Compare your answer in Problem 1, part m) with answer in Problem 2, part k). Is there a difference and why?
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- Compare your answer in Problem 1, part m) with answer in Problem 2, part k). Is there a difference and why?
Given
There are 2 95% CI which are as follows:
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- Suppose that you're regressing the Sales Price (the y-variable) of a home on the Square Footage (the x-variable) of that home, and the 95% Confidence Interval (CI) estimate of B1 is given by the interval: (-2.757,3031). Give the interpretation of this CI. Specifically, EXPLAIN how the CI answers the question about whether or not Home Prices and Square Footage are, or are not, significantly linearly related.So I generated this table with stat program, which is from a regression of temperature (in °C) on atmospheric concentrationof carbon dioxide (CO2), in ppm. Can you please construct a 95% confidence interval for the slope of the regression equation? Also, what are the chances of seeing a linear relationship at least as strong as observed from these data, when in fact there was none in the population? What would be the conclusion from this regression?Asset W has expected return of 11.6% and a beta of 1.23. If the risk free rate is 3.15%, complete the following table for portfolios of asset W, and a risk free asset. Illustrate the relationship between portfolio, expected return and portfolio beta by plotting the expected returns against the betas in a graph. What is the slope of the line that results? Results are in the table I just need the excel graph!!!! Percentage of portfolio asset in W. Portfolio expected return. Portfolio beta. 0% = (0 x 0.116) + (1 x 0.0315) = 3.15% 0 25% = (0.25 x 0.116) + (0.75 x 0.0315) = 5.26% = 1.23 x 0.25 = 0.3075 50% = (0.50 x 0.116) + (0.50 x 0.0315) = 7.37% = 1.23 x 0.50 = 0.615 75% = (0.75 x 0.116) + (0.25 x 0.0315) = 9.48% = 1.23 x 0.75 = 0.9225 100% = (1 x 0.116) + (0 x 0.0315) = 11.6% = 1.23 x 1 = 1.23 125% = (1.25 x 0.116) + (-0.25 x 0.0315) = 13.71% = 1.23 x 1.25 = 1.5375 150% = (1.50 x 0.116) + (-0.50 x 0.0315) = 15.82% = 1.23 x…
- Data were collected from a longitudinal study designed to investigate the relationship between blood sugar levels and brain shrinkage. The results of an analysis of the data for 22 observations are shown in the table below. Term Constant Coef -15.668 Blood sugar 0.161 SE Coef 6.154 0.073 Which of the following represents a 98 percent confidence interval for the slope of the least-squares regression line for brain shrinkage on blood sugar levels? Assume the conditions for inference are met. (A) -15.668 ± 2.528(6.154) (B) -15.668 ± 2.518(6.154) (C) 0.161 ± 2.528(0.073) (D) 0.161 ± 2.518(0.073) (E) 0.161 ± 2.197(0.073)calculate t value if df is 24 and 95% confidence intervalIf one explanatory Variable is correlated with the error term, is there any problem of heterosdasticity in the model?
- Do Construct a confidence interval on a slope?The 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: (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 Temp. Jan. 21 185.79 July. 68 621.55 Feb. 24 214.47 Aug. 74 675.06 Mar. 32 288.03 Sept. 62 562.03 Apr.o 50 452.93 47 424.84 Oct. May 50 454.58 Nov. 41 369.95 June 59 539.03 Dec. 30 a 273.98Data 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. Explain
- True or false 1) Chi-square test statistic is always positive. 2) When doing linear regression, the sign of the slope is always same as the sign of the correlation coefficient. 3) A slope that is not statistically significantly different than zero indicates no linear relation between two variables, X and Y. 4) Confidence interval of the estimated slope is always wider than prediction interval.Demand for greensaber in july 2019 is normal distributed with mean 100, standard dev 20.Demand in Aug 2019 is normally distributed with mean 120, standard dev 25. Demand in both months have a correlation of +0.2 1) what isi the expected total demand? 2) What is the stndard deviation of total demand? 3) If there are 220 units of Greensaber at the start of July, and is unable to replenish its inventory until end of august 2019, what is the probability that it fully stocks out of GreenSaber by end of August 2019?A simple linear regression model was created based on 25 samples. The 95% confidence interval for the estimate of B, is (1.5, 4.8). What is thet-statistic for B1? Report your answer in 3 decimal places.