4. A wacky stock analyst notices a pattern and comes up with the following model

Question

4. A wacky stock analyst notices a pattern and comes up with the following model:

S&PPtst = 1 + 2NFCPtst + et

Where

S&PPtst = the value of the Standard and Poors 500 stock index at the end of year t, measured in points

NFCPtst = the number of points scored in year t in the Super Bowl by the team representing the NFC (National Football Conference)

et is a random disturbance

1, 2 are parameters

The analyst collects data for 39 years and runs a simple regression with the following results:

b1 = 986.543 SEb1 = 345.678

b2 = 11.852 SEb2 = 2.345

Interpret these results. Does NFCPts have a statistically significant impact on S&PPts? Construct and interpret a 05% confidence interval for b2.

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

H0: The slope of the regression line is equal to zero.

Ha: The slope of the regression line is not equal to zero.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a linear regression t-test to determine whether the slope of the regression line differs significantly from zero.

Analyze sample data. To apply the linear regression t-test to sample data, we require thestandard error of the slope, the slope of the regression line, the degrees of freedom, the t statistic test statistic, and the P-value of the test statistic.

We get the slope (b1) and the standard error (SE) from the regression output.

b2 = 11.852, SE = 2.345, n = 39

We compute the degrees of freedom and the t statistic test statistic, using the following equations.

DF = n - 2 = 39- 2

D.F = 37

t = b2/SE

t = 5.054

where DF is the degrees of freedom, n is the number of observations in the sample, b2 is the slope of the regression line, and SE is the standard error of the slope.

The P-value = 0.000012

Interpret results. Since the P-value (0.000012) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we can say conclude that we have sufficient evidence that NFCPts have a statistically significant impact on S&PPts.

95% confidence interval for b2 is C.I = (7.101, 16.603).

C.I = b2+ t(1-alpha/2) × Sb2

t(1-alpha/2)= 2.026

C.I = 11.852 + 2.026 × 2.345

C.I = 11.852 + 4.75097

C.I = (7.101, 16.603)

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.