To analyze the year-to-year variation in temperature data for a given city, you regress the daily high temperature (Temp)for 100 randomly selected days in two consecutive years (1997 and 1998)for Phoenix. The results are (heteroskedastic-robust standard errors in parenthesis):

\widehat{\text { Temp } \begin{array} { l } \text { PHX } \\1998\end{array}}

= 15.63 + 0.80 ×

{\text { Temp } \begin{array} { l } \text { PHX } \\1998\end{array}}

; R²= 0.65, SER = 9.63
(0.10)
(a)Calculate the predicted temperature for the current year if the temperature in the previous year was 40°F, 78°F, and 100°F. How does this compare with you prior expectation? Sketch the regression line and compare it to the 45 degree line. What are the implications?
(b)You recall having studied errors-in-variables before. Although the web site you received your data from seems quite reliable in measuring data accurately, what if the temperature contained measurement error in the following sense: for any given day, say January 28, there is a true underlying seasonal temperature (X), but each year there are different temporary weather patterns (v, w)which result in a temperature

\hat { X }

different from X. For the two years in your data set, the situation can be described as follows:

\mathrm{X}_{1997}=\mathrm{X}+v_{1997} \text { and } \mathrm{X}_{1998}=\mathrm{X}+w_{1998}

\text { Subtracting } X_{1997} \text { from } X_{1998} \text {, you get } X_{1998}=X_{1997}+w_{1998}-v_{1997}

^{Hence the population parameter} for the intercept and slope are zero and one, as expected. It is not difficult to show that the OLS estimator for the slope is inconsistent, where

\hat { \beta } _ { 1 } \stackrel { p } { \longrightarrow } 1 - \frac { \sigma _ { v } ^ { 2 } } { \sigma _ { x } ^ { 2 } + \sigma _ { v } ^ { 2 } }

As a result you consider estimating the slope and intercept by TSLS. You think about an instrument and consider the temperature one month ahead of the observation in the previous year. Discuss instrument validity for this case.
(c)The TSLS estimation result is as follows:

\widehat{\text { Temp } \begin{array} { l } \text { PHX } \\1998\end{array}}

= -6.24 + 1.07×

{\text { Temp } \begin{array} { l } \text { PHX } \\1997\end{array}}

;
(0.06)
Perform a t-test on whether or not the slope is now significantly different from one.

(Requires Chapter 8)When Using Panel Data and in the Presence

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