1. Using Microsoft Excel to Conduct Regression Analysis

2. Example MarketsPriceQuantity 15020.0 25021.0 35519.0 45519.5 56020.5 66019.0 76516.0 86515.0 97014.5 107015.5 118013.0 128014.0 139011.5 149011.0 154017.0 -What do we know? Price and quantity of pens in 15 markets with similar characteristics. -What do we want to know? The relation between price and quantity, i.e., the demand curve. -Proposed Model Specification: Finding the demand curve where quantity (Q) is the dependent variable and price (P) is the explanatory variable. -What we want to know: (a)estimate of parameters (b)parameter testing (c)forecasting

3. Step 2: Run Regression in Excel 1.Click “Data Analysis” in “Tools” box. 2.Click “Regression” in “Data Analysis” box. 3.Setup in “Regression” window including: a)Input Y range: Quantity (P1:P16). b)Click “Label” is you want to have labels in output. c)Confidence level: 95%. d)Output options:

4. Step 3: Regression Results

5. Interpret Regression Output Estimated Linear Demand Function Intercept: Price coefficient: What proportion of variation of Y is explained by the regression? R² = 0.74; Estimates of standard error: Estimates of variance: = 3.02

6. Tests Based on Regression Results

7. The t-Test Example: H 0 : b = -0.3 Step 1: Step 2: - Degrees of Freedom = 15 – 2 = 13, 2-tailed test - One method: compare t-stat and c value t-stat: -3.51 < -c = -2.16 - Another method: finding out p-value p = tdist(3.51, 13, 2) = 0.00384 Step 3: - Reject the null hypothesis - Why? -3.51 < -2.16 or p = 0.00384 < 0.05

8. The t-Test Example: H 0 : b > -0.1 Step 1: Step 2: - DF = 15 – 2 = 13, 1-tailed test, significant level 5% - One method: compare t-stat and c value t-stat: -2.905 < -c = -t(0.10, 13, 1, 5%) = -1.771 - Another method: compare p-value and significant level p = tdist(2.905, 13, 1) = 0.006145 Step 3: - Reject the null hypothesis - Why? t stat = -2.905 < -c = -1.771 or p = 0.006145 < 0.05

9. Confidence Interval of a = 28.92 Calculate 95% CI of a = 28.92 How? - CI of coefficient: - The critical value of c is 2.16, given DF = 15 – 2 = 13, level of confidence = 5%, and two-tailed test. - se(estimate) = 2.09 is the standard error of the estimate. Interpretation: - Any number outside of the CI is statistically different than the estimated a = 28.92 - Any number within the CI is statistically no different than the estimated a = 28.92

10. Confidence Interval: b = -0.19 How? - CI of coefficient: - The critical value of c is 2.16, given DF = 15 – 2 = 13, level of confidence = 5%, and two-tail. - se(estimate) = 0.03 is the standard error of the estimate. Results: Interpretation: - Any number outside the CI is statistically different from -0.19 at the 5% level. - Any number within the CI is statistically no different than the estimated b = -0.19.