The management of a relatively new social networking website named BooglePlus is conducting a pilot study comparing use of its own site with use of a longer established social networking site named FaceList. Some articles published on the Internet give the reader the opportunity to register votes (called "likes") for the article on social networking sites to which the reader belongs. A BooglePlus employee selects from the Internet a random sample of 28 articles where the opportunity is given for registering votes for the article on both BooglePlus and Face List. Letting x be the number of votes on FaceList and y be the number of votes on the BooglePlus, the slope of the least squares regression line of y on x is found to be 0.0623, with a standard error of 0.0224. Required:
What could be used to compute a 95% confidence interval for the slope of the population regression line of y on x?

0.0623 ± ( 2.056 )( 0.0224 ) can be used to compute a 95% confidence interval for the slope of the population regression line of y on x

Step-by-step explanation:

Given the data in the question;

sample size n = 28

slope of the least squares regression line of y on x or sample estimate = 0.0623

standard error = 0.0224

95% confidence interval

level of significance ∝ = 1 - 95% = 1 - 0.95 = 0.05

degree of freedom df = n - 2 = 28 - 2 = 26

∴ the equation will be;

⇒ sample estimate ± ( t-test) ( standard error )

⇒ sample estimate ± ( ) ( standard error )

⇒ sample estimate ± ( ) ( standard error )

⇒ sample estimate ± ( ) ( standard error )

{ from t table; ( ) = 2.055529 = 2.056

so we substitute

⇒ 0.0623 ± ( 2.056 )( 0.0224 )

Therefore, 0.0623 ± ( 2.056 )( 0.0224 ) can be used to compute a 95% confidence interval for the slope of the population regression line of y on x

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!