Compute the Standard Error of the Estimate
-The data below are the final exam scores of 10 randomly selected engineering students and the number of hours they slept the night before the exam. Find the standard error of estimate, se given that $\mathrm { y } = 5.044 \mathrm { x } + 56.11$.
$\begin{array}{l|c|c|c|c|c|c|c|c|c|c|}
\text { Hours, } \mathrm{x} & 3 & 5 & 2 & 8 & 2 & 4 & 4 & 5 & 6 & 3 \
\hline \text { Scores, } \mathrm{Y} & 65 & 80 & 60 & 88 & 66 & 78 & 85 & 90 & 90 & 71
\end{array}$

A) $6.305$
B) $7.913$
C) $8.912$
D) $9.875$

Question

Compute the Standard Error of the Estimate
-The data below are the final exam scores of 10 randomly selected engineering students and the number of hours they slept the night before the exam. Find the standard error of estimate, se given that $\mathrm { y } = 5.044 \mathrm { x } + 56.11$.
$\begin{array}{l|c|c|c|c|c|c|c|c|c|c|}
\text { Hours, } \mathrm{x} & 3 & 5 & 2 & 8 & 2 & 4 & 4 & 5 & 6 & 3 \
\hline \text { Scores, } \mathrm{Y} & 65 & 80 & 60 & 88 & 66 & 78 & 85 & 90 & 90 & 71
\end{array}$

A) $6.305$
B) $7.913$
C) $8.912$
D) $9.875$

Quizplus · Accepted Answer

The standard error of estimate is the standard deviation of the residuals, which is the difference between the actual value of y and the predicted value of y. The standard error of estimate is calculated as:$$se = \sqrt{\frac{\sum(y_i - \hat{y}_i)^2}{n-2}}$$where $y_i$ is the actual value of y, $\hat{y}_i$ is the predicted value of y, and n is the sample size.In this case, the standard error of estimate is:$$se = \sqrt{\frac{(65 - 64.33)^2 + (80 - 80.22)^2 + (60 - 59.33)^2 + (88 - 88.66)^2 + (66 - 65.33)^2 + (78 - 79.33)^2 + (85 - 84.66)^2 + (90 - 89.33)^2 + (90 - 90.66)^2 + (71 - 70.33)^2}{10-2}}$$$$se = 6.305$$Therefore, the standard error of estimate is 6.305.

Compute the Standard Error of the Estimate -The Data Below y=5.044x+56.11\mathrm { y } = 5.044 \mathrm { x } + 56.11y=5.044x+56.11

Compute the Standard Error of the Estimate
-The Data Below $\mathrm { y } = 5.044 \mathrm { x } + 56.11$