Calculate Correlation, Slope, and Predict Sales Using Linear Regression

Math Problem Statement

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Topic 7 Homework (Nonadaptive)

Question 7 of 16 (1 point)**|**Question Attempt: 1 of Unlimited

Dennis

Question 7

You are the owner of Fast Break, a popular local place that sells drinks, snacks, and sandwiches. For inventory management purposes, you are examining how the weather affects the amount of hot chocolate sold in a day. You are going to gather a random sample of

days showing that day's high temperature (denoted by

, in

°C

) and the amount of hot chocolate sold that day (denoted by

, in liters). You will also note the product

·xy

of the temperature and amount of hot chocolate sold for each day. (These products are written in the row labeled "

").

(a)Click on "Take Sample" to see the results for your random sample.

Take Sample

High temperature,

(in

°C

)

Amount of hot chocolate sold,

(in liters)

120

126

253

112

145

180

184

180

Send data to calculator

Based on the data from your sample, enter the indicated values in the column on the left below. Round decimal values to three decimal places. When you are done, select "Compute". (In the table below,

is the sample size and the symbol

Σxy

means the sum of the values

142

102

1354

2472

Σxy

1434

Compute

Sample correlation coefficient (

−0.005

Slope (

−0.01

-intercept (

103.25

(b)Write the equation of the least-squares regression line for your data. Then on the scatter plot for your data, graph this regression equation by plotting two points and then drawing the line through them. Round each coordinate to three decimal places.

Regression equation:

Amount of hot chocolate sold (in liters)

High temperature (in

°C

)

(c)Use your regression equation to predict the amount of hot chocolate sold on a day with a high temperature of

19 °C

. Round your answer to the nearest whole number.

Predicted amount of hot chocolate sold:

liters19

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Solution

Let's proceed step by step to solve each part of the problem.

Given Data:

Sample size ( $n$ ): 9
Sum of temperatures ( $\Sigma x$ ): 142
Sum of hot chocolate sold ( $\Sigma y$ ): 102
Sum of squares of temperatures ( $\Sigma x^2$ ): 1354
Sum of squares of hot chocolate sold ( $\Sigma y^2$ ): 2472
Sum of the product of temperature and hot chocolate sold ( $\Sigma xy$ ): 1434

The formulas we'll use for calculating the required values are:

(a) Calculation of Sample Correlation Coefficient, Slope, and Intercept

Sample correlation coefficient ( $r$ ): [ r = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{\sqrt{[n(\Sigma x^2) - (\Sigma x)^2][n(\Sigma y^2) - (\Sigma y)^2]}} ]

Plugging in the values: [ r = \frac{9(1434) - (142)(102)}{\sqrt{[9(1354) - (142)^2][9(2472) - (102)^2]}} ]
Slope ( $b_1$ ): $b_1 = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{n(\Sigma x^2) - (\Sigma x)^2}$ $b_1 = \frac{9(1434) - (142)(102)}{9(1354) - (142)^2}$
Intercept ( $b_0$ ): $b_0 = \bar{y} - b_1 \bar{x}$ where: $\bar{x} = \frac{\Sigma x}{n} = \frac{142}{9}$ and $\bar{y} = \frac{\Sigma y}{n} = \frac{102}{9}$

(b) Regression Equation

The regression equation is: $y = b_0 + b_1 x$

(c) Prediction

To predict the amount of hot chocolate sold on a day when the temperature is 19°C, substitute $x = 19$ in the regression equation: $\hat{y} = b_0 + b_1 (19)$

Let's go ahead and calculate these values.

Calculation of $r$ , $b_1$ , and $b_0$ :

$\Sigma xy = 1434, \quad \Sigma x = 142, \quad \Sigma y = 102, \quad n = 9, \quad \Sigma x^2 = 1354, \quad \Sigma y^2 = 2472$

First, calculate the numerator and the denominator separately: $r = \frac{9 \cdot 1434 - 142 \cdot 102}{\sqrt{(9 \cdot 1354 - 142^2) \cdot (9 \cdot 2472 - 102^2)}}$

$b_1 = \frac{9 \cdot 1434 - 142 \cdot 102}{9 \cdot 1354 - 142^2}$

Calculate $\bar{x}$ and $\bar{y}$ : $\bar{x} = \frac{142}{9}, \quad \bar{y} = \frac{102}{9}$

Then: $b_0 = \bar{y} - b_1 \cdot \bar{x}$

I'll proceed with the calculation.

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Math Problem Analysis

Mathematical Concepts

Statistics
Regression Analysis
Correlation
Linear Equations

Formulas

Sample correlation coefficient formula: r = (nΣxy - ΣxΣy) / sqrt{[nΣx² - (Σx)²][nΣy² - (Σy)²]}
Slope formula: b₁ = (nΣxy - ΣxΣy) / (nΣx² - (Σx)²)
Intercept formula: b₀ = ȳ - b₁x̄
Regression equation: ŷ = b₀ + b₁x

Theorems

Least Squares Regression Theorem

Suitable Grade Level

Grades 10-12

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