3. The data below shows the annual salaries (in millions) and the number of view

Question

3. The data below shows the annual salaries (in millions) and the number of viewers (in millions) of eight television actors and actresses. Answer parts a-c. Salary (x) Viewers () 10 10.4 10 9.5 91 15 4.3 efficien 31 5.9 13.8 4.3 orrelatio a. Find the value of the linear correlation coefficient r (Round to three decimal places as needed.) b. Find the critical values of r from the table showing the critical values for the Pearson correlation coefficient using = 0.05 The critical values are c. Is there sufficient evidence to conclude that there is a linear correlation between the two variables? (Round to three decimal places as needed.) O A. Yes, because the absolute value of the correlation coefficient is less than the critical value O B. No, because the absolute value of the correlation coefficient is less than the critical value O C. Yes, because the absolute value of the correlation coefficient is greater than the critical value O D. No, because the absolute value of the correlation coefficient is greater than the critical value

Explanation / Answer

a) To find linear correlation coefficient r :

viewers(y)

formula=

n*sum(xy)-n*sum(x)*sum(y)/sqrt((n*sum(x^2)-[sum(X)]^2*((n*sum(y^2)-[sum(y)]^2))

n=no of observation=*

=8*1396.2-8*178*59.4/sqrt[(8*9876-(178)^2)*(8*562.08-(59.4)^2]

=596.4/sqrt[47324*968.28]

=596.4/sqrt[45822882.72]

=596.4/6769.2601

r=0.088 ........................calculated value

b) alpha =0.05

The critical values :

For two tailed test, degrees of freedom =n-2= 8-2 =6

r(n-2, alpha)= r(8-2,0.05)= r(6,0.05)= 0.811.......... from table value

rcal<rtable

accept H0 at 0.05 level of significance

c) A. because the absolute value of correlation coefficient is less than the critical value.

salary(x)

viewers(y)

x^2 y^2 x.y 91 10 8281 100 910 15 4.3 225 18.49 64.5 12 5.9 144 34.81 70.8 31 1.2 961 1.44 37.2 10 10.4 100 108.16 104 10 9.5 100 90.25 95 8 13.8 64 190.44 110.4 1 4.3 1 18.49 4.3 sum(x)= 178 sum(y)=59.4 sum(x^2)=9876 sum(y^2)=562.08 sum(x*y)=1396.2

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