Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.1 AMC 10/12 Find the answer

The 31st World University Summer Games will be held in Chengdu from July 28 to August 8, 2023, promoting sports for all. A sports equipment store has compiled data on the sales volume $y$ (in units of thousand sets) and the selling price $x$ (in units of yuan) of sports equipment from January to May 2023, as shown in the table below:

| Month | 1 | 2 | 3 | 4 | 5 |
|-------|-----|-----|-----|-----|-----|
| $x$ | 100 | 90 | 80 | 70 | 60 |
| $y$ | 5 | 7.5 | 8 | 9 | 10.5|

$(1)$ Calculate the correlation coefficient $r$ of $(x_i, y_i)$ for $i=1,2,3,\cdots,n$ , and determine if there is a strong linear correlation between the sales volume $y$ and the selling price $x$ (When $|r|\in [0.75,1]$ , it can be considered as a strong linear correlation; otherwise, there is no strong linear correlation) (Round to $0.001$ ); $(2)$ Establish the linear regression equation of $y$ with respect to $x$ (Round to $0.001$ ), and estimate the sales volume when the selling price of the equipment is $50$ yuan.

Reference formula: For a set of data $(x_i, y_i)$ for $i=1,2,3,\cdots,n$ , the correlation coefficient $r$ is calculated as $r=\frac{\sum_{i=1}^{n}(x_{i}-\overline{x})(y_{i}-\overline{y})}{\sqrt{\sum_{i=1}^{n}(x_{i}-\overline{x})^{2}}\sqrt{\sum_{i=1}^{n}(y_{i}-\overline{y})^{2}}}$ , and the least squares estimates of the slope and intercept of the regression line $\hat{y}=\hat{b}x+\hat{a}$ are $\hat{b}=\frac{\sum_{i=1}^{n}(x_{i}-\overline{x})(y_{i}-\overline{y})}{\sum_{i=1}^{n}(x_{i}-\overline{x})^{2}}$ and $\hat{a}=\overline{y}-\hat{b}\overline{x}$ , with reference data: $\sqrt{16500}\approx128.452$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

### Step-by-Step Solution

#### Part 1: Calculating the Correlation Coefficient $r$

First, we calculate the mean of $x$ and $y$ :

- $\overline{x} = \frac{100 + 90 + 80 + 70 + 60}{5} = 80$
- $\overline{y} = \frac{5 + 7.5 + 8 + 9 + 10.5}{5} = 8$

Next, we calculate the necessary sums for the correlation coefficient:

- $\sum_{i=1}^n(x_i-\overline{x})(y_i-\overline{y}) = -125$
- $\sum_{i=1}^n(x_i-\overline{x})^2 = 1000$
- $\sum_{i=1}^n(y_i-\overline{y})^2 = 16.5$

Now, we can calculate $r$ :

$r = \frac{\sum_{i=1}^{n}(x_{i}-\overline{x})(y_{i}-\overline{y})}{\sqrt{\sum_{i=1}^{n}(x_{i}-\overline{x})^{2}}\sqrt{\sum_{i=1}^{n}(y_{i}-\overline{y})^{2}}} = \frac{-125}{\sqrt{1000}\times\sqrt{16.5}} = \frac{-125}{128.452} \approx -0.973$

Therefore, there is a strong linear correlation between the sales volume $y$ and the selling price $x$ , since $|-0.973| \in [0.75,1]$ .

#### Part 2: Establishing the Linear Regression Equation

First, we calculate the slope $\hat{b}$ :

$\hat{b} = \frac{\sum_{i=1}^n(x_i-\overline{x})(y_i-\overline{y})}{\sum_{i=1}^n(x_i-\overline{x})^2} = \frac{-125}{1000} = -0.125$

Next, we calculate the intercept $\hat{a}$ :

$\hat{a} = \overline{y} - \hat{b}\overline{x} = 8 - (-0.125) \times 80 = 18$

Therefore, the linear regression equation of $y$ with respect to $x$ is:

$\hat{y} = -0.125x + 18$

Finally, we estimate the sales volume when the selling price of the equipment is $50$ yuan:

$y = -0.125 \times 50 + 18 = 11.75$

Encapsulating the final answers:

- The correlation coefficient $r$ is $\boxed{-0.973}$ , indicating a strong linear correlation.
- The linear regression equation is $\boxed{\hat{y} = -0.125x + 18}$ .
- The estimated sales volume for a selling price of $50$ yuan is $\boxed{11.75}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.