Given the simple regression analysis results shown below, find the 99% confidence interval and 99% prediction intervalfor the point where x = 13.ŷ₁ = 7+15x₁ S₂ = 10.14 x=7 n = 22nΣ (x − x)2 = 200i=1The 99% confidence interval for the expected value runs from(Round to three decimal places as needed.)to

The given regression equation is y^=7+15x, the remaining values are se=10.14,x¯=7,n=22 and…

Given the simple regression analysis results shown below, find the 99% confidence interval and 99% prediction interval for the point where x = 13. ₁ = 7+15x₁ S₂ = 10.14 x=7 n = 22 n Σ (x-x)2 = i=1 2 = 200 The 99% confidence interval for the expected value runs from to (Round to three decimal places as needed.)

Given the simple regression analysis results shown below, find the 99% confidence interval and 99% prediction interval for the point where x = 13. ₁ = 7+15x₁ S₂ = 10.14 x=7 n = 22 n Σ (x-x)2 = i=1 2 = 200 The 99% confidence interval for the expected value runs from to (Round to three decimal places as needed.)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

In the following case find the best predicted value of y given the following: x = 2.00, r = -0.123 (p-value = 0.62), y (bar)= 8.00, n = 30 and the equation of the regression line is: y (hat) = 7.00 - 2.00x
In the following case find the best predicted value of y given the following: x = 3.00, r = 0.052 (p-value = 0.85), y (bar)= 5.00, n = 20 and the equation of the regression line is: y (hat) = 6.00 + 4.00x
Can you contruct a 95% confidence interval for the Y intercept for the regression equation. Then explain what do you conclude from this regression?
The table below lists measured amounts of redshift and the distances (billions of light-years) to randomly selected astronomical objects. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 90% confidence level with a redshift of 0.0126. Find the explained variation. Redshift Distance OA, 0.157904 OB. 0.444904 OC. 0.000582 OD. 0.219582 0.0231 0.34 0.0536 0.76 0.0715 0.99 0.0391 0.56 0.0438 0.63 0.0109 0.15
The average midterm score in a large statistics class was 60 with an SD of 5. The average final score in the same class was 80 with an SD of 15. The correlation coefficient between midterm and final scores was r=0.6. Using the regression line, we predict the final score of a student with a midterm score of 70 to be but this prediction is likely to be off by about Fill in the blanks, rounding each answer to one decimal point.
Use the given data to find the best predicted value. If the prediction is unreliable, state so. 5) Eight pairs of data yield r = 0.742 and the regression equation y = 55.8 + 2.79x. Also, y = 71.125. What is the best predicted value of y for x = 8? Round to one decimal place.
If you have a b of 0.56 in a regression equation, what does this mean? For every one-unit increase in x, you get an increase of 0.56 in y r = .31 On average, the variability of real scores around the regression line is 0.56 For every 1 standard deviation increase in x, you get an increase of 0.56 standard deviations in y
8
12 A regression was run to determine if there is a relationship between hours of study per week (xx) and the final exam scores (yy).The results of the regression were: y=ax+b a=5.531 b=23.45 r2=0.378225 r=0.615 Use this to predict the final exam score of a student who studies 8.5 hours per week, and please round your answer to a whole numbe
In a regression model involving 30 observations, the following estimated regression equation was obtained:ŷ = 17 + 4x1 − 3x2 + 8x3 + 8x4For this model, SSR = 700 and SSE = 100.The critical F value at 95% confidence is _____. a. 2.76 b. 2.69 c. 2.53 d. 2.99
You run a regression analysis on a bivariate set of data (n = 106). You obtain the regression equation y = 4.241x – 10.085 =0.373 (which is significant at a value (on average) for the explanatory variable will give you a value of 70 on the response variable. with a correlation coefficient of r 0.01). You want to predict what What is the predicted explanatory value? X = (Report answer accurate to one decimal place.)
Given the correlation coefficient 0.794 and the linear regression equation ŷ = 16.4 +1.58x, compute the coefficient of determination. p² Round to 4 decimal places. =