A consumer's spending is widely believed to be a function of their income. To estimate this relationship, a university professor randomly selected 19 of his students and collected information on their spending (Y, in dollars) and income (X, in dollars) patterns in week 6 of the semester. Assuming a linear relationship between Y and X, the professor used the least-squares method and found that the Y intercept = 20.90 and the slope = 0.66. Also, the sum of squares total (SST) and the regression sum of squares (SSR) were equal to 65600.74 and 52831.23, respectively. What is the upper bound of a 95% prediction interval for a student's spending on a weekly income of $60. Assume the h Statistic equals 0.12. a. greater than $80 but less than $120 b. greater than $100 but less than $120 c. greater than $120 d. less than $100
Question
A consumer's spending is widely believed to be a function of their income. To estimate this relationship, a university professor randomly selected 19 of his students and collected information on their spending (Y, in dollars) and income (X, in dollars) patterns in week 6 of the semester. Assuming a linear relationship between Y and X, the professor used the least-squares method and found that the Y intercept = 20.90 and the slope = 0.66. Also, the sum of squares total (SST) and the regression sum of squares (SSR) were equal to 65600.74 and 52831.23, respectively. What is the upper bound of a 95% prediction interval for a student's spending on a weekly income of 80 but less than 100 but less than 120 d. less than $100
Solution
To answer this question, we first need to calculate the predicted spending (Y') using the given Y intercept and slope. The formula for a linear regression line is Y' = a + bX, where a is the Y intercept, b is the slope, and X is the independent variable (income in this case).
Given: a = 20.90 b = 0.66 X = 60
Substitute these values into the formula:
Y' = 20.90 + 0.66 * 60 = 59.90
Next, we need to calculate the standard error of the estimate (SEE), which is a measure of the accuracy of predictions. The formula for SEE is sqrt((SST - SSR) / (n - 2)), where n is the number of observations.
Given: SST = 65600.74 SSR = 52831.23 n = 19
Substitute these values into the formula:
SEE = sqrt((65600.74 - 52831.23) / (19 - 2)) = sqrt(12769.51 / 17) = 27.63
Finally, we calculate the upper bound of the 95% prediction interval using the formula Y' + t(0.025, n-2) * sqrt(1 + h) * SEE, where t(0.025, n-2) is the t-value for a 95% confidence level and n-2 degrees of freedom, and h is the h statistic.
Given: t(0.025, 17) = 2.11 (approximate value from t-table) h = 0.12
Substitute these values into the formula:
Upper bound = 59.90 + 2.11 * sqrt(1 + 0.12) * 27.63 = 59.90 + 2.11 * sqrt(1.12) * 27.63 = 59.90 + 2.11 * 1.06 * 27.63 = 59.90 + 62.42 = 122.32
So, the upper bound of a 95% prediction interval for a student's spending on a weekly income of 120. Therefore, the answer is (c) greater than $120.
Similar Questions
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