2016-2018 – S2 – Mathematics – TEST 1 – SOLUTIONS
IUT of Saint-Etienne – Sales and Marketing department
Mr. Ferraris Prom 2016-2018 14/04/2017
MATHEMATICS – 2
ndsemester, Test 1 length : 2 hours – coefficient 1/2
SOLUTIONS
Exercise 1 : MCQ (3 points) – tick the right boxes below
One correct answer only per question - 0 point in case of wrong/missing/multiple answer at a question 1) Fill the following sentence, using the correct item: "In a Chi-square testing, the significance level is the
probability that H0 may be ___[1]___ given that we ___[2]___ it."
[1] right [1] right [1] wrong [1] wrong
[2] reject [2] accept [2] reject [2] accept 2) From the series 3 ; 5 ; 4 ; 6 ; 4 ; 7 , the 2 by 2 moving means are:
4 ; 5 ; 5.5 3 ; 4 ; 4 4 ; 4.5 ; 5 ; 5 ; 5.5 8 ; 9 ; 10 ; 10 ; 11 3) Between a straight line and a point cloud, if all residues are very little, then:
|r| ≈ 1 cov(X,Y) ≈ 0 |r| ≈ 0 cov(X,Y) ≈ 1
Exercise 2 : χ² testing (5.5 points)
In the following table are gathered 418 women, sorted by their hair colour and their eye colour:
Hair colour
Black Brown Red Blond
Eye colour
Brown 82 118 20 35
Green 11 16 11 8
Blue 33 42 18 24
1) By the mean of a Chi-square testing, can we claim, with a 2% significant level, that hair colour and eye
colour are related in the population this sample comes from? 3 pts
theoretical frequencies array:
76.866 107.368 29.892 40.873 255 13.866 19.368 5.392 7.373 46 35.268 49.263 13.715 18.753 117
126 176 49 67 418
Chi-2s array:
0.3429 1.0527 3.2737 0.8439 0.5924 0.5858 5.8316 0.0533 0.1458 1.0709 1.3385 1.4677
16.6 Null hypothesis H0: Hair colour and eye colour are independent inside the population.
Calculated Chi-2, from the comparison between observed and theoretical frequencies: χ²calc = 16.6 Rejection area: 6 dof and α = 2% give χ²lim = 15
Comparison and decision: χ²calc > χ²lim , therefore H0 can be rejected at a 2% significance level.
2) Give a concrete explanation of the significance level. 1 pt
On claiming that both variables are dependent (people's distribution, per eye colour, depends on their hair colour), our chances to be wrong are less than 2%.
3) On having a closer look on the part chi squares, say for which hair colour people are not distributed by eye
colour like the rest of the population. 1.5 pt
The most important part Khi-2s correspond to red-headed; among them, people's eye colour distribution don't look like other people's distributions.
2016-2018 – S2 – Mathematics – TEST 1 – SOLUTIONS Exercise 3 : (6.5 points)
The table below displays the evolution of the French hourly minimum wage for the past 13 years. The corresponding scatter plot is also displayed.
year
year range
gross minimum
wage (€)
X Y
2005 1 8.03
2006 2 8.27
2007 3 8.44
2008 4 8.71
2009 5 8.82
2010 6 8.86
2011 7 9.10
2012 8 9.31
2013 9 9.43
2014 10 9.53
2015 11 9.61
2016 12 9.67
2017 13 9.76
Y
X
1) a. Give the expression of the Y on X fitting line, according to the least square method. 1 pt y = 0.1439x + 8.034
b. Draw this line on the scatter plot above. 1 pt
c. Using this linear fitting, calculate an estimate of the amount of this gross (fr.: "brut") minimum wage in
the year 2025. 1 pt
y = 0.1439×21 + 8.034 = €11.056
2) It seems that a linear fitting may not be the best way to model the growth of the minimum wage:
let's perform the variable change T = X .
a. Calculate the covariance of the pair (T, Y) and its linear correlation coefficient. Comment. 2 pts
( )
( )
( ) ( )
, ,
302.585
2.5273 9.04154 0.42533 13
0.42533
0.9955 0.78287 0.54572
i i
TY
Cov T Y t y t y n
Cov T Y
r T Y
=∑ − × = − × ≈
= ≈ ≈
× ×
σ σ
This is an excellent linear correlation coefficient because it's very close to 1 and much more than 0.95.
A linear regression would correctly summarise the relationship between T and Y.
b. Give the expression of the Y on T fitting line, according to the least square method. 0,5 pt y = 0.6940 t + 7.288
c. Then, give with this new model an estimation of this minimum wage in the year 2025. 1 pt y'0 = 0.6940 × 21 + 7.288 ≈ €10.468
2016-2018 – S2 – Mathematics – TEST 1 – SOLUTIONS Exercise 4 : (5 points)
A car consulting website has identified the resale values of several vehicles of the same model based on their age. The numerical results and the corresponding scatter plot are given below:
1) Here, a linear fitting would be irrelevant; but, a curve fitting would be. Let's perform the variable change:
T 1
= X , assuming a good linear correlation between T and Y. Give the expression of the Y on T regression
line, according to the least square method. 1 pt
y = 7241.36 t + 1904.96
2) We would like to have an estimate of the selling price of a 10 year-old vehicle of the same model.
a. Give the point estimate for this price. 1 pt
y'0 = 7241.36 × 1
10 + 1904.96 ≈ €2,629.10
b. Give its estimate by a 95% confidence interval (first, you will build a confidence interval for T using the rates method, and you will then translate it into an interval for X). 2.5 pts By the mean of the former expression, Y' values can be calculated;
then, we have to calculate the rates Z = Y/Y'.
X 1 1,5 2 3 3 4 5 7
Y 9200 6700 5500 4300 4200 3750 3400 3000
T 1 0.666667 0.5 0.333333 0.333333 0.25 0.2 0.142857
Y' 9146.33 6732.54 5525.65 4318.75 4318.75 3715.30 3353.24 2939.44 Z 1.005868 0.995167 0.995359 0.995658 0.972503 1.009339 1.013946 1.020601
Parameters of the variable Z: z≈1.001 and σZ ≈0.013926.
The 95% confidence interval is then: I =y0′
(
z−1.96σZ)
; y0′(
z −1.96σZ)
≈[
2560.1 ; 2703.6]
c. According to this confidence interval, what is the rate of such 10 year-old vehicles whose selling price
would be more than €2,560 ? 0,5 pt
Thus, approximately 97.5 % vehicles would show a selling price that is more than the lower bound of our interval.
vehicle's age (years) : X 1 1.5 2 3 3 4 5 7
resale value (€) : Y 9,200 6,700 5,500 4,300 4,200 3,750 3,400 3,000
