Residuals: Min 1Q Median 3Q Max -2.5629 -1.2581 -0.2550 0.8681 4.0581 Coefficien

Question

Residuals:
    Min      1Q Median      3Q     Max
-2.5629 -1.2581 -0.2550 0.8681 4.0581

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept) -6.33209    1.67005 -3.792 0.00353 **
x            9.20847    0.03382 272.255 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.946 on 10 degrees of freedom
Multiple R-squared: 0.9999,    Adjusted R-squared: 0.9999
F-statistic: 7.412e+04 on 1 and 10 DF, p-value: < 2.2e-16

Analysis of Variance Table

Response: y
          Df Sum Sq Mean Sq F value    Pr(>F)  
x          1 280590 280590   74123 < 2.2e-16 ***
Residuals 10     38       4                    
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

1) Test the hypothesis that = 0

I've got all the statistics above, I just don't know how to do the test with p = 0. Please help.

Explanation / Answer

1)

TS = r* sqrt((n-2)/(1-r^2))

here

r^2 = 0.9999

r = sqrt(0.9999) = 0.9999499

n = 1 +10 +1 = 12

hence

TS = 0.9999499*sqrt(10/(1-0.9999))

=316.2119230

which is clearly much more than critical value

hence we reject the null hypothesis

note that we also have table for critical value of r to test this hyothesis

http://www.statisticssolutions.com/table-of-critical-values-pearson-correlation/

Please rate

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

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