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, $p$ , and the number of marks achieved on the final exam, $f$ , for $34$ students and displayed them on the scatter diagram.

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

f=10.8+0.748p

For these students, the mean number of marks on the pre-test is $62.4$

(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 $0$ marks on the pre-test to score $10.8$ marks on the final exam.

(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 $p$ for which this regression model, $f=10.8+0.748p$ , 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 $98$ on the pre-test, scored $92$ not $29$ on the final exam. The summary statistics used for the model $f=10.8+0.748p$ are corrected to include this information and a new least squares regression line is found.

Given the original summary statistics were,

n=34\qquad \sum p=2120\qquad \sum pf=133486

S_{pp}=15573.76\qquad S_{pf}=11648.35

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

(5)

题目中文翻译

一门心理学课程的学生在课程开始时参加预试，在课程结束时参加期末考试。教师记录 $34$ 名学生的预试分数 $p$ 和期末考试分数 $f$ ，并绘制成散点图。

这些数据的最小二乘回归直线为 $f=10.8+0.748p$ ，预试平均分为 $62.4$ 。

(a) 用回归模型求期末考试的平均分。

(b) 解释回归直线斜率的含义。

Priya 根据回归直线方程认为，预试得 $0$ 分的人期末考试预计会得 $10.8$ 分。

(d) 写出比回归模型预测值高出最多分数的那名学生的期末考试分数。

(e) 求使模型预测的期末考试分数高于预试分数的 $p$ 的取值范围。

后来教师发现数据记录有误：预试得 $98$ 分的学生，期末考试实际得 $92$ 分，而不是 $29$ 分。教师据此修正原汇总统计量并求出新的最小二乘回归直线。

(f) 计算新回归直线的斜率，清楚写出计算过程。

解答

(a)

解法一

思路

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

答题过程

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\begin{align*} \bar{f} =&\,10.8+0.748(62.4)\\ =&\,57.4752. \end{align*}

So the mean number of marks on the final exam is

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

to $3$ significant figures.

(b)

解法一

思路

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斜率要结合两个变量的语境解释。

答题过程

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For each additional mark scored on the pre-test, the final exam mark increases by about $0.748$ on average.

(c)

解法一

思路

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从散点图看，数据没有接近 $p=0$ 的点，所以这是范围外估计。

答题过程

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

(d)

解法一

思路

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从散点图看，超过回归线最多的点对应 final exam mark 为 $76$ 。

答题过程

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The number of marks is

\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 $f>p$ 。

答题过程

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We need

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

\begin{align*} 10.8&>0.252p\\ p&<42.857\ldots \end{align*}

Therefore

p<42.9.

(f)

解法一

思路

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$p$ 没有改，所以 $S_{pp}$ 不变。错误只影响 $f$ 和 $pf$ ：把 $29$ 改为 $92$ ，即 $f$ 增加 $63$ ，而 $pf$ 增加 $98\times63$ 。

答题过程

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The corrected value of $\sum pf$ is

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

Use the original summary statistics to recover $\sum f$ :

\begin{align*} 11648.35=&\,133486-\frac{2120\sum f}{34},\\ \sum f=&\,\frac{34(133486-11648.35)}{2120}\\ =&\,1954.000\ldots \end{align*}

Hence the corrected value is $\sum f\approx1954+63=2017$ . Therefore

\begin{align*} S_{pf}=&\,139660-\frac{2120(2017)}{34}\\ =&\,13894.1\ldots \end{align*}

The new gradient is

\begin{align*} b =&\,\frac{S_{pf}}{S_{pp}}\\ =&\,\frac{13894.1\ldots}{15573.76}\\ =&\,0.892\ldots \end{align*}

So the gradient is approximately

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

解法二

思路

展开

直接计算修正对 $S_{pf}$ 的增量； $p$ 与 $S_{pp}$ 均不变。

答题过程

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The changes in the totals are

\begin{align*} \Delta\sum pf=&\,98(92-29)=6174,\\ \Delta\sum f=&\,92-29=63. \end{align*}

Therefore the change in $S_{pf}$ is

\begin{align*} \Delta S_{pf} =&\,6174-\frac{2120(63)}{34}\\ =&\,2245.764\ldots \end{align*}

So the corrected value of $S_{pf}$ and the new gradient are

\begin{align*} S_{pf}=&\,11648.35+2245.764\ldots\\ =&\,13894.114\ldots,\\ b=&\,\frac{13894.114\ldots}{15573.76}\\ =&\,0.892\ldots\approx0.9. \end{align*}

S1 2022 Jan Q6

题目

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(a)

解法一

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(b)

解法一

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(c)

解法一

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(d)

解法一

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(e)

解法一

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(f)

解法一

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解法二

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