Question

6. The following regression equation examines the relation between hotel prices (in dollars per night) and the number of short-term rental listings (e.g., Airbnb) in the city. hoteTprce = 150-0018 × ST-rentals

Answer 1

6. The regression is given as $\dpi{80} \widehat{hotelprice} = 150 - 0.018*\widehat{STrentals}$ .

(e) The predicted hotel price for 1500 short term listings would be $\dpi{80} \widehat{hotelprice} = 150 - 0.018*1500 = 123$ . The error in predicted hotel price would be $\dpi{80} hotelprice - \widehat{hotelprice} = 110 - 123 = -13$ (dollars).

(f) Assuming the inferential statistics are significant, we may conclude that short term rentals causes a deacrease in hotel prices on average. As the sign on coefficient in $\dpi{80} \widehat{hotelprice} = 150 - 0.018*\widehat{STrentals}$ is negative, it implies there is a negative association between hotelprice and STrenals, meaning that if STrental increases, hotelprice would decrease. The coefficient is the amount by which hotelprice would decrease on average for a unit increase in STrentals.

The new regression is $\dpi{80} ln(\widehat{hotelprice}) = 5.2 - 0.046*ln(\widehat{STrentals})$ .

(g) The coefficient is the elasticity of hotelprice with respect to STrentals, which is the percentage change in hotelprice on average for an increase in unit percent of STrentals.

We can see it as $\dpi{80} \mathrm{d} [ln(\widehat{hotelprice})] = \mathrm{d} 5.2 - 0.046* \mathrm{d} [ln(\widehat{STrentals})]$ or $\dpi{80} \frac{\mathrm{d} (\widehat{hotelprice})}{\widehat{hotelprice}} = - 0.046\frac{\mathrm{d} (\widehat{STrentals})}{\widehat{STrentals}}$ or $\dpi{80} \frac{\mathrm{d} (\widehat{hotelprice}) / \widehat{hotelprice}}{\mathrm{d} (\widehat{STrentals}) / \widehat{STrentals}} = - 0.046$ , which is the elasticity of hotelprice with respect to STrentals.

(h) For 1000 STrentals, we have $\dpi{80} ln(\widehat{hotelprice}) = 5.2 - 0.046*ln(1000)$ or $\dpi{80} ln(\widehat{hotelprice}) = 4.882243257$ or $\dpi{80} \widehat{hotelprice} = antilog(4.882243257) = e^{4.882243257}$ or $\dpi{80} \widehat{hotelprice} = 131.926276738$ or $131.93 (approx) on average.