The actual interpolated critical value at df 24 and a confidence interval of 95

The actual interpolated critical value at df 24 and a

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The actual interpolated critical value at df = 24 and a confidence interval of 95% will be slightly different from 2.060 but not enough to change the conclusion that the values of a and b are not random.2b. Lagging the predictor variable eliminates the inverse relationship between DMH and the value ofthe indirect cost pool. Y = $204,000 + $2.40 × 100,000 DMH = $444,000using the low value of the intercept value. But Howard cannot simply take this at face value because it is higher than the highest cost on the high–low graph. While the value of X = 100,000 is well within the relevant range of practical capacity, the estimate cost has never been experienced in almost nine years ofquarterly observed data [(26 × 4)/12  8.7 years]. 2c. Using the lagged cost driver DMH provides a better cost function that depicts the linear relationship between DMH and the value of the indirect cost pool. Notice, however, that it does not explain why this cost behaviour exists. While the scatter of blue dots indicating actual lagged values of X,Y are close to the red linear regression line plotted through the predicted values (X, y), Howard still does not know why. Until this is determined, the economic plausibility of the systematic change remains in doubt.The high–low method provides poor estimates in both instances. In the first example, the estimate of thesize of the cost pool is negative, which is impossible. In the second example, it is almost double most of the observed values of the cost pool over the past 25 quarters, which is also impossible.10-28(25 min.) OLS simple linear regression analysis
The basic data used in 1a. are:The chart of the OLS regression line and the summary of relevant statistics from the Excel results are:1a.Actual revenue from longstanding customers is $5,000 × 10 months = $50,000. Total revenue of $720,000 less $50,000 fixed = $670,000 variable. Variable revenue per advertising dollar = $670,000/$28,200 = $23.76. Therefore, Y = $5,000 + (23.76 × $1,900) = $50,144.1b.The high and low points of the revenue driver, advertising dollars, are $1,200 and $4,800 with corresponding revenues of $66,000 and $96,000. This results in Y = $56,000 + (8.33 × $1,900) = $71,827 1c.Y = 47,402 + (8.723 × $1,900) = $63,9763. Clean data is reliable data. The more reliable the input data, the better the partners' decisions will be. Reliability in an OLS simple linear regression requires far more than 10 data points. To clean up this data, the first thing the partners should do is collect more data points. They could consider using weekly instead of monthly data but then need to carefully assess the report on the residuals to assure there is no autocorrelation and no serial correlation in this time series data. With so few data points , there is a relatively large standard error for the values of a and b. This means there will be a wide range within which the observed future value of y when X = $1,900 could fall. While the explanatory power is reasonable at about 64.3%, this only refers to the explained portion of change in revenue. There remains unexplained revenue of approximately $42,170 or 78% of the actual revenue in September and 44% of actual revenue in October.

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