(e) Use the F test to test the hypotheses in part (d) at a 0.05 level of significance. Present the results in the analysis of variance table format.Set up the ANOVA table. (Round your values for MSE and F to two decimal places, and your p-value to three decimal places.)Sourceof VariationRegressionErrorTotalSumof Squares2531.25298.752830134Degreesof FreedomFind the p-value. (Round your answer to three decimal places.)p-value =MeanSquare2531.2599.58Find the value of the test statistic. (Round your answer to two decimal places.)25.42State your conclusion.O Reject Ho. We conclude that the relationship between x and y is significant.Do not reject Ho. We cannot conclude that the relationship between x and y is significant.Do not reject H. We conclude that the relationship between x and y is significant.Reject H. We cannot conclude that the relationship between x and y is significant.F0.015p-value You may need to use the appropriate technology to answer this question.Consider the data.X;Yi3 1260 40620 1465 5 15(a) Compute the mean square error using equation s² = MSE =SSEn-2(b) Compute the standard error of the estimate using equations = ✓MSE(c) Compute the estimated standard deviation of b, using equation 5b₁(d) Use the t test to test the following hypotheses (α = 0.05):Ho: B₁ = 0H₂: B₁ * 0(Round your answer to two decimal places.)=Σ(ΧSSEn-2Sx)²Find the value of the test statistic. (Round your answer to three decimal places.)8.396x(Round your answer to three decimal places.)(Round your answer to three decimal places.)Find the p-value. (Round your answer to four decimal places.)p-value = 0.0150State your conclusion.Reject H₁. We cannot conclude that the relationship between x and y is significant.Do not reject H. We cannot conclude that the relationship between x and y is significant.Reject H. We conclude that the relationship between x and y is significant.Do not reject Ho. We conclude that the relationship between x and y is significant.

The data shows x and y variables.

Answered: You may need to use the appropriate…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A Bivariate Regression was conducted to evaluate the predictive relationship between total years of schooling and annual income. The results of the regression model were F(1,88) = 4.1, p < .05. What can be concluded about these results? Group of answer choices total years of schooling is a significant predictor of annual income. total years of schooling is not a significant predictor of annual income.
Use the formula to find the standard error of the distribution of differences in sample means, x⎯⎯1−x⎯⎯2x¯1-x¯2.Samples of size 100 from Population 1 with mean 95 and standard deviation 10 and samples of size 80 from Population 2 with mean 76 and standard deviation 17Round your answer for the standard error to two decimal places.standard error =
To estimate the mean score μμ of those who took the Medical College Admission Test on your campus, you will obtain the scores of an SRS of students. From published information you know that the scores are approximately Normal with standard deviation about 6.56.5 . You want your sample mean ¯xx¯ to estimate μμ with an error of no more than 0.90.9 point in either direction. (a) What standard deviation must ¯xx¯ have so that 99.7%99.7% of all samples give an ¯xx¯ within 0.90.9 point of μμ ? Use the 68–95–99.768–95–99.7 rule. (Enter your answer rounded to four decimal places.) (b) How large an SRS do you need in order to reduce the standard deviation of ¯xx¯ to the value you found? (Enter your answer rounded to the nearest whole number.)
Use the formula to find the standard error of the distribution of differences in sample means, x⎯⎯1−x⎯⎯2x¯1-x¯2.Samples of size 40 from Population 1 with mean 3.7 and standard deviation 1.9 and samples of size 40 from Population 2 with mean 1.9 and standard deviation 1.2Round your answer for the standard error to two decimal places.standard error = Enter your answer in accordance to the question statement
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Consider the data. xi 1 2 3 4 5 yi 3 7 5 12 14 (a) Compute the mean square error using equation s2 = MSE = SSE n − 2 (Round your answer to two decimal places.) (b) Compute the standard error of the estimate using equation s = MSE = SSE n − 2 . (Round your answer to three decimal places.) (c) Compute the estimated standard deviation of b1 using equation sb1 = s Σ(xi − x)2 . (Round your answer to three decimal places.) (d) Use the t test to test the following hypotheses (? = 0.05): H0: β1 = 0 Ha: β1 ≠ 0 Find the value of the test statistic. (Round your answer to three decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value = Use the F test to test the hypotheses in part (d) at a 0.05 level of significance. Present the results in the analysis of variance table format. Set up the ANOVA table. (Round your values for MSE and F to two decimal…
Consider the data. 1 2 3 4 5 10 15 SSE (a) Compute the mean square error using equation s = MSE = (Round your answer to two decimal places.) n - 2 SSE (b) Compute the standard error of the estimate using equation S = V MSE = (Round your answer to three decimal places.) n - 2 (c) Compute the estimated standard deviation of b, using equation sS. = . (Round your answer to three decimal places.) E(x, - x)2 (d) Use the t test to test the following hypotheses (a = 0.05): Ho:B1 = 0 Hoißa Find the value of the test statistic. (Round your answer to three decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value = %3D State your conclusion. O Do not reject Ho. We cannot conclude that the relationship between x and y is significant. O Do not reject Ho. We conclude that the relationship between x and y is significant. O Reject H.. We cannot conclude that the relationship between x and y is significant. O Reject Ho. We conclude that the relationship between x and y is…
a b Absolute standard deviation = Coefficient of variation = y = 1.20 (+0.02) x 10-8-3.60(±0.2) × 10-⁹ x Result = Absolute standard deviation = Coefficient of variation = C Result = d y=90.95 (±0.08) - 89.40(±0.06) +0.200(+0.004) Absolute standard deviation == Coefficient of variation = Result = ± y= y = 0.0040 (+0.0005) x 10.28 (+0.02) × 395(+1) Result = ± Absolute standard deviation = Coefficient of variation = % 329(+0.03) x 10-14 1.47 (+0.04) x 10-16 % % H % ±
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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

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