Consider the data.2 3 4x, 1Y, 2 8 5 12 135(a) Compute the mean square error using equation s2 = MSE =SSEn-2(Round your answer to two decimal places.)(b) Compute the standard error of the estimate using equations = √MSE=SSEn-2Find the p-value. (Round your answer to four decimal places.).p-value=(c) Compute the estimated standard deviation of b, using equation Sp₁ √(x,x) (Round your answer to three decimal places.)(d) Use the t test to test the following hypotheses (-0.05):Ho B₁-0HB₁0Find the value of the test statistic. (Round your answer to three decimal places)(Round your answer to three decimal places.)State your conclusion.O Reject Ho 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 significantO Do not reject Ho We cannot conclude that the relationship between x and y is significant.O Do not reject H We conclude that the relationship between x and y is significant.(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. (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 SquaresDegreesof FreedomMeanSquareFind the value of the test statistic. (Round your answer to two decimal places.)Find the p-value. (Round your answer to three decimal places.)p-value=FState your conclusion.O Reject Ho. We cannot conclude that the relationship between x and y is significant.O Do not reject Ho. We cannot conclude that the relationship between x and y is significant.O Reject H. We 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.p-value

The given data is as follows: x y 1 2 2 8 3 5 4 12 5 13

Answered: Consider the data. 1 2 3 4 5 y 28 5 12…

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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Consider the data. x 12 3 Y₁ 4 6 51 4 5 10 14 (a) Compute the mean square error using equation s² = MSE = SSE n-2 (Round your answer to two decimal places.)
A regression of average monthly expenditure (AME, measured in euros) on average monthly income (AMI, measured in euros) using a random sample of college-educated full-time workers earnings €100 to €1.5 million yields the following: AME = 710.7 + 8.8 × AMI, R2 = 0.030, SER = 540.30 d. What does the regression predict will be the expenditure of a person with an income of €100? With an income of €200? e. Will the regression give reliable predictions for a person with an income of €2 million? Why or why not? ( f. Given what you know about the distribution of earnings, do you think it is plausible that the distribution of errors in the regression is normal? (Hint: Do you think that the distribution is symmetric or skewed? What is the smallest value of earnings, and is it consistent with a normal distribution?).
A paper gives data on x = change in Body Mass Index (BMI, in kilograms/meter2) and y = change in a measure of depression for patients suffering from depression who participated in a pulmonary rehabilitation program. The table below contains a subset of the data given in the paper and are approximate values read from a scatterplot in the paper. BMI Change (kg/m²) Depression Score Change S = The accompanying computer output is from Minitab. Depression score change 15- 10- -0.5 S Fitted Line Plot Depression score change = 6.577 +5.440 BMI change 20- 5.30586 Coefficients T 0.0 0.5 -0.5 R-sq 25.96% - 1 Term Coef Constant 6.577 BMI change 5.440 % 0.5 BMI change 1.0 SE Coef 2.28 2.90 9 0 0.1 0.7 0.8 1 1.5 4 T-Value 2.88 1.87 Interpret this estimate. s is the typical amount by which the ---Select--- line. 4 5 Regression Equation Depression score change = 6.577 +5.440 BMI change P-Value 0.0164 0.0906 S 5.30586 25.96% R-Sq R-Sq (adj) 18.56% 8 (b) Give a point estimate of o. (Round your answer to…
A survey collected data from a random sample of 121 people living in Jade city. The sample average of the distance people travel to reach their workplaces (Y) is 23.28 km and the standard deviation (sy) is 7.48 km. The standard error of the sample average of the distance people travel to reach their workplaces is 68 km. (Round your answer to two decimal places.) Let μy denote the mean of the distance all the people in Jade city travel to reach their workplaces. The p-value of the test Ho: #y #22 km vs. H₁: Hy #22 km is (Round your answer to two decimal places.)
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=6.22 b=21.6 r2=0.477481 r=0.691 Use this to predict the final exam score of a student who studies 8 hours per week, and please round your answer to a whole number.
Please help!
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 % ±
Write an absolute value function f(t) to describe an individual variance from normal body temperature where t is the individual current temperature
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The regression line for Y vs X is given by Y = 0.82X + 59.1. The standard deviations for X and Y are 1.5 and 2.2 respectively. Suppose now we construct a regression line that uses Y to predict X. The predicted average increase of X when Y is increased by 1 unit is ______________. (Give your answer correct to 2 decimal places.)
In the model in the table below, how is "1.3" interpreted? Model R2 Standard error y 5 + 1.3x 0.68 2.56 A) 130% of the variation in y is accounted for by x. B) On the average, the deViation of the estimated value from the actual value is 1.3. C) Whenever the value of x is 0, the value of y is 1.3. D) For every unit of change in x, there is a corresponding change in y by 1.3.

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