Use the following information for questions:
A sample of 19 female bears was measured for chest girth (y) and neck girth (x), both in inches. The least squares regression equation relating the two variables is $\hat{y}$= 5.32 + 1.53 x. The standard deviation of the regression is s = 3.692.
-What is a 95% prediction interval for the chest girth of a bear with a neck girth of 20 inches? The standard error of the fit, s.e.(fit), = 0.880
A) (27.9, 43.9)
B) (28.1, 43.7)
C) (34.1, 37.8)
D) None of the above

Question

Use the following information for questions:
 A sample of 19 female bears was measured for chest girth (y) and neck girth (x), both in inches. The least squares regression equation relating the two variables is $\hat{y}$= 5.32 + 1.53 x. The standard deviation of the regression is s = 3.692.
-What is a 95% prediction interval for the chest girth of a bear with a neck girth of 20 inches? The standard error of the fit, s.e.(fit), = 0.880
A) (27.9, 43.9)
B) (28.1, 43.7)
C) (34.1, 37.8)
D) None of the above

Quizplus · Accepted Answer

To calculate the 95% prediction interval for the chest girth of a bear with a neck girth of 20 inches, we first use the regression equation to predict the chest girth: $\hat{y} = 5.32 + 1.53(20) = 36.32$. The prediction interval is given by $\hat{y} \pm t_{\alpha/2, n-2} \times s \times \sqrt{1 + \frac{1}{n} + \frac{(x - \bar{x})^2}{\sum(x_i - \bar{x})^2}}$. Since we're not given $\bar{x}$, $\sum(x_i - \bar{x})^2$, or the exact t-value, we simplify using the provided standard error of the fit for a single prediction, which is 0.880. Thus, the interval is $36.32 \pm (1.96 \times 3.692)$ (using 1.96 for a 95% confidence level from a standard normal distribution, assuming large sample size or t-distribution approximates to normal as n > 30). This calculation directly doesn't match any of the options, indicating a need for correction in calculation or interpretation. The correct approach involves using the standard error of the fit specifically for the prediction interval calculation, which is $36.32 \pm (1.96 \times 0.880)$, but without the exact values and steps to reach the options provided, the correct answer involves understanding that a prediction interval includes the predicted value plus a margin of error that accounts for the standard error of the prediction and the variability in the y-values around the regression line (captured by s). Given the options, none directly follow from the simplified explanation, suggesting a more detailed calculation or a different interpretation is needed. However, the closest approach based on the given information and standard practices in statistics would involve calculating the margin of error using the standard error of the fit and applying it to the predicted value, which typically results in a range around the predicted value. The correct choice should reflect an understanding of these principles, but without the exact calculations or additional information (like the t-value for the specific degrees of freedom), pinpointing the exact option is challenging. The response assumes a general understanding of statistical principles for constructing prediction intervals.

Use the Following Information for Questions: a Sample of 19

Use the Following Information for Questions:
a Sample of 19