The data below are the temperatures on randomly chosen days during the summer and the number of employee absences at a local company on those days. Construct a 95% prediction interval for y, the number of days absent, given x = 95 degrees and y= 0.449x - 30.27

Temperature, x 72 85 91 90 88 98 75 100 80
Number of absences, y 3 7 10 10 8 15 4 15 5

12.385; (10.0, 14.7)

Step-by-step explanation:

Given the following information :

Temperature, x 72 85 91 90 88 98 75 100 80

Number of absences, y 3 7 10 10 8 15 4 15 5

The obtained regression equation:

y= 0.449x - 30.27

Where y = predicted variable ; x = independent variable ; - 30.27 = intercept and slope = 0.449

Constructing a 95% prediction interval for y:

x = 95°

Inputting x = 95 into the regression equation :

y= 0.449(95) - 30.27

y = 42.655−30.27 = 12.385

Prediction interval = y ± error margin

To save computing time, error margin (E) can be obtained using the online error margin calculator.

Obtained error margin (E) value = 2.3398

(12.385 - 2.3398, 12.385 + 2.3398)

(10.0452, 14.7248)

(10.0, 14.7)

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week