Standard Error of Estimate A random sample of 118 different fe...

Question

Standard Error of Estimate A random sample of 118 different female statistics students is obtained and their weights are measured in kilograms and in pounds. Using the 118 paired weights (weight in kg, weight in lb), what is the value of se? For a female statistics student who weighs 100 lb, the predicted weight in kilograms is 45.4 kg. What is the 95% prediction interval?

Accepted Answer

Welcome back, everyone. In this problem, a random sample of 130 male engineering students is obtained, and their heights are measured in inches and in centimeters. From the regression analysis, the predicted height in centimeters for a student who is 70 inches tall is found to be 177.8 centimeters. The sum of squared residuals from the regression is 42.25 square centimeters. First, what is the value of the standard error of estimate? And second, for a student whose height is 70 inches, what is the 95% prediction interval? It says the standard error of estimate is 0.46 centimeters and the 95% prediction interval is from 176.45 to 179.15 centimeters. B says it's 0.57 centimeters with an interval from 176.66 to 178.94 centimeters. C says it's 0.46 centimeters with an interval from 176.45 centimeters to 178.94 centimeters and the D says it's 0.57 centimeters with an interval from 176.66 to 179.15 centimeters. Now let's start with the first part of our problem. What is the value of the standard error of estimate? Well, thinking about the information that we have so far, we know that the sample size N is 130, OK, because it's 130 male engineering students. We also know that the predicted height. Add 70 inches. OK, is Y minus 177.8 centimeters, OK. And we know that the sum of squared residuals, OK. Is going to be equal to 42.25 square centimeters. So how can that help us? Well, let me put this in blue, recall that the standard error of estimate is equal to the square root of the sum of squared residuals. Divided by n minus 2. We have those values because we know that the sum of squared residuals is 42.25 square centimeters and N is 130. So when we substitute those values into our formula, OK. Then we should find that we get the standard error of estimate to be equal to 0.5745 or approximately 0.57 to two decimal places, OK, like the way the rest of our answers are written. So, our standard error of estimate is approximately 0.57 centimeters. Let's move on to the second part of our problem. For a student whose height is 70 inches, what is the 95% prediction interval? Well, since we have N data points and we are estimating two parameters, then the number of degrees of freedom, OK, the degrees of freedom. Would be equal to n minus 2. That's 130 minus 2, which equals 128. Nor recall that the prediction interval OK. The values in the prediction interval. It's going to be the predicted value. OK. Plus R minus T or T value from our table multiplied by the standard error of estimate and we choose the value of T based on the confidence level and the degrees of freedom. No, all we need, since we have the standard error of estimate is and the predicted value, sorry, is our T value from our table from the tea table. And no In a two-tailed test, OK. In a two-tailed test. With. A 95% confidence level. 95% confidence level. And At 128, OK. rather than at 11, it's level. And 128 degrees of freedom. OK. Leaving 2.5% in each tail, then our T value, OK, 0.025, 2.5% in each tail, to 128 degrees of freedom, is going to be approximately equal. 2 1.98. I know that we have the T value. This implies then that our prediction interval. Is going to be equal to our predicted value 177.8 plus or minus 1.98 multiplied by 0.5745. And when we saw it here, then we should get it to be 177.8 plus or minus 1.13751. Now, if we add or subtract it to get both sides or to get the upper and lower uh limit of the interval, then that means the lower bound, OK, the lower bound of our interval is going to be 177.8 minus 1.13751. Which equals 176.66249 or approximately 176.66 centimeters. While the upper bound It's gonna be equal to 177.8 plus 1.13751, which equals 178.93751, or approximately 178.94 centimeters. Thus our standard error of estimate is 5 0.57 centimeters while the limits of our interval, or rather our interval, or 95% prediction interval goes from 176.66 centimeters to 178.94 centimeters. Therefore, B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Standard Error of Estimate (SE)

Prediction Interval

Linear Regression

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