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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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a) Null hypothesis:\mu \leq 30.2  

Alternative hypothesis:\mu > 30.2  

b) X \sim N(\mu=32.12, \sigma=4.83)

And the distribution for the random sample is given by:

\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)

c) t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81    

p_v =P(t_{(49)}>2.81)=0.0035  

d) If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.  

e) The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation  

\bar X=32.12 represent the sample mean  

s=4.83 represent the sample standard deviation

n=50 sample size  

\mu_o =30.2 represent the value that we want to test

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

a. Define the parameter and state the hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 30.2 mpg, the system of hypothesis would be:  

Null hypothesis:\mu \leq 30.2  

Alternative hypothesis:\mu > 30.2  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

b. Define the sampling distribution (mean and standard deviation).

Let X the random variable who represent the variable of interest. And we know that the distribution for X is:

X \sim N(\mu=32.12, \sigma=4.83)

And the distribution for the random sample is given by:

\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)

c. Perform the test and calculate P-value

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=50-1=49  

Since is a one right tailed test the p value would be:  

p_v =P(t_{(49)}>2.81)=0.0035  

d. State your conclusion.

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.  

e. Explain what the p-value means in this context.

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

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