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Assertiveness and leadership. Management professors at Columbia University examined the relationship between assertiveness and leadership (Journal of Personality and Social Psychology, February 2007). The sample represented 388 people enrolled in a full-time MBA program. Based on answers to a questionnaire, the researchers measured two variables for each subject: assertiveness score (x) and leadership ability score (y). A quadratic regression model was fit to the data, with the following results:

a. Conduct a test of overall model utility. Useα=0.05 .

b. The researchers hypothesized that leadership ability increases at a decreasing rate with assertiveness. Set up the null and alternative hypotheses to test this theory.

  1. Use the reported results to conduct the test, part b. Give your conclusion(atα=0.05 )in the words of the problem.

Short Answer

Expert verified

a) At 95% significance level, β1≠β2≠0

b) The null hypothesis is whether β2=0while the alternate checks if the value of β2<0Mathematically, whileHa:β4<0

c) Therefore,β2=0,This means that the parabola doesn’t have a curvature and it essentially is a straight line.

Step by step solution

01

Overall goodness of fit

H0:β1=β2=0

Ha:At least one of the parameters β1or β2is non zero

Here, F test statistic =SSEn-(k+1)

H0is rejected if p-value < α. For α=0.05, since for x and x2p - value is less than 0.01

Sufficient evidence to rejectH0at 95% confidence interval.

Thus,β1≠β2≠0.

02

Significance of  β2

The null hypothesis is whether β2=0while the alternate checks if value of t <t0.05,199

Mathematically, H0:β4=0whileHa:β4<0

03

Consequence of  β2

H0:β2=0

Ha:β2<0

Here, t-test statistic=βÁ圲õβ2=-0.088-3.97=0.0221

Value of t0.05,385 is 1.645

H0 is rejected if t statistic > t0.05,385. For α=0.05, since t <t0.05,199

Not sufficient evidence to reject H0 at 95% confidence interval.

Hence,β2=0.

This means that the parabola doesn’t have a curvature and it essentially is a straight line.

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