IUT de Saint-Etienne - département Techniques de Commercialisation

2017-2019 – S3 – Mathematics – TEST 2 SOLUTIONS Page 1 / 3

IUT de Saint-Etienne - département Techniques de Commercialisation

M. Ferraris Promotion 2017-2019 12/12/2018

Semestre 3 - MATHEMATIQUES – DEVOIR 2 durée : 2 heures – coefficient 1/2

SOLUTIONS

Exercise 1 sampling (6 points)

In a farm, the harvested apples have masses distributed by a normal law which mean is 145 g and standard deviation is 15 g.

1) What is the probability that an apple's mass would be more than 150 g? 1 pt Let X be the variable "mass of an apple". p(X > 150) = 0.3694.

2) You bought a case containing 50 apples chosen at random. Calculate the probability that their average mass would be:

a. higher than 150 g 1 pt

X , variable "average mass of an apple in a sample of 50", is distributed by:

N

50≈ ).

p(X > 150) = 0.009211.

b. less than 142 g 0.5 pt

p(X < 142) = 0.07865.

c. How many apples would have to be bought in order to take less than 1% chance that their average mass

would be less than 142 g? 1.5 pt

By a direct use of the form (standard normal law table), we can deduce that p(U < –2.33) = 0.01.

Moreover, the variable X is distributed by

N

n

15 ) and the variables X et U are related this way:

U X

n

= −145

15 . Therefore:

( )

2 15 2.33

15 145 15 15

135.72

15 145 145 145 142 145

X n U U U

n n n

U X X X

n

 × − 

− ×  × 

= ⇔ = ⇔ = ⇔ =  ⇔ =  =

− −  −   −  .

At least 136 apples would have to be bought in order to reduce this chance at 1%.

3) a. What is, in the total harvest, the proportion of apples that weigh more than 150 g ? 0.5 pt The question 1 shown us that π = 0.3694.

b. What is the probability that, in a case of 50 apples, the proportion of those weighing more than 150 g

would exceed 50%? 1.5 pt

The law for P, "proportion of apples weighing more than 150 g in a sample of 50", is:

N

(

)

n

 π − π 

π 

 

 

=

N

Exercise 2 estimation (5.5 points)

A market research was drawn on 150 companies around Saint-Etienne, in relation to their needs in temporary workers. The following table displays the results:

needs (hours/month) [0 ; 200[ [200 ; 400[ [400 ; 600[ [600 ; 800[ [800 ; 1200[

freq. (companies) 12 25 68 37 8

2017-2019 – S3 – Mathematics – TEST 2 SOLUTIONS Page 2 / 3

1) Give the average needs and their standard deviation. 1 pt

x=510.67 and s = 206.28

2) Give the point estimates of the mean and standard deviation of the needs for the overall population of

companies around Saint-Etienne. 1 pt

ˆ=

µ 510.67 and ˆ 150

206.28 206.97

1 149

n s

σ = n = =

−

3) Give the 95% confidence interval of the estimated mean (considering that the standard deviation is known,

taking the estimated value from question 2). 2 pts

With a known standard deviation, the following formula has to be used: I x u ;x u

n n

 

= − + 

 

σ σ

.

[ ]

; ;

206.97 206.97

510.67 1.96 510.67 1.96 477.5 543.8

150 150

I  

= − + =

 

4) To what confidence level would correspond an interval which size would be 40 hours? 1.5 pt The number u

n

σ , half-size of the interval, has to take the value 20, so: 206.97 150 20

u = . Thus, u = 1.183.

This standard normal law coefficient corresponds to a 76.3% confidence level.

Exercise 3 adequacy Chi-squared test (4.5 points)

A car dealership is talking to you about Kiroul vehicles (which have been circulating for many years). He maintains that when these vehicles age a year, 50% of them suffer so many breakdowns that they are removed from circulation. You want to test this assertion and patiently, collect data on the age and condition of some Kiroul vehicles, and compile your results in the following table:

vehicle's age (years) 1 2 3 4 5

observed number of circulating

Kiroul vehicles 78 56 30 18 4

1) Justify that the dealership's hypothesis implies the theoretical frequencies – to be compared with your

figures – given in the second row of the following table: 0.5 pt

vehicle's age (years) 1 2 3 4 5

theoretical number of

circulating Kiroul vehicles 96 48 24 12 6

The "dividing by two, for each additional year" principle is met. Plus, the total number of vehicles corresponds to the total that can be calculated in the first table: 186.

2) Perform a Chi-2 test to decide whether the observed distribution can, or can't, contradict the dealership's assertion, with a 5% significance level. Explain this level's meaning. 4 pts Let's match the observed and theoretical frequencies, and calculate the part Chi-2 and the total Chi-2, Chi2calc :

obs 78 56 30 18 4

th 96 48 24 12 6

χ² 3.375 1.333333 1.5 3 0.666667

Chi2calc = 9.875

In the Chi2 table, with 4 dof, we can see that 9.875 is located between the significance levels 2% and 5%.

Therefore, with a 5% level, we can reject the adequacy between our observations and the theoretical distribution: we are at least 95% confident on denying the car dealership's assertion (between 95% and 98% confident).

2017-2019 – S3 – Mathematics – TEST 2 SOLUTIONS Page 3 / 3 Exercise 4 conformance testing of a proportion (4 points)

A computer manufacturer says that "more than 90% of the tablets of the new a-Tab model do not fail for the first 900 hours of use". This model is a large-scale production. One company ordered him and used 25 a-Tab tablets.

After 900 hours of service, 4 tablets experienced at least one failure.

From the results obtained in this sample of 25 tablets, build a conformance test on whether to reject the supplier's declaration, at a 5% significance level.

Hypothesis:

The value to be tested is a proportion: π⁰ = 90% = 0.9.

We would reject the manufacturer's assertion in case we believe that in fact less than 90% of the a-Tabs don't experience any failure in the first 900 hours. This is then a one-sided test.

H0 : π = 0.9 H1 : π < 0.9 Statistics:

Under the null hypothesis, the proportions that can be measured from 25-sized samples form a variable F described by the law ^{0.9; 0.9 0.1}

(

)

25

 × =

 

 

 

N N

Significance level: α = 5%

Comparison:

According to the standard normal law, the value u_th such that p(U < u_th) = 5% is –1.645 (below this value, we will consider our sample as too far from the expected proportion, and we will reject H0).

The proportion p (of non-defective tablets) noted in the sample is 21/25 = 0.84. A variable change bound for

N

Decision: At a 5% significance level, H0 can't be rejected: even though the proportion was 84% in ours ample, we can't deny the assertion of the manufacturer (proportion = 90% in the whole production) with 95% confidence.

____________________ TEST END ____________________