Question

The table below gives the number of hours five randomly selected students spent studying and their corresponding midterm exam grades. Using this data, consider the equation of the regression line, y = bo + 6 x, for predicting the midterm exam grade that a student will earn based on the number of hours spent studying. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant. 1 2 Hours Studying 0 Midterm Grades 73 1 84 89 45 92 99 | tog Table Copy Data Step 1 of 6: Find the estimated slope. Round your answer to three decimal places.

Step 2 of 6: Calculate the estimated variance of errors, s^2e. Round your answer to three decimal places.

Step 3 of 6: Calculate the estimated variance of slope, s^2b1. Round your answer to three decimal places.

Step 4 of 6: Construct the 95% confidence interval for the slope. Round your answers to three decimal places.

Lower and upper endpoints.

Step 5 of 6: Construct the 98% confidence interval for the slope. Round your answers to three decimal places. Lower and upper endpoints

Step 6 of 6: Construct the 90% confidence interval for the slope. Round your answers to three decimal places. Lower and upper endpoints.

Answer 1

Hous y x-24 Grades (x-7) (x-X² (4-7 (7-7" xix) (4-7) 73 -2.4 5-76 -4.4 207.36 34.56 84 - 1-4 1.96. - 3.4 11.56 4.76 89 -0- 4 0-16116 2.56 -0.64 92 1.6 2.56 4.6 21.16 7-36. 99 9.6 6.76 11.6 134.56 30.16 437 17.20 37726 76. 20 19 Ex =19 x = 12 5 TX = 2.4 -2.4 {y=437 Y = 437 Ty = 87.24 r = {( xi-x) (Yi-Y) Jeux)2 Jel4i - 2 r = 76.20 76.20 76.20 / 6487.84 80-55 N 17.20 X 37720 r=0.95 equation of Y- y = regression of youx. r E(917) 2 (x-2). y-87.4 = 0.958 377.20 (xi-2.4 17-20 y 87.4 = 20.834 (x-2.4) -y-87- 4 20.834 34 - 5000 y = 20.834 - 50+87-4

x 9) 73 84 Y = 20.834x + 37.4 e 4 37-4 35-6 58.234 1. 25.766 179-068 9.932 120.736 - 28.736 141.57 - 42,57 | e² 1267-36 663.86 98.64 825.76 1812.20 4667.85 92 99 cef eerer 1555.95 Estimated variance of error Variance of enrer- {ep² 4667.85 (1555.95 удал Висор Тере — B, = 20.834 n-a 3) SER - Vse { (26-2)2 Var Biz Eex 1555.95-190-46] 4) 95% CI for slope ß t trs SEO 28.834 + dos 90.46 2018 344 3.182 9.5m - 120,834 + 30.264 , 1-9.430,51.098 2:27. =0-02 98% CI for slope - ß £ t x 9.5) - Ê t 4.541 X 9.571

6 x 2log. =ool 90% CI for slope ß t t o X S.EB - 20.834 +2353x9.511 - 20.834 + 22.379 -1,545,43.213 2