Using Microsoft Excel to Conduct

Regression Analysis

ExampleMarkets Price Quantity

1 50 20.0

2 50 21.0

3 55 19.0

4 55 19.5

5 60 20.5

6 60 19.0

7 65 16.0

8 65 15.0

9 70 14.5

10 70 15.5

11 80 13.0

12 80 14.0

13 90 11.5

14 90 11.0

15 40 17.0

- 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

Step 2: Run Regression in Excel1. 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:

Step 3: Regression Results

Interpret Regression OutputEstimated 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:

92.28ˆ 19.0ˆ

72.02 R

74.1ˆ

74.1ˆ 22 = 3.02

Tests Based on Regression Results

The t-Test Example: H0: b = -0.3 51.3

0313.03.01.6

sa

sb

sabt

bbbStat

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) = 0.00384Step 3:

- Reject the null hypothesis- Why?-3.51 < -2.16 or p = 0.00384 < 0.05

The t-Test Example: H0: b > -0.1Step 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) = 0.006145Step 3:

- Reject the null hypothesis- Why?tstat = -2.905 < -c = -1.771 or p = 0.006145 < 0.05

905.20313.0

1.01.6

Statt

Confidence Interval of a = 28.92Calculate 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

]33.43 41.24[51.492.2809.2*16.292.28 CI

)(* estimatesecestimateCI

Confidence Interval: b = -0.19How?

- 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.

)(* estimatesecestimateCI

]0.1245- 2577.0[0666.01911.00313.0*16.21911.0 CI