CalcGospel 國際數學圖譜
返回

IAL 2022 Jan Q6

A Level / Edexcel / S1

IAL 2022 Jan Paper · Question 6

题目

Problem

Students on a psychology course were given a pre-test at the start of the course and a final exam at the end of the course. The teacher recorded the number of marks achieved on the pre-test, pp, and the number of marks achieved on the final exam, ff, for 3434 students and displayed them on the scatter diagram.

Scatter diagram

The equation of the least squares regression line for these data is found to be

f=10.8+0.748pf=10.8+0.748p

For these students, the mean number of marks on the pre-test is 62.462.4

(a) Use the regression model to find the mean number of marks on the final exam.

(2)

(b) Give an interpretation of the gradient of the regression line.

(1)

Considering the equation of the regression line, Priya says that she would expect someone who scored 00 marks on the pre-test to score 10.810.8 marks on the final exam.

(c) Comment on the reliability of Priya’s statement.

(1)

(d) Write down the number of marks achieved on the final exam for the student who exceeded the expectation of the regression model by the largest number of marks.

(1)

(e) Find the range of values of pp for which this regression model, f=10.8+0.748pf=10.8+0.748p, predicts a greater number of marks on the final exam than on the pre-test.

(3)

Later the teacher discovers an error in the recorded data. The student who achieved a score of 9898 on the pre-test, scored 9292 not 2929 on the final exam. The summary statistics used for the model f=10.8+0.748pf=10.8+0.748p are corrected to include this information and a new least squares regression line is found.

Given the original summary statistics were,

n=34∑p=2120∑pf=133486n=34\qquad \sum p=2120\qquad \sum pf=133486 Spp=15573.76Spf=11648.35S_{pp}=15573.76\qquad S_{pf}=11648.35

(f) calculate the gradient of the new regression line. Show your working clearly.

(5)

解答

(a)

解法一

思路

展开

回归线经过 (pˉ,fˉ)(\bar{p},\bar{f})，所以把 pˉ=62.4\bar{p}=62.4 代入即可。

答题过程

展开 fˉ= 10.8+0.748(62.4)= 57.4752.\begin{align*} \bar{f} =&\,10.8+0.748(62.4)\\[3mm] =&\,57.4752. \end{align*}

So the mean number of marks on the final exam is

57.5\begin{align*} 57.5 \end{align*}

to 33 significant figures.

(b)

解法一

思路

展开

斜率要结合两个变量的语境解释。

答题过程

展开

For each additional mark scored on the pre-test, the final exam mark increases by about 0.7480.748 on average.

(c)

解法一

思路

展开

从散点图看，数据没有接近 p=0p=0 的点，所以这是范围外估计。

答题过程

展开

Priya’s statement is not reliable because p=0p=0 is outside the range of the data.

(d)

解法一

思路

展开

从散点图看，超过回归线最多的点对应 final exam mark 为 7676。

答题过程

展开

The number of marks is

76.\begin{align*} 76. \end{align*}

(e)

解法一

思路

展开

要求模型预测 final exam marks 大于 pre-test marks，即 f>pf>p。

答题过程

展开

We need

10.8+0.748p>p.\begin{align*} 10.8+0.748p>p. \end{align*}

So

10.8>0.252pp<42.857…\begin{align*} 10.8&>0.252p\\[3mm] p&<42.857\ldots \end{align*}

Therefore

p<42.9.p<42.9.

(f)

解法一

思路

展开

pp 没有改，所以 SppS_{pp} 不变。错误只影响 ff 和 pfpf：把 2929 改为 9292，即 ff 增加 6363，而 pfpf 增加 98×6398\times63。

答题过程

展开

The corrected value of ∑pf\sum pf is

133486+98(92−29)=139660.\begin{align*} 133486+98(92-29)=139660. \end{align*}

The original value of ∑f\sum f can be found from fˉ=57.4752\bar{f}=57.4752:

∑f=34(57.4752)=1954.1568.\begin{align*} \sum f=34(57.4752)=1954.1568. \end{align*}

So the corrected value of ∑f\sum f is

1954.1568+63=2017.1568.\begin{align*} 1954.1568+63=2017.1568. \end{align*}

Therefore

Spf= ∑pf−∑p∑fn= 139660−2120(2017.1568)34= 13894.1…\begin{align*} S_{pf} =&\,\sum pf-\frac{\sum p\sum f}{n}\\[3mm] =&\,139660-\frac{2120(2017.1568)}{34}\\[3mm] =&\,13894.1\ldots \end{align*}

The new gradient is

b= SpfSpp= 13894.1…15573.76= 0.892…\begin{align*} b =&\,\frac{S_{pf}}{S_{pp}}\\[3mm] =&\,\frac{13894.1\ldots}{15573.76}\\[3mm] =&\,0.892\ldots \end{align*}

So the gradient is approximately

0.9.\begin{align*} 0.9. \end{align*}