Question

2. Wendy has collected a random sample of 50 observation on consumption of orange juice. She has run a regression where the dependent variable is LNSales (the natural log of the number of half-gallon cartons sold per weekend) and the independent variable is LNPrice (the natural log of the price per half-gallon in $). The output for that regression is provided in a separate sheet and labeled REGRESSION 1.

a) Based on the REGRESSION 1 output, construct a 95% confidence interval for the slope coefficient and explain in words what it means.

b) Using REGRESSION 1 output, what is your guess for the quantity of half-gallon cartons sold when the price is $2.72? Give a 95% confidence interval for your guess.

REGRESSION 1 Dependent Variable: Independent Variables: LNPrice LNSales Regression Statistics R-Squared 0.755 Adj.R-Sqr 0.750 Std.Err.Re 0.386 # Fitted 50 Std.Dep.Var 0.772 Coefficient Estimates: Variable Constant LNPrice Coefficient 4.812 1.752 Std.Err. 0.148 0.144 t-Statistic 32.504 12.173 P-value 0.000 0.000 Forecasts: Forecast 3.234 3.058 2.881 StErrFcst LNPrice 0.390 0.390 0.390 0.90 1.00 1.10

Answer 1

1) Slope coefficient in Regression 1 model = -1.752

Std Error of the slope coefficient = 0.144 (from the Std.Err. column in regression output)

Now, degrees of freedom, df = n-2 = 48

Now, for 95% confidence interval, critical probability, p^* = 1 - 0.05/2 = 0.975. Hence, the critical value is the t-statistic with 48 degrees of freedom and cum. probability = 0.975

So, critical value (from t-value calculator/table) comes to be 0.8328

$\therefore$ Margin of Error, ME = critical value * Std Error = 0.8328 * 0.144 = 0.1199

Hence, 95% confidence interval for slope is: -1.752 +/- 0.1199 = (-1.8719, -1.6321)

2) The regression model above is: LN(Sales) = =1.752 * LN(Price) + 4.812

For Price = $2.72, LN(Sales) = -1.752 * LN(2.72) + 4.812 = -1.752 * 1 + 4.812 = 3.06 ------ (i)

=> Sales = e^3.06 = 21.33 half-gallon cartons

95% confidence interval for LNSales can be derived using 48 dof and the Std Errors of slope and intercept from the regression output as: CI(LNSales): (3.06 +/- (0.8328 * 0.144 + 0.8328 * 0.148) ) [using the std error both for the slope and the intercept]

= (3.06 +/- 0.243) = (2.817, 3.303)