IUT of Saint-Etienne – Sales and Marketing department Mr Ferraris Prom 2019-2021 03/2020 MATHEMATICS – 2

2019-2021 – S2 – Mathematics – TEST 1 SOLUTIONS – page 1 / 3

IUT of Saint-Etienne – Sales and Marketing department

Mr Ferraris Prom 2019-2021 03/2020

MATHEMATICS – 2

semester, Test 1 length: 2 hours – coefficient 1/2

SOLUTIONS

Exercise 1: Chi-square test (4 points)

A market study crosses two characteristics of a tool supplier's clientele: the type of clientele (individual, professional) and the category of goods purchased (tools, materials, products, electricity). The following table shows, at the end of the study, the number of customers for each type/category crossover (e.g. 22 professional customers purchased materials).

tools materials products electricity

individual 24 25 15 38 102

professional 14 22 25 17 78

38 47 40 55 180

After conducting a Chi-square test, discuss how much confidence can be placed in the statement "The category of goods purchased depends on the type of clientele".

Null hypothesis (H0): category of goods purchased and type of clientele are independent Calculated Chi-square:

From the subtotals in the observation table, a theoretical average table is derived:

21.5333333 26.6333333 22.6666667 31.1666667 102 16.4666667 20.3666667 17.3333333 23.8333333 78

38 47 40 55 180

Then, the comparison of the two previous tables allows us to obtain the partial Chi-2:

0.28255934 0.10016688 2.59313725 1.49821747 0.36950067 0.13098745 3.39102564 1.95920746

10.3248022

The bottom right total is our χ²calc. Limit Chi-square: with 3 dof; at 1%, χ²lim = 11.3 and at 2%, χ²lim = 9.84.

Comparison and decision: At a 1% significance level, χ²lim > χ²calc . We cannot reject the null hypothesis at this level (there is more than 1% risk to be wrong on claiming the correlation).

At a 2% significance level, χ²lim < χ²calc . The null hypothesis can be rejected at this level (the risk to be wrong is less than 2%, on claiming the correlation).

2019-2021 – S2 – Mathematics – TEST 1 SOLUTIONS – page 2 / 3 Exercise 2: linear regression (8.5 points)

A study was carried out to compare the spending habits of individuals on high-tech equipment compared to their income. Each column in the table below represents, in a given French region, the median monthly income of working people (X) and the average monthly expenditure (Y) on high-tech equipment.

region A B C D E F

income X (€) 1550 1620 1770 1850 1930 2000

expenditure Y (€) 48 51 56 60 62 66

1) a. Calculate the covariance, then the linear correlation coefficient, of the pair

(

)

Give a comment for both parameters. 2 pts

(

)

Cov X Y = 6 − × ≈ , positive, so when income rises, spending tends to increase..

995.55556

0.9971 160.2775 6.2294

ρ≈ ≈

× , very close to 1, hence an excellent linear correlation between X and Y.

b. Using the calculator, give the Y in X regression line expression. 0.5 pt y’ = 0.03875x – 12.07

c. Determine the 99% confidence interval of the estimated high-tech equipment expenditure for

individuals with a median income of 2500 €. 2.5 pts

y' 47.9948097 50.7076125 56.5207612 59.6211073 62.7214533 65.4342561 z 1.00010814 1.00576615 0.99078637 1.00635501 0.9884975 1.00864599 z ≈1.00002653; we’ll take 1. σ_z≈0.007806.

x0 = 2500, thus y’0 = 0.03875×2500 – 12.07 = 84.81.

confidence level: 99%, associated with u = 2.58.

Interval: I=y0′

(

) (

)

[

]

d. Of these, what percentage would spend more than €83.1? 1 pt

Environ 99,5% (99% dans l’intervalle + 0,5% au-delà de 86,52€.

2) a. Determine Mayer's line expression relative to the table T. 1.5 pt Let's divide the statement table into two groups: {A, B, C} and {D, E, F} (indeed, the values of X are given in ascending order). The coordinates of their midpoints are:

G1(1646.6667 ; 51.666667) and G2(1926.66667 ; 62.666667)

The Mayer’s line, (G1G2), has an equation that can be written y’ = ax + b.

( )

2 1

1 1

2 1

0.03929 and 13.02

0.03929 13.02

G G

G G

G G

M

y y

a b y a x

x x

D y x

= − ≈ = − × ≈ −

−

′ = −

b. What point estimate of Y would be given by this line for individuals with a median income of €2500?

What is the percentage difference with the point estimate calculated in question 1)c.? 1 pt

:

0.03929 2500 13.02 85.19 85.19 84.81

percentage difference 100 0.4469

84.81

y′ = × − =

− × ≈

The least-square and Mayer point estimates show (for x = 2500) a difference of 0.4469%.

2019-2021 – S2 – Mathematics – TEST 1 SOLUTIONS – page 3 / 3 Exercise 3: crossed table (3.5 points)

Mrs Betty Cash, sales representative, analyses her activity and efficiency. She has classified her customers' orders according to two variables: the duration of a visit (X, in minutes) leading to an order, and the number of items ordered (Y) at the end of this visit. The numbers of visits are visible in the intersections of the table opposite.

X : duration of a visit

(minutes)

Y : number of items ordered per visit

1 2 3 4

[0 ; 30[ 3 1 1 0

[30 ; 60[ 2 4 3 1

[60 ; 90[ 0 4 5 3

1) What is the meaning of the frequency "5" in the table? 1 pt 5 visites ont duré entre 60 et 90 minutes et donné lieu à trois articles commandés.

2) Calculate manually the average time spent per order placed. 1.5 pt average time: 5 15 10 45 12 75 1425 minutes

52.78 minutes/order

5 10 12 27 orders

× + × + × = ≈

+ + .

3) What process should be followed to estimate the number of items sold during a visit, for a group of visits

lasting two hours? 1 pt

The relationship between the two variables should be modelled by a linear regression (if the scatterplot shows that this regression is relevant) and then a confidence interval of the number of items ordered per visit should be established. Given that the distribution of frequencies in rows 2 and 3 of the table is relatively symmetrical around a maximum, one could simply position oneself in the centre of the interval (i.e. on the point estimate) to give an estimate of the number of items sold during a visit.

Exercise 4: variable change (4 points)

The table opposite gives the price list of a private company, in metropolitan France, for the delivery of letters or packages sent, according to their mass.

Our goal is to analyse and model, thanks to this table, the evolution rate of Y according to X; then to propose a suitable price to be applied in the case of packages heavier than those proposed in this table.

X : mass (g) Y : price (€)

[0 ; 20[ 0.72

[20 ; 50[ 1.15

[50 ; 100[ 1.65 [100 ; 250[ 2.50 [250 ; 500[ 3.33 [500 ; 1000[ 4.38 [1000 ; 2000[ 5.50 [2000 ; 3000[ 6.55 1) Graphing this table would make it clear that a linear regression would not be appropriate here. It is

proposed to perform the following variable change: T =X^0.3.

a. After entering the values of T and Y into your calculator, calculate the linear correlation coefficient

between these two variables, and then interpret. 1.5 pt

(

)

Cov 5.7367 3.2225 5.5169

T Y = 8 − × ≈ ; 5.5169

r 0.9994

2.797 1.9737

= ≈

×

TY .

This coefficient is very close to 1. The linear correlation between T and Y is very strong.

b. Thanks to your calculator, give the least square line’s expression of Y with respect to T. Deduce a

relationship between Y and X. 1 pt

y’ = 0.7052t – 0.8230 ⇔ y’ = 0.7052X^0.3 – 0.8230

2) According to this last relationship, what price should be applied for packages weighing 6 kg? 1.5 pt y’0 = 0.7052 ×6000^0.3 – 0.8230 ≈ ^€8.77

____________________ TEST END ____________________