# alex puts his spare change in a jar every night. if he has \$11.09 at the end of january \$22.27 at the end of february \$44.35 in april \$75.82 july \$89in august and \$114.76 at the end of october...

alex puts his spare change in a jar every night. if he has \$11.09 at the end of january \$22.27 at the end of february \$44.35 in april \$75.82 july \$89

in august and \$114.76 at the end of october perform a linear regression on this data to complete the following items. what does the value of the correlation coefficient tell you about the correlation of the data? write the equation of the best-fitting line (round to the nearest thousandths) on average how much money does alex add to the jar each month? alex wants to buy a video game console at the end of december for \$140. will he have enough for this purchase?

embizze | High School Teacher | (Level 1) Educator Emeritus

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We have the "points" (1,11.09),(2,22.27),(4,44.35),(7,75.82),(8,89),(10,114.76).

Using a graphing utility we find the linear regression:

`y~~-.900+11.334x` with `r~~.99917` and `r^2~~.99834`

(1) The correlation coefficient is positive and very close to 1. So the amount of money at the end of each month is strongly, positively correlated with the month. (Very close to a perfect correlation).

(2) The best-fitting line is y=-.900+11.334x

(3) The slope of this line (constant rate of change) is approximately 11.33, so he adds approximately \$11.33 each month.

(4) The value of the equation when x=12 is 135.108 so he will have \$135.11; not quite enough to buy the game console.

** If you use a median-median line the equation is y=11.36x-1.04