Math Problem Statement

The table below gives the list price and the number of bids received for five randomly selected items sold through online auctions. Using this data, consider the equation of the regression line, yˆ=b0+b1x , for predicting the number of bids an item will receive based on the list price. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant.

Price in Dollars 23 34 44 46 50 Number of Bids 1 2 4 9 10 Table

Step 2 of 6 : Find the estimated y-intercept. Round your answer to three decimal places.

Solution

To find the estimated y-intercept b0b_0 of the regression line y^=b0+b1x\hat{y} = b_0 + b_1x, we will use the formula:

b0=yˉb1xˉb_0 = \bar{y} - b_1 \bar{x}

Where:

  • yˉ\bar{y} is the mean of the number of bids (the dependent variable),
  • xˉ\bar{x} is the mean of the list prices (the independent variable),
  • b1b_1 is the slope of the regression line, which we will calculate first.

Step 1: Calculate the means of xx (list price) and yy (number of bids).

The values for xx (price in dollars) are: 23,34,44,46,5023, 34, 44, 46, 50

The values for yy (number of bids) are: 1,2,4,9,101, 2, 4, 9, 10

First, let's compute the means:

xˉ=xn=23+34+44+46+505=1975=39.4\bar{x} = \frac{\sum x}{n} = \frac{23 + 34 + 44 + 46 + 50}{5} = \frac{197}{5} = 39.4

yˉ=yn=1+2+4+9+105=265=5.2\bar{y} = \frac{\sum y}{n} = \frac{1 + 2 + 4 + 9 + 10}{5} = \frac{26}{5} = 5.2

Step 2: Calculate the slope b1b_1

The formula for the slope b1b_1 is:

b1=(xixˉ)(yiyˉ)(xixˉ)2b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}

We need to compute the following:

  1. (xixˉ)(x_i - \bar{x}) for each xx,
  2. (yiyˉ)(y_i - \bar{y}) for each yy,
  3. The product (xixˉ)(yiyˉ)(x_i - \bar{x})(y_i - \bar{y}) for each pair,
  4. The squared differences (xixˉ)2(x_i - \bar{x})^2.

Let me now calculate b1b_1 and then b0b_0. numpy as np

Given data

x = np.array([23, 34, 44, 46, 50]) y = np.array([1, 2, 4, 9, 10])

Calculate means

mean_x = np.mean(x) mean_y = np.mean(y)

Calculate slope (b1)

numerator = np.sum((x - mean_x) * (y - mean_y)) denominator = np.sum((x - mean_x)**2) b1 = numerator / denominator

Calculate y-intercept (b0)

b0 = mean_y - b1 * mean_x

Round the y-intercept to three decimal places

round(b0, 3)The estimated y-intercept b0b_0 of the regression line is approximately -7.784.

Would you like to proceed with calculating the slope or any other details? Here are a few related questions:

  1. How would you calculate the slope b1b_1 for this data?
  2. How can you assess whether the correlation coefficient is statistically significant?
  3. What would be the interpretation of the y-intercept value in this context?
  4. How do you use the regression line to make predictions based on the list price?
  5. What are the assumptions of linear regression that we need to consider in this analysis?

Tip: Always check the residuals (the differences between actual and predicted values) to assess the goodness of fit and whether the linear model is appropriate.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Regression
Statistics
Correlation

Formulas

yˆ = b0 + b1x
b0 = ȳ - b1x̄

Theorems

Least Squares Method
Properties of Linear Regression

Suitable Grade Level

Grades 10-12