Inference for coefficients

In depth

The conditions

The t distribution of (b₁ − β₁) ÷ SE(b₁) rests on four conditions about the errors, often remembered by Penn State’s STAT 462 course as LINE:

• Linearity: the average of y really is a straight line in x. If it curves, the slope is an average of different rates and its interval describes the wrong thing.
• Independence: each error is unrelated to the others. Violations make the standard error wrong, usually too small.
• Normality: the errors are roughly normal. With dozens of rows the slope’s t test is fairly robust to skewed errors; small samples, and prediction intervals for single cases, depend on it much more.
• Equal variance: the errors have the same spread at every x. If the spread grows with x, the usual standard error is off, and robust standard errors are the remedy.

OpenStax’s list adds a fifth: the data come from a random sample or a randomised experiment. This lesson names the conditions; assumptions-and-diagnostics checks each one with residual plots and tests. The case’s generator is multiplicative, so its spread in transactions grows with the average; over Ana’s 49 weekends the effect is too small to see (the Breusch–Pagan test that module 4 introduces gives p = 0.42), though across all 306 of Harbourside’s days, weekdays and weekends together, the same test on transactions against the high gives p = 0.006.

Where the standard error comes from

The least-squares slope is a weighted sum of the y values, $b_1 = \sum (x_i - \bar{x})\,y_i \div S_{xx}$. If the errors are independent with common variance $\sigma^2$, the variance of that sum is $\sigma^2 / S_{xx}$. The residual standard error s estimates $\sigma$ on $n - 2$ degrees of freedom, and replacing $\sigma$ by s is what turns a normal distribution into a t distribution. Both steps lean on independence: the weights only add up this way if each day’s error is unconnected with the next.

Adding a predictor that accounts for part of the error lowers s and with it every standard error. Rain is the obvious one for a waterfront café: with the day’s rain in the model as well (multiple regression, module 3), s falls from 15.24 to 12.91 transactions and the temperature slope’s standard error from 0.131 to 0.111.

The intercept

The intercept has its own standard error, $SE(b_0) = s\sqrt{1/n + \bar{x}^2/S_{xx}}$, here $15.24 \times \sqrt{1/49 + 64.35^2/13{,}529} = 8.71$. Its t is 26.10 ÷ 8.71 = 3.00 (p = 0.004) and its 95% interval runs from 8.6 to 43.6 transactions. Software prints this test for every model, but read it only when x = 0 means something. Here it is the average weekend at 0°F, colder than any day in the data (the coldest was 33°F), so the test answers a question nobody asked. The least-squares-line lesson shows the same problem in a starker form: Harbourside’s sales on transactions has an intercept of −$76.58.

Testing a value other than 0

The t statistic works for any hypothesised slope: $t = (b_1 - \beta_{1,0}) \div SE(b_1)$. The useful question is often not “is there a relationship?” but “is it the size the plan assumes?”. Ana’s budget (a fictional planning figure) assumes each extra transaction at Harbourside brings $6.40 of sales. Across the café’s 306 days, sales on transactions give a slope of $6.641 per transaction with a standard error of $0.0479, so $t = (6.641 - 6.40) \div 0.0479 = 5.03$ on 304 degrees of freedom, p below 0.001. The worked examples follow it through.

The F test and the ANOVA table

The ToolPak’s ANOVA block splits the total sum of squares as the residuals lesson did, SST = SSR + SSE, and turns the split into a test. Each sum of squares divided by its degrees of freedom is a mean square; the residual mean square is $s^2$. The F statistic is their ratio:

F tests whether all the slopes are 0 at once. With one predictor that is the same hypothesis as the t test, and F is exactly t²: 7.671² = 58.85. Significance F is therefore the same p-value as the slope’s, 7.92E-10. The F test earns its keep in multiple regression, where it tests several slopes together and the t tests no longer add up to it.

Reading the coefficient table

• Coefficients are b₀ and b₁; Standard Error is SE(b₀) and SE(b₁).
• t Stat is the coefficient divided by its standard error: the test of H₀: β = 0 only. To test another value, subtract it first.
• P-value is two-sided. E-notation such as 7.92E-10 means 7.92 × 10⁻¹⁰; report it as “below 0.001”.
• Lower 95% and Upper 95% are the coefficient ± T.INV.2T(0.05, n − 2) × its standard error. The ToolPak’s Confidence Level box adds a second pair at another level.

LINEST gives the same numbers in a different arrangement: =LINEST(C2:C50,B2:B50,TRUE,TRUE) returns the slope and intercept in its first row and their standard errors in its second, so =INDEX(LINEST(C2:C50,B2:B50,TRUE,TRUE),2,1) is SE(b₁). LINEST prints no t or p; divide and use T.DIST.2T.

Statistical and practical significance

A p-value says whether the slope is distinguishable from 0, not whether it matters. At Millbrook, the office-district café, footfall on the day’s high across all 306 days gives a slope of 0.85 people per °F, t = 2.77, p = 0.006: statistically significant. But R² is 0.025, and a day 10°F warmer goes with about 8.5 more people on an average of 226, roughly 4% more. Whether that has practical significance depends on the decision: it is too small to change a roster, whose half-hour steps cover about 3.5 peak transactions: 8.5 more people at Millbrook’s 55% conversion is about 4.7 transactions, 2.6 of them in the peak.

The reverse mistake is as common. Millbrook’s footfall on rain gives a slope of −42.9 people per inch, p = 0.074: not significant at 5%. Its interval, −90.0 to +4.2, still includes losing 90 people on a one-inch day. “Not significant” means the data can’t rule out 0, not that the effect is 0; the interval shows what they can’t rule out on either side.

When rows aren’t independent

Independence is the condition that business data break most often, and the one the output gives no warning about. Two forms appear in Ana’s log.

Days in sequence. Harbourside’s 220 weekdays (festival days left out) give a slope of footfall on the day’s high of 1.36 people per °F, with a formula standard error of 0.094 and t = 14.4. But the high follows the season, with a correlation of 0.93 between one day’s high and the next, and the residuals do too: one weekday’s residual and the next correlate at 0.23. Neighbouring days repeat each other, so the 220 days carry less independent information than 220 separate draws. Newey–West standard errors, which allow for residuals correlated over a chosen number of lags (beyond this course to derive), put the slope’s standard error at 0.121 allowing 5 lags, 0.147 allowing 10 and 0.185 allowing 20: the formula’s figure is up to about half what it should be, and its interval up to about twice too narrow. The weekdays also include the café’s build-up over its first four months, which coincided with the spring’s warming (over the first 120 days, the high and the build-up correlate at 0.79); holding the build-up fixed (multiple regression) moves the slope to 1.15. Ana’s weekend sample avoids both problems: it starts after the build-up, and its residuals in date order correlate at only 0.03. It is not the seasonal x that matters, but whether the errors depend on each other.

Leaving a pattern out can create dependence of its own. All 306 Harbourside days together give residuals that correlate at 0.30 from one day to the next and 0.43 a week apart: that is the weekly pattern of busy weekends, which a line in temperature alone leaves in the residuals. The cure there is to model the weekday pattern (categorical predictors, module 3), not only to adjust the standard error.

Rows that share a day. Pooling both cafés, footfall on rain across 612 rows gives a slope of −62.0 people per inch with a formula standard error of 18.0. But the 612 rows hold only 306 days of weather: each day’s rain is entered twice, and the formula treats the two rows of a day as independent. They aren’t: on the same day the two cafés’ residuals correlate at −0.43, because Harbourside is busy at weekends and Millbrook on weekdays, so one café’s busy day is the other’s quiet one. Here that makes the formula’s standard error too large rather than too small: allowing for the pairing by day (a day-clustered standard error, beyond this course) gives 9.68, about half. Dependence doesn’t only make standard errors too small; it makes them wrong, in a direction that depends on how the rows are related. Rows that resemble each other (two branches in the same town on a sunny day) make the formula too optimistic; rows that offset each other, as here, make it too cautious. Fit each café separately (Harbourside’s slope is −81.1, Millbrook’s −42.9): each café’s 306 rows have one row a day, and the question “what does rain cost this café?” is the one Ana can act on.