Math Problem Statement

The following data give the experience of machine operators and their performance rating as given by the number of good parts turned out per 100 pieces. Operator 1 2 3 4 5 6 Performance rating (x) 23 43 53 63 73 83 Experience (y) 5 6 7 8 9 10 Calculate the regression line of performance rating on experience and also estimate the probable performance if an operator has 11 years of experience.

Solution

To calculate the regression line of performance rating on experience, we need to follow the formula for the linear regression equation:

y=a+bxy = a + bx

Where:

  • yy is the dependent variable (performance rating),
  • xx is the independent variable (experience),
  • aa is the y-intercept, and
  • bb is the slope of the regression line.

The formulas for aa (intercept) and bb (slope) are as follows:

b=n(xy)(x)(y)n(x2)(x)2b = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2} a=yb(x)na = \frac{\sum y - b(\sum x)}{n}

Where:

  • nn is the number of data points (here n=6n = 6),
  • xx is the experience,
  • yy is the performance rating.

Step 1: Compute the necessary sums

Given data: x=[23,43,53,63,73,83](Performance rating)x = [23, 43, 53, 63, 73, 83] \quad (\text{Performance rating}) y=[5,6,7,8,9,10](Experience)y = [5, 6, 7, 8, 9, 10] \quad (\text{Experience})

We now compute:

  • x\sum x
  • y\sum y
  • x2\sum x^2
  • y2\sum y^2
  • xy\sum xy

Step 2: Plug the values into the formulas

Let’s perform the calculations. numpy as np

Given data

x = np.array([23, 43, 53, 63, 73, 83]) # Performance rating y = np.array([5, 6, 7, 8, 9, 10]) # Experience

Calculate necessary sums

n = len(x) sum_x = np.sum(x) sum_y = np.sum(y) sum_x_squared = np.sum(x2) sum_y_squared = np.sum(y2) sum_xy = np.sum(x * y)

Calculate the slope (b) and intercept (a) of the regression line

b = (n * sum_xy - sum_x * sum_y) / (n * sum_x_squared - sum_x**2) a = (sum_y - b * sum_x) / n

Predicted performance when experience (x) = 11

predicted_y = a + b * 11

b, a, predicted_yThe regression line of performance rating on experience is given by:

y=2.671+0.086xy = 2.671 + 0.086x

Where:

  • yy is the performance rating,
  • xx is the experience in years.

To estimate the performance of an operator with 11 years of experience, we substitute x=11x = 11 into the regression equation:

y=2.671+0.086×11=3.614y = 2.671 + 0.086 \times 11 = 3.614

Thus, the estimated performance rating for an operator with 11 years of experience is approximately 3.61 good parts per 100 pieces.

Would you like to dive deeper into the steps, or is there another related concept you'd like to explore?

Here are 5 follow-up questions:

  1. How would the regression change if the roles of experience and performance rating were swapped?
  2. What is the significance of the slope and intercept in this context?
  3. How can we use correlation to validate the linear regression model?
  4. What are the assumptions made in linear regression analysis?
  5. How can we assess the goodness of fit for this regression model?

Tip: Always check if the data is approximately linear before using linear regression, as this method assumes a linear relationship between the variables.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Regression
Statistics
Algebra

Formulas

y = a + bx
b = (n(Σxy) - (Σx)(Σy)) / (n(Σx²) - (Σx)²)
a = (Σy - b(Σx)) / n

Theorems

Least Squares Method

Suitable Grade Level

Grades 10-12