Question

The Regression Equation Suppose in this semester, our Exam 1 average was about 91 with an SD of about 10.5. Suppose the correlation between our Exam 1 and Exam 2 scores will be similar to what it has been in the past, about 0.75, and finally, suppose our Exam 2 scores will be similar to previous semesters Exam 2 scores with an average of 85 and a SD of 8.2. Use this information
1. What the slope of the regreso equation for predicting our Exam 2 scores from Exam 1 scores Round to 3 decimal places Tries 0/5
2. What the yntercept of the regression ecuation for predicting our team scores from Exam 1 scores found to 3 decimal places Tres 0/5
3. A student got a 96 on our Exam 1 Use the regression equation you just calculated to predict her Exam 2 score (round to the rounds to 29.) rest whole number. For example, 79.5 rounds to 80, n . But 79.49 Tres 0/5
4. Another student got a 65 on Fram 1. What would you predict her Exam 2 score to be? Round your answer to the nearest whole num
5. What is the SD of the prediction errors (same as RMSE) of our regression ecuation Round to 2 decimal places.

Answer 1

Answer:

1. Slope = 0.5857

2. Intercept = 31.7

3. Y = 88

4. X = 70

5. 5.4234

Step-by-step explanation:

We have x and y

Mean of x = 91

Mean of y = 85

Sd of x = Sx = 10.5

Sd of y = Sy = 8.2

The registration regression equation

Y - ybar = r(Sy/Sx)(x-xbar)

y - 85 = 0.75(8.2/10.5)(x-91)

Y - 85 = 0.75(0.7809)(x-91)

Y-85 = 0.5857X - 53.3

Y = 0.5857X - 53.3+85

Y = 0.5857X + 31.7

1. Slope = 0.5857

2. Intercept = 31.7

3. If x = 96,

Y = 0.5857(96) + 31.7

Y = 87.927

Y = 88

4. If x = 65,

Y = 0.5857(65)+31.7

Y = 69.77

Y = 70

5. Sy * √(1-r²)

= 8.2 x √(1-0.75²)

= 8.2 x √(1-0.5625

8.2 x √0.4375

= 8.2x 0.6614

= 5.4234