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pogonyaev
3 years ago
13

A statistics practitioner in a large university is investigating the factors that affect salary of professors. He wondered if ev

aluations by students are related to salaries. To this end, he collected 100 observations on:
y = Annual salary (in dollars)
x = Mean score on teaching evaluation

To accomplish his goal, he assumes the following relationship:

y = β(0) + β(1)x + ε

Then, using Excel’s Data Analysis, he obtained the following result.

R2=0.23

Coefficient Standard Error
Intercept 25675.5 11393
x 5321 2119

What is the p-value?

Select one:

A. 0.02
B. 0.01
C. 0.17
D. 0.34
Mathematics
1 answer:
Ulleksa [173]3 years ago
3 0

Answer:

Step-by-step explanation:

Hello!

Given the linear regression of Y: "Annual salary" as a function of X: "Mean score on teaching evaluation" of a population of university professors. It is desired to study whether student evaluations are related to salaries.

The population equation line is

E(Y)= β₀ + β₁X

Using the information of a n= 100 sample, the following data was calculated:

R²= 0.23

                Coefficient    Standard Error  

Intercept    25675.5           11393

x                  5321                  2119

The estimated equation is

^Y= 25675.5 + 5321X

Now if the interest is to test if the teaching evaluation affects the proffesor's annual salary, the hypotheses are:

H₀: β = 0

H₁: β ≠ 0

There are two statistic you can use to make this test, a Student's t or an ANOVA F.

Since you have information about the estimation of  β you can calculate the two tailed t test using the formula:

t= \frac{b - \beta }{\frac{Sb}{\sqrt{n} } } ~t_{n-2}

t= \frac{ 5321 - 0 }{\frac{2119}{\sqrt{100} } } = 25.1109

The p-value is two-tailed, and is the probability of getting a value as extreme as the calculated t_{H_0} under the distribution t_{98}

p-value < 0.00001

I hope it helps!

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