题目

Problem

Soapern sells washing machines. When a customer buys a washing machine from Soapern, the customer is also invited to buy a guarantee policy to cover breakdowns and repairs for the next three years.

The manager of Soapern believes that the relationship between the number of washing machines sold ( $m$ ) and the number of guarantee policies sold ( $p$ ) can be modelled by a straight line.

She collected data each month for 10 months. The scatter diagram below illustrates these data.

The data are summarised by the following statistics.

\sum m=1124\qquad \sum p=281\qquad \sum mp=32958\qquad S_{mm}=6046.4\qquad S_{pp}=382.9

(a) Show that $S_{mp}=1373.6$

(1)

(b) Find the value of the product moment correlation coefficient for these data.

(2)

(1)

The manager noticed that the total number of washing machines sold was $k$ times the total number of guarantee policies sold and suggests a model of the form $p=\frac1k m$ , where $k$ is an integer.

(d) Find the value of $k$ .

(2)

Jiang works for Soapern and thought that this model oversimplified the situation and suggested that a linear regression of $p$ on $m$ may be more appropriate.

(e) Find the equation of the linear regression of $p$ on $m$ , giving your answer in the form $p=a+bm$ , where $a$ and $b$ should be given to 3 significant figures.

(4)

(f) Use Jiang’s model to estimate the number of guarantee policies sold when 70 washing machines are sold in a month.

(1)

Usually about 70 washing machines are sold in January. Soapern decides to offer a bonus to staff during January based on the number of guarantee policies sold. If the number of guarantee policies sold is greater than the number estimated by the model, the bonus will be paid.

(g) State, giving your reasons, whether you would recommend that the staff use the manager’s model or Jiang’s model.

(2)

解答

(a)

解法一

思路

展开

用 $S_{mp}=\sum mp-\frac{\sum m\sum p}{n}$ ，并算到题目给出的值。

答题过程

展开

\begin{aligned} S_{mp} =&\,\sum mp-\frac{\sum m\sum p}{n}\\ =&\,32958-\frac{1124(281)}{10}\\ =&\,1373.6. \end{aligned}

Hence $S_{mp}=1373.6$ , as required.

(b)

解法一

思路

展开

直接代入相关系数公式。

答题过程

展开

\begin{aligned} r =&\,\frac{S_{mp}}{\sqrt{S_{mm}S_{pp}}}\\ =&\,\frac{1373.6}{\sqrt{6046.4(382.9)}}\\ =&\,0.9027\ldots. \end{aligned}

Therefore

\begin{align*} r=0.903\quad\text{to 3 significant figures}. \end{align*}

(c)

解法一

思路

展开

$r$ 接近 1，说明强正相关，支持用直线模型。

答题过程

展开

Yes. The data are consistent with the manager’s belief because $r=0.903$ , showing strong positive correlation.

(d)

解法一

思路

展开

题目说总 washing machines 是总 guarantee policies 的 $k$ 倍，所以用两个总和相除。

答题过程

展开

\begin{align*} k=\frac{\sum m}{\sum p}=\frac{1124}{281}=4. \end{align*}

(e)

解法一

思路

展开

回归线 $p$ on $m$ 的斜率是 $b=\frac{S_{mp}}{S_{mm}}$ ，然后用 $(\bar m,\bar p)$ 求截距。

答题过程

展开

b=\frac{S_{mp}}{S_{mm}} =\frac{1373.6}{6046.4} =0.2271\ldots.

Also,

\bar m=\frac{1124}{10}=112.4,\qquad \bar p=\frac{281}{10}=28.1.

\begin{align*} a=\bar p-b\bar m. \end{align*}

Hence

\begin{align*} a=28.1-(0.2271\ldots)(112.4)=2.565\ldots. \end{align*}

Therefore the regression line is

\begin{align*} p=2.57+0.227m. \end{align*}

(f)

解法一

思路

展开

把 $m=70$ 代入 Jiang 的回归线。

答题过程

展开

\begin{align*} p=2.565\ldots+0.2271\ldots(70)=18.467\ldots. \end{align*}

So the estimate is approximately

\begin{align*} 18.5 \end{align*}

guarantee policies.

(g)

解法一

思路

展开

奖金条件是实际销量大于模型估计值。员工角度希望门槛较低。manager 的模型在 $m=70$ 时给 $p=70/4=17.5$ ，比 Jiang 的约 18.5 更低。

答题过程

展开

The manager’s model gives

\begin{align*} p=\frac{70}{4}=17.5. \end{align*}

Jiang’s model gives approximately $18.5$ .

The staff should use the manager’s model, because it gives the lower estimate and so makes it easier for the bonus to be paid.

IAL 2020 Jan Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

(d)

解法一

思路

答题过程

(e)

解法一

思路

答题过程

(f)

解法一

思路

答题过程

(g)

解法一

思路

答题过程