Question

Problems 5 through 9 only please

C. Regression Analysis In this section, you will analyze the effects of weather conditions on workers' productivity. You collect a random sample of 500 workers. Your data includes the average time it takes each worker to complete the same task (minutes), the temperature that day (measured in Farenheit), the natural log of temperature that day, whether it was raining that day (dummy variable), and the workers' age (years). An extract of the data is shown below: temperature| log_temperature | rainy worker_id date average_time age Jan-1 10 1.61 1 20 A 41 12 A Jan-2 37 1.57 20 A Jan-3 34 1.53 21 Jan-1 13 41 B 1.61 1 40 15 37 40 B Jan-2 1.57 0 34 1.53 0 Jan-3 40 What type of dataset is this? 1. You run a regression of average time on log temperature, rainy and age, and get the following results (standard errors are shown below each coefficient in parenthesis): avg time = 12 + 6.0 x log(temperature) - 2 x rainy - 0.05 x age (2.5) (2.0) (1.2) (0.01) 2. Interpret the coefficient on rainy. Be precise about units, magnitude, direction! 3. Interpret the coefficient on log_temperature. Be precise about units, magnitude, direction! 4. Suppose you measured the dependent variable "avg time" in seconds instead of minutes. How would the estimated intercept of the regression equation change? 5. What needs to be the percent change in temperature between a rainy and a non-rainy day for worker productivity to remain unaffected? Be clear about the magnitude and direction of the change. 6. Do you reject the null hypothesis that rain does not have an effect on average time it takes to complete the task against the alternative that rain does have an effect at a 5% significance level? 7. Given your answer to question 6 above, is the p-value associated with the null hypothesis that rain does not have an effect on average time it takes to complete the task greater or smaller than 0.05? What is the predicted level of productivity for worker A on January 1st? [Use the table extract shown above.] Calculate the 90% confidence interval for the "age" coefficient 8. 9.

Answer 1

5)

6 * log (T_rain) - 2 = 6 log(T_not_rain)

6 log(T_rain) - 6 log(T_not_rain) = 2

6 log ( T_rain / T_not_rain) = 2

log ( T_rain / T_not_rain) = 2/6 = 0.3333

hence

33.333 % more Temperature during rainy days

6)

TS = -2/1.2 = - 1.66666

since n = 500

df = n-k-1 = n-4 = 496

t-critical values are -1.96 and 1.96 for 0.05 level

since TS does not lie in critical region

we fail to reject the null hypothesis

7)

p-value= 2 P( t < -1.6666)

= 0.096225    {approximate by normal distribution}

p-value > 0.05

8)

y^= 12 + 6 * 1.61 - 2 * 1 - 0.05 * 20
= 18.66

9)

t = 1.645 for 90%

(-0.05 - 1.645 * 0.01 , -0.05 + 1.645 * 0.01)

= ( -0.06645 ,-0.03355)

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