The table shows a meteorologists predicted temperatiures for an April day in Washington D.C. Use quadratic regression to find a quadratic model for this data. Use the 24 hour clock to represent times after noon).
time predicted temperature(F)
8a.m. 50.68
10a.m. 62.24
12p.m. 69.75
2p.m. 73.2
4p.m. 72.59
6p.m. 67.93
Would really appreciate if someone could please explain.
To find a quadratic model for the given data, we can use quadratic regression. Quadratic regression is a method used to find the best-fit curve for a set of data points.
Step 1: Assign variables to the given data.
Let the hours be represented by x, and the predicted temperatures be represented by y.
x = [8, 10, 12, 14, 16, 18]
y = [50.68, 62.24, 69.75, 73.2, 72.59, 67.93]
Step 2: Create a table for the quadratic regression formulas.
Create a new table with three columns: x, x^2 (x squared), and y.
x | x^2 | y
8 | 64 | 50.68
10 | 100 | 62.24
12 | 144 | 69.75
14 | 196 | 73.2
16 | 256 | 72.59
18 | 324 | 67.93
Step 3: Use the quadratic regression formulas.
The quadratic regression formulas are:
Σx - sum of all x values
Σx^2 - sum of all x^2 values
Σy - sum of all y values
Σxy - sum of all (x * y) values
n - number of data points
The formulas to calculate the coefficients of the quadratic model (a, b, and c) are:
a = (Σy * Σx^2 - Σx * Σxy) / (n * Σx^2 - (Σx)^2)
b = (n * Σxy - Σx * Σy) / (n * Σx^2 - (Σx)^2)
c = (Σy - a * Σx - b * n) / n
Using these formulas, we can calculate the values of a, b, and c.
Σx = 8 + 10 + 12 + 14 + 16 + 18 = 78
Σx^2 = 64 + 100 + 144 + 196 + 256 + 324 = 1084
Σy = 50.68 + 62.24 + 69.75 + 73.2 + 72.59 + 67.93 = 396.39
Σxy = (8 * 50.68) + (10 * 62.24) + (12 * 69.75) + (14 * 73.2) + (16 * 72.59) + (18 * 67.93) = 6714.07
n = 6
a = (396.39 * 1084 - 78 * 6714.07) / (6 * 1084 - 78^2)
b = (6 * 6714.07 - 78 * 396.39) / (6 * 1084 - 78^2)
c = (396.39 - a * 78 - b * 6) / 6
Calculating the values:
a = (-17152.1644) / (-3084) ≈ 5.5628
b = (4028.92) / (-3084) ≈ -1.3053
c = (396.39 - 5.5628 * 78 - (-1.3053) * 6) / 6 ≈ 47.6072
Hence, the quadratic model for the given data is:
y = 5.5628x^2 - 1.3053x + 47.6072
Note: The units for x are hours (after noon) and the units for y are degrees Fahrenheit.
To find a quadratic model for the given data, we need to use quadratic regression. Quadratic regression involves finding a quadratic equation of the form y = ax^2 + bx + c that best fits the given data.
First, let's create a table using the given data with the time in the 24-hour clock format:
time (hours) predicted temperature (F)
8 50.68
10 62.24
12 69.75
14 73.2
16 72.59
18 67.93
Next, we have to assign variables to the time and temperature values. Let's use x for time and y for temperature.
x = [8, 10, 12, 14, 16, 18]
y = [50.68, 62.24, 69.75, 73.2, 72.59, 67.93]
Now, we will square each x value to get the x^2 values:
x^2 = [8^2, 10^2, 12^2, 14^2, 16^2, 18^2]
= [64, 100, 144, 196, 256, 324]
Our quadratic equation is y = ax^2 + bx + c. We need to find the values of a, b, and c that will minimize the error when fitting the data.
To find these values, we will use the method of least squares. We need to solve the following system of equations:
aΣx^4 + bΣx^3 + cΣx^2 = Σxy
aΣx^3 + bΣx^2 + cΣx = Σy
aΣx^2 + bΣx + cN = Σx^2y
where Σ denotes the sum and N is the number of data points.
Now, let's calculate the necessary sums:
Σx^2 = 64 + 100 + 144 + 196 + 256 + 324 = 1084
Σx^3 = 512 + 1000 + 1728 + 2744 + 4096 + 5832 = 15912
Σx^4 = 4096 + 10000 + 20736 + 38416 + 65536 + 104976 = 233656
Σxy = 8 * 50.68 + 10 * 62.24 + 12 * 69.75 + 14 * 73.2 + 16 * 72.59 + 18 * 67.93 = 2288.72
Σy = 50.68 + 62.24 + 69.75 + 73.2 + 72.59 + 67.93 = 396.39
Σx^2y = 64 * 50.68 + 100 * 62.24 + 144 * 69.75 + 196 * 73.2 + 256 * 72.59 + 324 * 67.93 = 161903.48
Now we can substitute these values into the system of equations:
1084a + 46b + 6c = 2288.72
512a + 28b + 6c = 396.39
64a + 4b + c = 161903.48
Solving this system of equations will give us the values of a, b, and c.
By solving this system of equations using a calculator or software, we get:
a ≈ 0.056
b ≈ 1.088
c ≈ 47.487
Therefore, the quadratic model for the data is:
y = 0.056x^2 + 1.088x + 47.487
This equation can be used to predict the temperature (in Fahrenheit) at any time during the day.
put 8,10,12,14,16,18 in for L1
put the times in for L2
do quadratic reg on calc