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Estimate the slope of the regression line manufacturer of cases for sound equipment requires drilling holes for metal screws. The drill bits wear out and must be replaced; there is expense not only in the cost of the bits but also for lost production. Engineers varied the rotation speed of the drill and measured the lifetime in thousands of holes drilled of bits at different speeds in surface feet per minute (sfm). Suppose you want to create a model that estimates the drill bit lifetime ( y) based on rotation speed ( x). Summary statistics for 20 observations collected are given below.

20 20 20 20
Σ xi = 2,000 ; Σ yi = 86.6 Σ xi^2= 216,000 Σ xiyi= 8,338
i= 1 i=1 i=1 i=1

Estimate the slope of the regression line.

a. -0.017
b.-0.037
c. -0.057
d.-0.077

Sagot :

MrRoyalidn

Answer:

[tex]b_i = -0.020125[/tex]

Step-by-step explanation:

Given

[tex]\sum x_i= 2000[/tex]

[tex]\sum y_i= 86.6[/tex]

[tex]\sum x_i^2= 216000[/tex]

[tex]\sum x_iy_i = 8338[/tex]

[tex]n = 20[/tex]

Required

Determine the slope (b) of the regression line

This is calculated as:

[tex]b_i = \frac{\sum xy - \frac{\sum x\sum y}{n}}{\sum x^2 - \frac{(\sum x)^2}{n}}[/tex]

Substitute values for each term, we have:

[tex]b_i = \frac{8338 - \frac{2000 * 86.6}{20}}{216000 - \frac{(2000)^2}{20}}[/tex]

Simplify the numerator

[tex]b_i = \frac{8338 - 8660}{216000 - \frac{(2000)^2}{20}}[/tex]

Simplify the denominator

[tex]b_i = \frac{8338 - 8660}{216000 - 200000}[/tex]

[tex]b_i = \frac{-322}{16000}[/tex]

[tex]b_i = -0.020125[/tex]

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