In a study of the performance of a new engine design the...

In a study of the performance of a new engine design, the weight of 22 aircrafts (in tons) and the top speed (in mph) were recorded. A regression line was generated and shown to be an appropriate description of the relationship. The results of the regression analysis are below. Depend Variable: Top Speed Variable Constant Weight Coefficient 11.6559 3.47812 s.e. of Coeff 0.3153 0.294 t-ratio 37 11.8 prob ≤ 0.0001 ≤ 0.0001 R squared = 87.5% R squared (adjusted) = 86.9% s = 0.6174 with 22 - 2 = 20 degrees of freedom Part A: What is the LSRL based on the analysis provided? Make sure to identify what the variables represent in the context of the problem. (4 points) Part B: What is the predicted value for the top speed of an aircraft if its weight is 100 tons? Show your work. (6 points)

Transcript

00:01 Okay so let me here let me make a let me make a table okay so let this is variable okay and this is weight this is r square weight and r square this is correlation okay correlation this is s e up estimate okay s e up estimate and this is n okay this is coefficient this is s e this is coefficient of t score t score and this is p value let me write it with another color okay so this is 11 .6559 this is 0 .3153 37 and this is less than equal to 0 .001 okay and this is also this then equal to 0 .001 now this is 11 .8 3 .8 3 .7 812 okay okay and this is 3 .47812 okay and the r square is sorry sorry sorry let me so let me write it here okay so the r square is 86 .9 percent okay and the core is 0 .932 and this is 22 sorry this is 0 .617 this is 0 .617 4 okay and n is 22.

01:55 Now let in part a to provide the regression equation best done the analysis provided okay so the regression equation will be equal to a plus b x okay and these values helps to find the response variable speed for a given value of the predictor weight we can find these under the coefficients column of the table a is the constant and b is the coefficient of x okay so the y intercept is y intercept is 11 .6 5 59 and b is the slope which is 3 .4781 and the regression equation will now be equal to 11 .655 9 plus 3 .47 .81 x okay now in part b here let in part b so here we need to test if there is a relationship between weight and top speed so part this we can conduct a slope test where slope shows the magnitude of the change in y due to the unit change in x okay so this is a linear regression slope test so the conditions to be made to conduct the test are the y values should be independent for each x and the y should be roughly normally normally distributed and the variance of y should be same for given value of x and there must be a linear relationship between x and y okay now the null slope b is zero and the alternative slope b is not equal to zero okay and this is the null and this is the null and this is the alternative now from the coefficient uptis let the test statistic are the test core is b bar minus b not divided by s e not okay so from that let the test statistic is 11 .8303 and the p value is less than 0 .001 so it is less than the level of significance so we reject the null hypothesis reject h0 okay at 1 % so there is a sufficient evidence of the relationship between weight in the top speed okay now let in part see here let this is the final part so for a 1 minus alpha person confidence interval we have the following formula this is b plus minus t sub alpha by 2 n minus 2 s eb okay and the confidence value is t sub alpha by 2 n minus 2 is t sub 0 .02 pipe and 20 alpha is equal to 1 minus 0 .9 pipe which is 0 .05 and the degrees of freedom is n minus 1 which is 20 okay now using the table i this value is a 3 .47 .81 plus minus 2 .08596 times 0 .294...

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