slope is large. Doubling the interval from Example 6, 2 more seedlings per square meter go with between 2(0.177) \approx 0.35 and 2(1.511) \approx 3.02 more burrows per square meter on average, a wide range.
Boundaries
- The p-value is honest only under the five conditions. If the trend curves, a test of \beta = 0 can miss a strong curved relationship, since \beta measures straight-line trend only.
- The printed P is two-sided. A one-sided p-value is half of it only when b lies on the side H_a names, and only when that side was chosen before the data were seen.
- The printed T and P test \beta_0 = 0. Any other null slope needs t = (b - \beta_0)/SE_b.
- The Constant row’s test concerns the intercept, the mean of y at x = 0, which is often outside the data and of no interest.
- Rejecting shows a linear association, not cause, unless the x values were randomly assigned. Failing to reject does not show \beta = 0.
- \text{df} = n - 2 is for one explanatory variable.
Where this goes next
With the test and Interval for a slope together, the questions of Inference for slopes are answered for one explanatory variable: whether there is a linear relationship, and how steep it plausibly is. The same Coef, SE Coef, T and P columns appear row by row when a model has several explanatory variables, each T testing one slope with the others held in the model.
Exercises
Anong records, for a random sample of 20 of the 320 plots, the years since planting, 1 to 10, and the crab burrows per square meter (made-up data). Output A:
Niran records, for a random sample of 14 plots, the distance from the plot to the nearest tidal channel, 5 to 140 meters, and the crab burrows per square meter (made-up data). Output C:
Assume the conditions hold unless an exercise says otherwise. Find p-values with technology.
Day 1: tests from output
- For Output A, define \beta in context, state the hypotheses for “is there any linear relationship”, and give the degrees of freedom.
- Carry out the test of Exercise 1 at \alpha = 0.05 and conclude in context.
- Recompute T for Output A from Coef and SE Coef, and the two-sided p-value from your T.
- Anong had predicted, before sampling, that older plots have more burrows. State her hypotheses and give the p-value for her test.
- Niran predicted, before sampling, fewer burrows farther from a channel, and chose \alpha = 0.01. Carry out his test from Output C.
- Had Niran’s alternative been \beta > 0, what would the p-value have been?
- Test H_0\colon \beta = 0 against H_a\colon \beta \neq 0 from Output C at \alpha = 0.05.
- A printout from 22 plots gives the slope row Coef 0.415, SE Coef 0.180. Find T and the two-sided p-value.
- On a smudged printout from 15 plots, the slope row shows Coef -0.26 and T -1.85. Find SE Coef and P.
- Find the correlation r between years and burrows from Output A, and say what R-Sq says.
Day 2: standard error, calculator, other null values
- Anong measures 6 plots flooded 2, 3, 4, 5, 6 and 7 hours a day, with 9.2, 7.1, 8.4, 6.6, 5.1 and 5.3 burrows per square meter (made-up data). Find b and a.
- For the plots in Exercise 11 the squared residuals add to 2.808. Find s, s_x and SE_b.
- Test H_0\colon \beta = 0 against H_a\colon \beta \neq 0 at \alpha = 0.05 for Exercises 11 and 12.
- Construct a 95\% interval for \beta from Exercises 11 and 12 and check that it agrees with Exercise 13.
- Niran runs LinRegTTest with \beta \neq 0 on 10 plots at another site. The screen shows t = 2.24642, p = 0.05487, \text{df} = 8, a = 3.76145, b = 0.59639, s = 0.86533, r^2 = 0.38681, r = 0.62194. Recover SE_b and decide at \alpha = 0.05.
- For the screen in Exercise 15, what p-value would \beta > 0 have given, and when would using it be allowed?
- A planting guide says burrows rise by 1 per square meter per year after planting. Test H_0\colon \beta = 1 against H_a\colon \beta < 1 from Output A at \alpha = 0.05.
- Construct a 95\% interval for \beta from Output C. Which of \beta_0 = 0, -0.01 and -0.04 would a two-sided test at \alpha = 0.05 reject? Confirm the answer for -0.04 with a p-value.
Reasoning
- Explain why the P on a printout is two-sided, and why halving it is wrong when b falls on the side opposite to H_a.
- Output A rejects \beta = 0. Does it show that the passing years cause more burrows? Explain, and say what kind of study could.
- A friend says Example 3 shows salinity has no effect on crabs. Explain what it shows instead.
- A study of 400 plots on another coast prints P = 0.000 for the slope and R-Sq = 3\%. Explain what the test does and does not tell you.
- Explain why the T column cannot be used to test H_0\colon \beta = 1.
Answers
- \beta is the change in mean crab burrows per square meter for each additional year since planting, among all 320 plots. H_0\colon \beta = 0, H_a\colon \beta \neq 0; \text{df} = 20 - 2 = 18.
- t = 3.47, \text{df} = 18, P = 0.003 \le 0.05: reject H_0. Convincing evidence of a linear relationship between years since planting and mean burrows per square meter in the plots; the positive slope says older plots have more.
- 0.568/0.164 \approx 3.46, matching the printed 3.47 up to the rounding of Coef and SE Coef. With \text{df} = 18, both tails beyond 3.46 give about 0.0028.
- H_0\colon \beta = 0, H_a\colon \beta > 0. b is positive, so the p-value is half the two-sided one: 0.003/2 \approx 0.0015 (about 0.0014 before the printout rounds P).
- H_0\colon \beta = 0, H_a\colon \beta < 0. b < 0, so the p-value is half of P: 0.029/2 \approx 0.0145 (about 0.0143 before rounding). Since 0.0143 > 0.01, fail to reject: at the 0.01 level there is not convincing evidence that mean burrows fall with distance from a channel.
- b is on the wrong side, so the p-value is the upper tail, about 1 - 0.0143 = 0.986.
- P = 0.029 \le 0.05: reject. Convincing evidence of a linear relationship between distance to a channel and mean burrows, falling with distance.
- t = 0.415/0.180 \approx 2.31, \text{df} = 20, two-sided p-value about 0.032.
- SE Coef = 0.26/1.85 \approx 0.141; with \text{df} = 13, P is about 0.087.
- \displaystyle r = +\sqrt{0.400} \approx 0.632, positive because the slope is. The line accounts for 40.0\% of the variation in burrows among the 20 plots.
- \bar{x} = 4.5, \bar{y} = 6.95; \sum (x - \bar{x})^2 = 17.5 and \sum (x - \bar{x})(y - \bar{y}) = -13.65, so b = -13.65/17.5 = -0.78 burrows per hour and a = 6.95 + 0.78(4.5) = 10.46.
- \displaystyle s = \sqrt{2.808/4} \approx 0.8379; \displaystyle s_x = \sqrt{17.5/5} \approx 1.8708; \displaystyle SE_b = \dfrac{0.8379}{1.8708\sqrt{5}} \approx 0.2003.
- t = -0.78/0.2003 \approx -3.89, \text{df} = 4, two-sided p-value about 0.018. Reject: convincing evidence of a linear relationship, mean burrows falling with hours flooded.
- t^* = 2.776 at \text{df} = 4; margin 2.776 \times 0.2003 \approx 0.556; from -1.336 to -0.224 burrows per hour. It leaves out 0, as a two-sided rejection at 0.05 requires.
- SE_b = b/t = 0.59639/2.24642 \approx 0.2655. Since p = 0.05487 > 0.05, fail to reject: no convincing evidence of a linear relationship at that site.
- About 0.0274, half of 0.05487, since b > 0. It may be used only if \beta > 0 was chosen before the data were seen; choosing it afterward to get under 0.05 halves the p-value with no new evidence.
- t = (0.568 - 1)/0.164 \approx -2.63, \text{df} = 18, lower-tail p-value about 0.0084. Reject: convincing evidence that mean burrows rise by less than 1 per square meter per year.
- t^* = 2.179 at \text{df} = 12; margin 2.179 \times 0.0082 \approx 0.0179; from -0.0382 to -0.0024. It rejects 0 and -0.04, which lie outside, but not -0.01. For -0.04: t = (-0.0203 + 0.04)/0.0082 \approx 2.40, two-sided p-value about 0.033 < 0.05.
- The program does not know which alternative was chosen, so it reports the one that covers both directions, both tails beyond |t|. A one-sided p-value is the tail on the side H_a names. When b falls on the other side, that tail holds the bulk of the curve, 1 minus half of P, never half of P.
- No. The plots were sampled, not assigned ages: older plots were planted earlier, perhaps on different ground or by other methods, and any such difference could explain the burrows. The output shows a linear association. Showing cause needs a study in which the explanatory variable is assigned at random, which cannot be done for age; a randomized trial of some treatment could.
- Failing to reject means the data are consistent with \beta = 0, not that \beta = 0. With 12 plots and this scatter, a real negative slope of moderate size could easily produce such a sample. More plots could settle it.
- The test says there is convincing evidence of some linear association; with 400 plots, SE_b is small, so even a slight slope stands out. R-Sq of 3\% says x accounts for very little of the variation, and the interval would show how small the slope is. A small p-value measures evidence, not size.
- Each T entry is Coef divided by SE Coef, the distance of b from 0 in standard errors. Testing \beta = 1 needs the distance from 1, (b - 1)/SE_b, and the printed P belongs to the printed T.