2015-2017 – S2 – Mathematics – TEST 1 – page 8 on 7

IUT of Saint-Etienne – Sales and Marketing department Mr. Ferraris Prom 2015-2017 01/04/2016

MATHEMATICS – 2nd semester, 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) Mayer’s criterion is:

� 0ie =∑ � 2 0ie =∑ � is minimumie∑ � 2 is minimumie∑

2) Fill the gap in the following sentence: « A Chi2 testing allows us to […] of two variables »

� calculate the � perform a � establish the � estimate the level

covariance fitting independence of dependence

3) In order to enter the data of a contingency table in your calculator, you must fill:

� one list � two lists � three lists � four lists � one fish

4) A linear correlation coefficient found between 0 and 0.5 implies that the cloud’s points…

� are almost distri- � follow a curve � are located far � don’t follow a � draw a fish

buted at random from each other straight line

5) Given the list (5, 3, 5, 4, 3, 2, 7), the last but one moving mean, taking the values 3 by 3, is:

� 4 � 3 � 5 � 3.5

6) The size of a confidence interval increases as…

� r increases � the confidence level � x0 decreases � y’ 0 decreases � the fish

increases goes up

Exercise 2 : χ² (5 points)

A survey has given the possibility to sort 1,000 respondents according to two criteria: the level of the highest

degree obtained (respondents are aged 25 years minimum; level 1 = master2/engineer, level 5 = BEP/CAP)

and the number of their contacts on a popular social network. The results are as follows:

level of the highest degree

1 2 3 4 5

Number of

contacts

[0 ; 40[ 32 72 85 57 55

[40 ; 150[ 25 70 95 70 90

> 150 19 56 94 80 100

After practicing a chi-square test on these results, decide, at a 2% significance level, whether a person's

degree level is related to his/her number of contacts on this social network. In the end, you will give a

concrete explanation of your conclusion and a clear explanation of what the significance level does mean.

Calculation of the subtotals (margin frequencies) and the overall total (total frequency):

32 72 85 57 55 301

25 70 95 70 90 350

19 56 94 80 100 349

76 198 274 207 245 1000

Calculation of the frequencies theoretically expected in case of independence between both criterions:

22.876 59.598 82.474 62.307 73.745 301

26.6 69.3 95.9 72.45 85.75 350

26.524 69.102 95.626 72.243 85.505 349

76 198 274 207 245 1000

2015-2017 – S2 – Mathematics – TEST 1 – page 9 on 7

Calculation of the detailed Chi2s and their sum, χ²calc:

3.6391 2.5808 0.0774 0.4520 4.7647

0.0962 0.0071 0.0084 0.0829 0.2106

2.1343 2.4842 0.0276 0.8329 2.4572

19.86 = χ²calc

H0 : there is no link between the degree level and the number of contacts on this social network.

χ²calc = 19.86

dof = 4×2 = 8 ; significance level: 2%. In the Chi-square table, we can read: χ²lim = 18.2.

As our calculated value is more than the latter one, the null hypothesis can be rejected at a 2% significance

level. This means that we can assess that “there is a link between the degree level and the number of

contacts on this social network” with a more than 98% confidence level.

(a closer look at the distribution of the observed frequencies shows us that, globally, the higher people’s

degree level is, the lower the number of their contacts is).

Exercise 3 : (6 points)

The table below shows the air concentrations of carbon dioxide (CO2), since the beginning of the industrial

era (1850).

Year’s rank xi 0 50 100 140 150

CO2 concentration 275 290 315 350 370

CO2 concentration is expressed in parts per million (ppm).

The rank 0 is the year 1850, the rank 50 is the year 1900, etc.

Clearly, the CO2 concentration does not increase at a constant rate in time, that is to say: a straight line is

unable to model its evolution correctly.

1) We set the variable change T = X

70e where « e » is the exponential number. After getting the T values on

your calculator, calculate the covariance of the pair (T, Y), and then the linear correlation coefficient of

the same couple. Comment. 2 pts

values of T 1 2.0427 4.1727 7.3891 8.5238

( ) ., . .

7921 76Cov 4 625655 320 104 14

5

tyT Y t y

n∑= − × ≈ − × ≈

( ), ..

. .

Cov 104 140 9982

2 92758 35 637T Y

T Yr

σ σ= ≈ ≈

× ×. This coefficient is very close to 1, far from 0.95. There is then

an excellent linear correlation between both variables.

2) Give the expression of the Y on T regression line, according to least square method. Deduce a modeled

expression of the relation between Y and X. 1.5 pt

. .12 15 263 8y t= + , hence . .7012 15 e 263 8x

y = × + .

3) Considering that the latter expression is correctly modeling the evolution of the CO2 concentration in the

atmosphere, make the following predictions:

a. What will be the concentration in 2020? 1 pt

. . .170

7012 15 e 263 8 401 6 ppmy = × + ≈ .

b. In what year this concentration will exceed 450 ppm? 1.5 pt

( )

( )

. . . . . ln .

ln . .

70 70 70450 12 15 e 263 8 186 2 12 15 e 15 325 e 15 32570

70 15 325 191 In the year 2041, 450 ppm will be exceeded.

x x x x

x

= × + ⇔ = × ⇔ = ⇔ =

⇔ = ≈

2015-2017 – S2 – Mathematics – TEST 1 – page 10 on 7

Exercise 4 : (6 points)

A statistics analyst wished to compare the number of vehicles registered in each town of a set of eight, to the

annual volume, in euro, of fines for lack of parking payment, in each one of these cities. The observations are

recorded in the table below.

registered vehicles (thousands) 3 5 6 10 15 16 19 22

annual amount of fines (k€) 45 73 89 153 220 245 279 336

1) Prove that the Mayer’s line expression of this series is y = 15x. 2 pts

Mayer’s method consists in dividing the point cloud into two half-size clouds, separated by their X values;

in other words: the first half-cloud is formed with the first four columns of the table and the second one

with the last four. Let’s calculate the coordinates of both mean points:

xG1 = 24/4 = 6 ; yG1 = 360/4 = 90 ; xG2 = 72/4 = 18 ; yG2 = 1080/4 = 270.

Slope : a = (yG2 - yG1) / (xG2 - xG1) = 180 / 12 = 15.

y-intercept : let’s use the point G1 : yG1 = a×xG1 = + b, hence 90 = 15×6 + b, therefore b = 0.

Indeed, the expression of the Mayer’s line is: y = 15x.

2) On the graph page 6, represent the scatter plot and draw the Mayer’s line. 1 pt

3) a. By using the expression of this line, determine the 99% confidence interval of the annual amount of

fines that could be predicted in a town where 30,000 vehicles would be registered. 2 pts

We have to calculate the eight Y’ values from the X values, relying on our Mayer’s line, and then to

calculate the eight Z values, Z = Y/Y’. We get: the mean of Z is about 0.9972 and its standard

deviation is about 0.01896.

The point estimate of the annual amount of fines is y’0 = 15x0 = 15×30 = 450 (k€).

The coefficient u is 2.58, corresponding to a 99% confidence level.

( ) ( ) [ ]; . ; .0 0 426 7 470 8Z ZI y z u y z uσ σ′ ′= × − × × + × =

b. Represent this interval on the graph page 6, at its correct location. 0.5 pt

c. What is the probability that, in such a town, this annual amount would exceed 427 k€?

0.5 pt

Approximately 99.5%.

2015-2017 – S2 – Mathematics – TEST 1 – page 11 on 7

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