Exam SRM — Generalized Linear Models Practice Flashcards

**Answer: (C).** First form the linear predictor: $\hat\eta=-2.0+0.5(5)=-2.0+2.5=0.5$. Apply the inverse logit (logistic) link: $\hat p=\dfrac{1}{1+e^{-\hat\eta}}=\dfrac{1}{1+e^{-0.5}}=\dfrac{1}{1+0.606531}=\dfrac{1}{1.606531}\approx 0.6225$. Using the wrong sign, $\hat p=\dfrac{1}{1+e^{+0.5}}\approx 0.3775$, gives distractor (A). Reporting the **odds** $e^{0.5}\approx 1.6487$ instead of the probability gives (D).

**Answer: (A).** Linear predictor: $\hat\eta=-0.7+0.4(3)-0.5(2)=-0.7+1.2-1.0=-0.5$. $\hat p=\dfrac{1}{1+e^{-\hat\eta}}=\dfrac{1}{1+e^{0.5}}=\dfrac{1}{1+1.648721}=\dfrac{1}{2.648721}\approx 0.3775$. Dropping the negative sign on the $x_2$ term gives $\hat\eta=+0.5$ and $\hat p\approx 0.6225$ (distractor C). Reporting $e^{-0.5}\approx 0.6065$ — the value $1-\hat p$ is *not* $0.6065$, that is just $e^{-\hat\eta}$ — gives (D).

**Answer: (A).** For a $10$-point increase the log-odds change by $\hat\beta_1\times 10=-0.02(10)=-0.2$, so the odds ratio is $e^{-0.2}\approx 0.8187$. Since $0.8187<1$, the odds of default **fall** by $1-0.8187=0.1813$, about $18.1\%$. Using $e^{-0.02}\approx 0.9802$ (the per-*point* ratio, not per $10$ points) gives distractor (B). Treating the sign as positive gives (C)/(D).

**Answer: (C).** The odds ratio for the indicator is $e^{\hat\beta_1}=e^{1.1}\approx 3.0042$: a smoker's odds are about $3.0\times$ a non-smoker's. Smoker linear predictor ($x=1$): $\hat\eta=-1.0+1.1=0.1$, so $\hat p=\dfrac{1}{1+e^{-0.1}}=\dfrac{1}{1+0.904837}\approx 0.5250$. Reporting $\hat\beta_1=1.1$ itself as the odds ratio gives (A); using the non-smoker probability $1/(1+e^{1.0})\approx 0.2689$ gives (B). Distractor (D) uses $e^{0.9}\approx 2.46$, a different coefficient.

**Answer: (A).** Odds $=\dfrac{\hat p}{1-\hat p}=\dfrac{0.40}{0.60}\approx 0.6667$ (i.e. $2$-to-$3$ in favor). Log-odds $=\ln(0.6667)\approx -0.4055$, which equals the model's linear predictor $\hat\eta$. Check: $\hat p=\dfrac{1}{1+e^{0.4055}}=\dfrac{1}{1+1.5}=0.40$ ✓. Inverting the odds to $\dfrac{1-\hat p}{\hat p}=1.5$ gives distractor (B); using $\ln(\hat p)=\ln 0.4\approx -0.9163$ instead of $\ln(\text{odds})$ gives (C).

**Answer: (A).** $\hat\eta=-1.5+0.9(2)-0.4(1)=-1.5+1.8-0.4=-0.1$. $\hat p=\dfrac{1}{1+e^{-(-0.1)}}=\dfrac{1}{1+e^{0.1}}=\dfrac{1}{1+1.105171}\approx 0.4750$. Dropping the sign and using $\hat\eta=+0.1$ gives $\hat p=\dfrac{1}{1+e^{-0.1}}\approx 0.5250$ (distractor B). Reporting $e^{-0.1}\approx 0.9048$ instead of the probability gives (C).

**Answer: (C).** Wald statistic: $z=\dfrac{\hat\beta_1}{SE(\hat\beta_1)}=\dfrac{0.96}{0.32}=3.00$. Since $|3.00|>1.96$, $\beta_1$ is significant at the $5\%$ level. Point odds ratio: $e^{\hat\beta_1}=e^{0.96}\approx 2.612$. Inverting the ratio ($SE/\hat\beta=0.333$) gives the wrong $z$ in (A); $e^{-0.96}\approx 0.382$ gives the inverted odds ratio in (B); reporting $\hat\beta_1$ itself as the odds ratio gives (E).

**Answer: (A).** In logistic regression $e^{\hat\beta_1}$ is the **odds ratio**, not a probability ratio. Here $e^{0.693}\approx 2.0$, so being in the agent channel multiplies the **odds** of lapse by $2$ (holding other predictors fixed). It does **not** double the probability — the probability change depends on the baseline $p$ via $\partial p/\partial x=\beta_1 p(1-p)$, so (B) is the classic odds-vs-probability error. The coefficient $\hat\beta_1=0.693$ is the additive change in log-odds (not a multiplicative factor of $2$ on the log-odds), ruling out (C); and (D) confuses the log-odds change with a probability change.