Concept explainers
Suppose that a researcher, using data on class size (CS) and average testscores from 50 third-grade classes, estimates the OLS regression "TestScore" = 640.3 - 4.93 X CS, R2 = 0.11, SER = 8.7. (23.5) (2.02)a. Construct a 95% confidence interval for β1, the regression slopecoefficient.b. Calculate the p-value for the two-sided test of the null hypothesisH0 : β1 = 0. Do you reject the null hypothesis (i) at the 5% level(ii) at the 1% level?c. Calculate the p-value for the two-sided test of the null hypothesisH0 : β1 = -5.0. Without doing any additional calculations, determinewhether -5.0 is contained in the 95% confidence interval for β1.d. Construct a 90% confidence interval for β0.
Trending nowThis is a popular solution!
Step by stepSolved in 3 steps with 3 images
- The table below lists measured amounts of redshift and the distances (billions of light-years) to randomly selected astronomical objects. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. 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. Redshift Distance a. Find the explained variation. 0.0237 0.34 (Round to six decimal places as needed.) 0.0541 0.75 0.0723 0.98 C 0.0397 0.57 0.0444 0.62 0.0103 0.13arrow_forwardThe number of pounds of steam used per month by a chemical plant is thought to be related to the average ambient temperature (in °F) for that month. The past year's usage and temperatures are in the following table by using excel: (a) Find a regression relating steam usage to temperature. (b) Test for significance of regression using a=0.05. (c) Find a 99% confidence interval for B1 Usage/ 1000 Usage/ 1000 Month Temp. Month Тemp. Jan. 21 185.79 July. 68 621.55 Feb. 24 214.47 Aug. 74 675.06 62 562.03 Sept. Oct. Mar. 32 288.03 Apr. 47 424.84 50 452.93 May 50 454.58 Nov. 41 369.95 June 59 539.03 Dec. e 30 273.98arrow_forwardThe accompanying table lists systolic blood pressures (mm Hg) and diastolic blood pressures (mm Hg) of adult females. Find the (a) explained variation, (b) unexplained variation, and (c) prediction interval for a systolic blood pressure of 120 mm Hg using a 99% confidence level. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. Click the icon to view the blood pressure data. Blood Pressures a. The explained variation is 650 (Round to two decimal places as needed.) b. The unexplained variation is (Round to two decimal places as needed.) Systolic Diastolic 126 69 102 67 128 75 107 63 c. The 99% prediction interval for a systolic bloo (Round to one decimal place as needed.) 156 74 95 52 154 89 110 71 120 69 116 75 103 60 126 69 Print Done Xarrow_forward
- You estimated a single regression relating the annual bonus given to employees of big corporations to the age of those employees. Your simple regression shows a positive value for the slope coefficient. You are absolutely convinced that the true effect of age on bonus amount is negative, so you think you have an omitted variable bias. If you are correct, which of the following is true about the omitted variable? There is some (omitted) variable that positively affects bonus amount and that is positively correlated with age. There is some (omitted) variable that negatively affects bonus amount and that is completely uncorrelated with age. There is some (omitted) variable that positively affects bonus amount and that is negatively correlated with age. There is some (omitted) variable that positively affects bonus amount and that is completely uncorrelated with age.arrow_forwardThe table below lists measured amounts of redshift and the distances (billions of light-years) to randomly selected astronomical objects. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. 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. D Redshift Distance 0.0237 0.32 0.0545 0.75 0.0724 1.02 0.0397 0.56 0.0442 0.61 0.0103 0.16 a. Find the explained variation. (Round to six decimal places as needed.) b. Find the unexplained variation. (Round to six decimal places as needed.) c. Find the indicated prediction interval. billion light-yearsarrow_forwardListed below are altitudes (thousands of feet) and outside air temperatures (°F) recorded during a flight. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. 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 95% confidence level with the altitude of 6327 ft (or 6.327 thousand feet). Altitude 8 15 22 28 31 33 Temperature 56 39 24 - 28 - 41 - 60 a. Find the explained variation. (Round to two decimal places as needed.)arrow_forwardwhat is a confidence coefficinent for a simple linear regression line whoes alpha =0.05 and line is y(hat)=-7.45-10xarrow_forwardData on 220 reported crimes is collected from district X in 2016. Suppose CS denotes the total cost to the state of offering crime protection services to this district (in thousand dollars), LEOP denotes the number of law enforcement officers on patrol, DTP denotes the damage to public and private property (in thousand dollars), CCTV denotes the number of CCTV cameras installed in the district, and Prison denotes the number of prison inmates The following table shows the results of a few regressions of the total cost to the state. Dependent variable: total cost to the state (in thousand dollars) Regressor (1) (2) (3) (4) 12.32 17.99 14.55 18.1 LEOP (0.52) (0.84) (2.25) (0.82) 0.73 0.59 0.75 DTP (0 12) (0.04) 0.73 (0.06) CCTV (0.13) 2.11 2.12 (0.39) 288.5 (4 14) Prison (0.5) 182.5 191.6 219.95 Intercept (11.52) (6.68) (5.26) 0.64 0.75 0.12 0.75 R 220 220 220 220 Heteroskedasticity-robust standard errors are given in parentheses under coefficients. Which of the following statements…arrow_forwardData was collected from 60 students from the SAT, the average sat score was 909 with a standard deviation of 178. The average ACT score was 19 with a standard deviation of 3.5. The correlation between the two variables is 0.811. a) To predict the SAT score from the ACT score, what is the equation of the leasts-square regression line. b) What fraction of the variation in the values of the SAT scores is accounted for by the linear relationship between SAT and ACT scores. Explainarrow_forward4. The following regression is fitted using variables identified that could be related to tuition charges ($) of a university. TUITION = a+ B ACCEPT + y MSAT + 1 VSAT Where ACCEPT = the percentage of applicants that was accepted by the university, MSAT = Median Math SAT score for the freshman class and VSAT = Median English SAT score for the freshman class. The data was processed using MNITAB and the following is an extract of the output obtained: Predictor Coef StDev T P Constant -26780 6115 -4.38 0.000 ACCEPT 116.00 37.17 * 0.003 MSAT -4.21 14.12 VSAT 70.85 15.77 -0.30 4.49 0.767 ** S = 2685 R-Sq = 69.6% Analysis of Variance R-Sq (adj) = 67.7% Source DF SS MS F P Regression 3 808139371 Residual Error 49 353193051 269379790 7208021 37.37 0.000 Total 52 1161332421 a) Write out the regression equation. b) State the dependent and independent variable(s) c) Fill in the blanks identified by ** and ****. d) Is B significant, at the 10% level of significance? e) State one limitation of using…arrow_forwardListed below are altitudes (thousands of feet) and outside air temperatures (°F) recorded during a flight. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. 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 95% confidence level with the altitude of 6327 ft (or 6.327 thousand feet). Altitude Temperature a. Find the explained variation. (Round to two decimal places as needed.) 2 55 8 40 13 25 20 - 3 28 - 26 31 - 41 34 - 53arrow_forwardThe 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.arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman