Interval Estimation, Hypothesis Testing and ANOVA

How the exam asks it: four core exam templates

Midterm exams on interval estimation and hypothesis testing consistently feature four question types: building confidence intervals, conducting t-tests of significance, applying the confidence interval decision rule, and constructing ANOVA tables for F-tests.

Exam template 1: Constructing a 95% confidence interval for beta2 and beta1

Question: Using the consumption regression output ($n = 10, \alpha = 0.05, df = 8$), $b_2 = 0.509$ ($SE = 0.035$) and $b_1 = 24.45$ ($SE = 6.41$). From the t table, $t_{0.025, 8} = 2.306$. a) Compute the margin of error and the 95% confidence interval for the true slope $\beta_2$.
b) Compute the margin of error and the 95% confidence interval for the true intercept $\beta_1$.
c) State the precise textbook statistical interpretation of the 95% confidence interval for $\beta_2$.

Exam template 2: Two-sided hypothesis test (Test of significance approach)

Question: For the same model ($b_2 = 0.509091, SE(b_2) = 0.035743, n = 10, \alpha = 0.05, t_{0.025, 8} = 2.306$): a) State the null and alternative hypotheses to test whether weekly income has a statistically significant effect on consumption ($H_0: \beta_2 = 0$).
b) Calculate the sample t-statistic and state the test decision.
c) Test someone’s claim that the true MPC is 0.30 ($H_0: \beta_2 = 0.30$ vs $H_1: \beta_2 \neq 0.30$) at the 5% level.

Exam template 3: Confidence interval approach to hypothesis testing

Question: The board’s 95% confidence interval for the marginal propensity to consume $\beta_2$ is $[0.428, 0.590]$ (or rounded as $[0.42, 0.58]$). a) State the decision rule for testing $H_0: \beta_2 = c$ against $H_1: \beta_2 \neq c$ using the confidence interval approach.
b) Test $H_0: \beta_2 = 0.60$ against $H_1: \beta_2 \neq 0.60$ at $\alpha = 0.05$.
c) Test $H_0: \beta_2 = 0.55$ against $H_1: \beta_2 \neq 0.55$ at $\alpha = 0.05$.

Exam template 4: ANOVA decomposition and F-test

Question: An EViews regression output for $n = 10$ reports $RSS = 337.2727$ and $R^2 = 0.962062$. The 5% critical value of the F-distribution is $F_{0.05; 1, 8} = 5.32$. a) Compute $TSS$ and $ESS$.
b) Build the ANOVA table and compute the sample F-statistic $F = \dfrac{ESS/1}{RSS/(n-2)}$.
c) Test $H_0: \beta_2 = 0$ using the F-statistic and verify that $F = t^2$.

Cheat-sheet & the 4-step method

Why one number is not enough

The EViews regression of weekly consumption on weekly income (Gujarati Table 2.4, $n = 10$) gave the slope $b_2 = 0.5091$. That is a point estimate: our single best guess of the true marginal propensity to consume $\beta_2$.

But $b_2$ comes from one sample of ten families. A different sample would give a different number, maybe 0.47, maybe 0.55. The point estimate alone tells us nothing about how close it is likely to be to $\beta_2$.

So instead of relying on the point estimate alone, we construct an interval around it that has a known probability, say 95 percent, of containing the true parameter value.

The confidence interval idea

• $1-\alpha$ is the confidence coefficient. With $\alpha = 0.05$ it is 0.95, a 95% interval.
• $\alpha$ is the level of significance: the probability that the procedure produces an interval that misses $\beta_2$.
• The probability of constructing an interval that contains $\beta_2$ is $1-\alpha$.

Building the interval with the t distribution

From the previous lessons, under the classical assumptions (with normally distributed disturbances) the standardised estimate follows a t distribution with $n-2$ degrees of freedom:

The t table gives a critical value $t_{\alpha/2}$ that leaves $\alpha/2$ in each tail. Then:

Worked example: the EViews consumption function

The output from the OLS lesson:

The question from the board: if income increases by one unit, in which interval will consumption increase? That is a 95% confidence interval for $\beta_2$.

The second board question: when income is equal to zero, in which range will average consumption be? That is a 95% confidence interval for the intercept $\beta_1$.

Hypothesis testing: the setup

A hypothesis test asks whether a specific claimed value of $\beta_2$ is compatible with the data.

• Null hypothesis $H_0$: the claim being tested, written with an equals sign: $H_0: \beta_2 = c$.
• Alternative hypothesis $H_1$: what we conclude if $H_0$ is rejected. Its form decides the type of test:

The most common case is $c = 0$: “X has no effect on Y”. Rejecting $H_0: \beta_2 = 0$ means the coefficient is statistically significant.

Approach 1: the test of significance (t test)

Decision rules:

• Two-sided: reject $H_0$ if $|t| \gt t_{\alpha/2}$.
• One-sided, $H_1: \beta_2 \gt c$: reject $H_0$ if $t \gt t_\alpha$.
• One-sided, $H_1: \beta_2 \lt c$: reject $H_0$ if $t \lt -t_\alpha$.

Approach 2: the confidence-interval approach

The interval and the two-sided test are two views of the same calculation:

With the 95% interval $[0.4267, 0.5915]$:

• $c = 0$ is outside → reject: income is significant.
• $c = 0.3$ is outside → reject, the same answer as the t test ($t = 5.85$).
• $c = 0.5$ is inside → do not reject, the same answer as the t test ($t = 0.25$).

Try it: move the null, watch the decision

Slide $c$ and watch both views. In the two-sided test, the computed t enters the red rejection region exactly when $c$ leaves the confidence interval. Switch to a one-sided alternative to see the rejection region move into a single tail.

The one-sided test: Keynes’s claim

Keynes argued that the MPC is less than 1: people spend part, not all, of an extra unit of income. That is a one-sided claim.

p-values: what Prob. means in EViews

The p-value is the probability, if $H_0$ were true, of getting a t statistic at least as extreme as the one observed. It is the smallest significance level at which $H_0$ would be rejected.

• Rule: reject $H_0$ when p-value $\lt \alpha$.
• EViews’ Prob. column is the two-sided p-value for $H_0$: coefficient $= 0$.
• In the consumption output: X has Prob. $= 0.0000$, so income is significant at any usual level. C has Prob. $= 0.0051$, so the intercept is significant at 5% and at 1%.

Type I and Type II errors

Any test can go wrong in two ways:

• $\alpha$, the level of significance, is the probability of a Type I error, chosen in advance.
• Lowering $\alpha$ (say from 5% to 1%) makes Type I errors rarer but Type II errors more likely: the two trade off for a given sample size.

ANOVA: splitting the total variation

Analysis of variance (ANOVA) looks at the regression as a whole: how much of the movement of $Y$ around its own mean does the line explain?

• TSS (total sum of squares) $= \sum(Y_i - \bar Y)^2$: the total change around $Y$‘s own mean.
• ESS (explained sum of squares) $= \sum(\hat Y_i - \bar Y)^2$: the part of that change explained by the regression.
• RSS (residual sum of squares) $= \sum e_i^2$: the part that cannot be explained by the regression.

Decision rule: reject $H_0: \beta_2 = 0$ if $F$ exceeds the critical value from the F table, or if Prob(F-statistic) $\lt 0.05$. Here $F_{0.05;\,1,8} = 5.32$ and $F = 202.87 \gt 5.32$ (EViews: Prob(F-statistic) $= 0.000001$), so we reject: income explains consumption.

Switch the lab below to TSS = ESS + RSS and drag the last family’s consumption: a point far from the line inflates RSS, shrinks ESS’s share, and pulls $R^2$ and F down.

Normality: histogram and Jarque–Bera

Classical normal linear regression assumes the disturbance term is normally distributed: $u_i \sim N(0, \sigma^2)$. Why does this matter? The OLS estimators $b_1$ and $b_2$ are linear combinations of the disturbances. If $u_i$ is normal, then $b_1$ and $b_2$ are strictly normally distributed in any sample size, which makes the $t$ and $F$ tests exact. If $u_i$ is not normal, $t$ and $F$ tests are only large-sample approximations.

Histogram of residuals

A histogram of residuals is a simple graphical check of the shape of the probability density function (PDF). We divide the range of OLS residuals $e_i$ into equal intervals and plot their frequencies. Mentally superimposing a normal bell curve on the histogram gives an immediate visual check for:

• Skewness (symmetry): does one tail stretch out much further than the other?
• Kurtosis (peakedness): are values too clustered at the centre (tall peak, heavy tails) or too flat?

Jarque–Bera (JB) test of normality

The Jarque–Bera (JB) test is an asymptotic (large-sample) test based directly on the OLS residuals. It computes the skewness coefficient $S$ and the kurtosis coefficient $K$:

For a normally distributed variable:

1. Skewness: $S = 0$ (symmetric around the mean).
2. Kurtosis: $K = 3$ (mesokurtic).

Therefore, the JB test is a joint test of the null hypothesis: $H_0: S = 0 \text{ and } K = 3 \quad (\text{residuals are normally distributed})$ $H_1: S \neq 0 \text{ or } K \neq 3 \quad (\text{residuals are not normally distributed})$

Under $H_0$, as sample size grows, the JB statistic asymptotically follows a chi-square distribution with 2 degrees of freedom: $JB \sim \chi^2(2)$ Degrees of freedom is 2 because there are two restrictions ($S = 0$ and $K = 3$).

Prediction: mean prediction and its interval

Once we estimate a regression line, we often want to forecast the dependent variable $Y$ for a chosen value of the explanatory variable, denoted $X_0$. Econometrics distinguishes between two prediction tasks:

1. Mean prediction: predicting the conditional average $E(Y \mid X_0) = \beta_1 + \beta_2 X_0$.
2. Individual prediction: predicting an actual individual outcome $Y_0 = \beta_1 + \beta_2 X_0 + u_0$.

Here we focus on the mean prediction.

Point prediction

The point estimator of the mean value $E(Y \mid X_0)$ is obtained directly from the sample regression function (SRF): $\hat{Y}_0 = b_1 + b_2 X_0$ For example, from Gujarati Table 2.4 consumption regression: $\hat{Y}_i = 24.4545 + 0.5091 X_i$ To predict average weekly consumption for families with weekly income $X_0 = 100$: $\hat{Y}_0 = 24.4545 + 0.5091(100) = 75.3645 \text{ dollars}$

Variance and standard error of mean prediction

Because $b_1$ and $b_2$ vary from sample to sample, $\hat{Y}_0$ is an estimator with its own sampling distribution. It is normally distributed with mean $E(\hat{Y}_0) = \beta_1 + \beta_2 X_0$ and variance:

Replacing the unknown $\sigma^2$ with the unbiased residual variance $\hat{\sigma}^2$: $\operatorname{se}(\hat{Y}_0) = \sqrt{\hat{\sigma}^2 \left[ \frac{1}{n} + \frac{(X_0 - \bar{X})^2}{\sum x_i^2} \right]}$

Confidence interval for E(Y | X0)

A $100(1 - \alpha)\%$ confidence interval for the true mean value $E(Y \mid X_0)$ is: $\hat{Y}_0 - t_{\alpha/2, n-2}\,\operatorname{se}(\hat{Y}_0) \le E(Y \mid X_0) \le \hat{Y}_0 + t_{\alpha/2, n-2}\,\operatorname{se}(\hat{Y}_0)$

Exam practice

Summary and review

• A point estimate says nothing about precision; an interval estimate adds a margin of error.
• $\Pr(\text{lower} \leq \beta_2 \leq \text{upper}) = 1-\alpha$: $1-\alpha$ is the confidence coefficient, $\alpha$ the level of significance.
• The interval is random, $\beta_2$ is fixed: 95% of intervals built this way contain $\beta_2$.
• $t = (b_2 - \beta_2)/SE(b_2)$ follows a t distribution with $n-2$ df; this gives the interval $b_2 \pm t_{\alpha/2}SE(b_2)$, and likewise $b_1 \pm t_{\alpha/2}SE(b_1)$.
• Consumption example: 95% interval for the MPC about $[0.43, 0.59]$; for the intercept $[9.66, 39.24]$.
• Higher confidence or larger SE → wider interval; more data → narrower.
• t test: $t = (b_2 - c)/SE(b_2)$. Two-sided: reject if $|t| \gt t_{\alpha/2}$. One-sided: reject if $t \gt t_\alpha$ (right) or $t \lt -t_\alpha$ (left).
• Two-sided test at level $\alpha$ and the $100(1-\alpha)\%$ interval always agree: reject exactly when $c$ is outside the interval.
• p-value $\lt \alpha$ → reject. EViews’ Prob. tests coefficient $= 0$.
• The confidence-interval approach works only for two-sided tests. Board answers: the MPC is different from 0.60 (outside $[0.42, 0.58]$) and not significantly different from 0.55 (inside).
• “Individually significant” means coefficient $= 0$ is rejected: here both the constant (Prob. 0.0051) and the slope (Prob. 0.0000) are.
• Type I error = rejecting a true $H_0$ (probability $\alpha$); Type II = not rejecting a false $H_0$. Say “do not reject”, never “accept”.
• ANOVA: $TSS = ESS + RSS$; $R^2 = ESS/TSS = 1 - RSS/TSS$ (coefficient of determination). Example: $8890 = 8552.73 + 337.27$, $R^2 = 0.962$.
• $F = (ESS/1)/(RSS/(n-2))$; with one regressor $F = t^2$ ($202.87 = 14.24^2$), the same verdict as the two-sided t test.

Step-by-step examples

Work through the ladder one level at a time: at every step, try to say what comes next before you reveal it.

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Medium

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