Math Problem Statement

An experiment has shown the relationship between Voltage (V) and Current (A). The problem asks to calculate the regression line y = a + bx using given data, determine specific values like n, tabulate results, and find the regression line equation with corresponding calculations.

Solution

Let's solve this step by step based on the image. Here's how we'll proceed:


(a) Determine the value of nn:

The value of nn is the number of data points provided. From the table, there are 5 rows (pairs of xx and yy), so: n=5n = 5


(b) Tabulate the results for xx, yy, x2x^2, and xyxy:

We'll calculate x2x^2 (the square of xx) and xyxy (the product of xx and yy) for each data point:

xxyyx2x^2xyxy
00.1000
30.1390.39
50.14250.70
80.17641.36
100.191001.90

(c) Calculate the values of aa and bb:

To calculate aa and bb, use the formulas: b=nxyxynx2(x)2b = \frac{n \sum xy - \sum x \sum y}{n \sum x^2 - (\sum x)^2} a=ybxna = \frac{\sum y - b \sum x}{n}

Step 1: Compute the sums:

x=0+3+5+8+10=26\sum x = 0 + 3 + 5 + 8 + 10 = 26 y=0.10+0.13+0.14+0.17+0.19=0.73\sum y = 0.10 + 0.13 + 0.14 + 0.17 + 0.19 = 0.73 x2=0+9+25+64+100=198\sum x^2 = 0 + 9 + 25 + 64 + 100 = 198 xy=0+0.39+0.70+1.36+1.90=4.35\sum xy = 0 + 0.39 + 0.70 + 1.36 + 1.90 = 4.35

Step 2: Substitute into bb:

b=(5)(4.35)(26)(0.73)(5)(198)(26)2b = \frac{(5)(4.35) - (26)(0.73)}{(5)(198) - (26)^2} b=21.7518.98990676b = \frac{21.75 - 18.98}{990 - 676} b=2.773140.0088b = \frac{2.77}{314} \approx 0.0088

Step 3: Substitute into aa:

a=0.73(0.0088)(26)5a = \frac{0.73 - (0.0088)(26)}{5} a=0.730.22885a = \frac{0.73 - 0.2288}{5} a=0.501250.1002a = \frac{0.5012}{5} \approx 0.1002


(d) Write the regression line:

The regression line is: y=0.1002+0.0088xy = 0.1002 + 0.0088x

In terms of current (II) and voltage (VV): I=1RV+aI = \frac{1}{R}V + a 1R=0.0088anda=0.1002\frac{1}{R} = 0.0088 \quad \text{and} \quad a = 0.1002

Thus: I=0.0088V+0.1002I = 0.0088V + 0.1002


Would you like a detailed explanation of any specific part or further calculations?


Follow-up Questions:

  1. Why is calculating the regression line important in this experiment?
  2. How does 1R\frac{1}{R} relate to the physical properties of the circuit?
  3. What does the slope of the regression line represent in the context of the experiment?
  4. How would the regression line change if another data point was added?
  5. What is the significance of rounding values to 4 decimals in this calculation?

Tip: Always double-check your summations and calculations to ensure accuracy, especially when dealing with decimals.

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Math Problem Analysis

Mathematical Concepts

Linear Regression
Summation
Slope and Intercept Calculation
Data Tabulation

Formulas

b = (nΣxy - ΣxΣy) / (nΣx² - (Σx)²)
a = (Σy - bΣx) / n
y = a + bx

Theorems

Least Squares Regression

Suitable Grade Level

Grade 11-12 or introductory college level