Question

The data below are the temperatures on randomly chosen days during the summer and the number of employee absences at a local company on those days. What is the 95% confidence interval about the slope of the true least-squares regression line, for the data given below?

Temperature (x)	Number of absences (y)
72	3
85	7
91	10
90	10
88	8
98	15
75	4
100	15
80	5

• A. \((0.371, 0.527)\)
• B. \((0.367, 0.530)\)
• C. \((-1.760, 2.658)\)
• D. \((0.385, 0.513)\)

Hint:

Find the least-squares slope \(b=S_{xy}/S_{xx}\), then use \(b \pm t_{n-2}\cdot SE(b)\) with \(t_{0.025,7}=2.365\).

Answer 1

The Correct Option is B

Step 1: List the paired data and compute the basic sums needed for regression, with \(n = 9\).

\[ \sum x = 779, \quad \sum y = 77, \quad \sum x^2 = 68163, \quad \sum y^2 = 813, \quad \sum xy = 6995 \]

Step 2: Compute the corrected sums of squares and cross products.

\[ S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 68163 - \frac{779^2}{9} = 68163 - 67426.78 = 736.22 \]\[ S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 813 - \frac{77^2}{9} = 813 - 658.78 = 154.22 \]\[ S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 6995 - \frac{779 \times 77}{9} = 6995 - 6664.78 = 330.22 \]

Step 3: Compute the slope of the least-squares line.

\[ b = \frac{S_{xy}}{S_{xx}} = \frac{330.22}{736.22} = 0.4486 \]

Step 4: Compute the residual (error) sum of squares and the standard error about the regression line.

\[ SSE = S_{yy} - b \cdot S_{xy} = 154.22 - 0.4486 \times 330.22 = 154.22 - 148.16 = 6.07 \]\[ s = \sqrt{\frac{SSE}{n-2}} = \sqrt{\frac{6.07}{7}} = \sqrt{0.867} = 0.931 \]

Step 5: Compute the standard error of the slope estimate.

\[ SE(b) = \frac{s}{\sqrt{S_{xx}}} = \frac{0.931}{\sqrt{736.22}} = \frac{0.931}{27.13} = 0.0343 \]

Step 6: For \(n - 2 = 7\) degrees of freedom, the two-tailed critical value at 95% confidence is \(t_{0.025,7} = 2.365\). The margin of error for the slope is:

\[ E = t \times SE(b) = 2.365 \times 0.0343 = 0.0811 \]

Step 7: The 95% confidence interval for the true slope is:

\[ b - E \text{ to } b + E = 0.4486 - 0.0811 \text{ to } 0.4486 + 0.0811 = 0.367 \text{ to } 0.530 \]

The correct answer is (B) (0.367, 0.530).