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Exercises in Statistics
Each theme begins with an abstract of the lecture and an exercise with detailed solution. The computations have been made with a software; due to rounding errors, there may be some minor differences with computations from statistical tables.
- 1 Data and models Contents
- 1.1 Empirical distributions
- 1.2 Probabilities and conditional probabilities
- 1.3 Binomial distribution
- 1.4 Hypergeometric distribution
- 1.5 Normal distribution
- 1.6 Approximation of a binomial to a normal distribution
- 2 Parametric estimation
- 2.1 Estimating a parameter
- 2.2 Confidence intervals for a Gaussian sample
- 2.3 Confidence interval for the expectation on a large sample
- 2.4 Confidence interval of a probability for a large sample
- 3 Statistical testing
- 3.1 Decision rule, threshold and p-value
- 3.2 Tests on a sample
- 3.3 Comparison of two independent samples
- 3.4 The chi-squared adjustment test
- 3.5 The chi-squared independence test
- 4 Linear regression
- 4.1 Regression line and prediction
- 4.2 Confidence and prediction intervals
- 4.3 Tests on a regression
1 Data and models
1.1 Empirical distributions
Let ( x 1 ,... , xn ) be a sample, i.e. a series of numerical values for a certain variable in a set of n individuals.
- The modalities are the different values.
- The empirical mean is x =
n
∑^ n
i =
xi.
- The empirical variance is s^2 x =
( 1 n
∑^ n
i =
x^2 i
) − x^2.
- The empirical standard deviation is the square root of the empirical variance.
- A sample is centered and reduced if its mean is 0 and its variance 1. In order to center and reduce a sample, substract the mean from each modality, then divide by the standard deviation.
- The empirical frequency of an interval is the ratio of the number of values in that interval, to the total number of individuals.
- The median is the smallest modality such that at least 50% of the values are smaller or equal.
- The lower quartile is the smallest modality such that at least 25% of the values are smaller or equal.
- The upper quartile is the smallest modality such that at least 75% of the values are smaller or equal.
- A statistical character is considered as continuous when (almost) all values are different. When for most modalities, several individuals have the same value, the character is discrete.
Exercise 1.1.1. Here are numbers by age of non-smoking mothers at delivery.
age 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 number 7 8 9 10 12 3 2 5 4 5 2 4 2 0 1
- What are the modalities?
The modalities are the whole numbers between 21 and 35_._
The standard deviation is the square root of the variance:
sx =
that is approximately 3 years and 7 months.
- Find the values of the empirical distribution function.
The values of the empirical distribution function are the cumulated sums of fre- quencies.
age 21 22 23 24 25 26 27 cum. freq. (^747157424743474467449745174) rounded 0.095 0.203 0.324 0.459 0.622 0.662 0.
56 74
60 74
65 74
67 74
71 74
73 74
73 74
74 74 0.757 0.811 0.878 0.905 0.959 0.986 0.986 1
- What is the empirical frequency of the interval [22 ; 25]?
It is the sum of empirical frequencies for the modalities 22 , 23 , 24 , 25 , or else the increment of the empirical distribution function F (25) − F (21) , that is 39 / 74 ' 0_._ 527_. More than half of the women in the sample are between_ 22 and 25 years old.
- Draw a graphical representation of the empirical distribution function. Determine from the graph the median and the quartiles of the sample.
0.0 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
The median is 25 years; the first quartile is 23 years, the last quartile is 28 years.
- Compare the mean with the median, then the standard deviation with the dis- tances between the median and the quartiles.
The mean is larger than the median, which is normal for a distribution skewed to the right. For the same reason, the gap between the last quartile and the median is larger than that between the median and the first quartile. Both are lower than the standard deviation: this is the case for most distributions, whether they are symmetrical or skewed.
Exercise 1.1.2. Here are numbers by age of smoking birth mothers at delivery.
age 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 number 5 5 4 3 3 5 1 4 3 2 3 2 1 1 1
- What are the modalities?
- Is this a discrete or a continuous variable?
- Find the empirical frequencies of the modalities.
- Represent the empirical frequencies on a bar chart.
- Find the empirical mean, variance, and standard deviation of the sample.
- Find the values of the empirical distribution function.
- What is the empirical frequency of the interval [22 ; 25]?
- Draw a graphical representation of the empirical distribution function. Determine from the graph the median and the quartiles of the sample.
- Compare the mean with the median, then the standard deviation with the dis- tances between the median and the quartiles.
Exercise 1.1.3. Consider the sample (1 , 0 , 2 , 1 , 1 , 0 , 1 , 0 , 0).
- What is its empirical mean?
- What is its empirical variance?
- Center and reduce this sample.
- If you had to propose a model for these data: would you choose a discrete or a continuous model?
Exercise 1.1.4. Consider the sample
(1_._ 2 , 0_._ 2 , 1_._ 6 , 1_._ 1 , 0_._ 9 , 0_._ 3 , 0_._ 7 , 0_._ 1 , 0_._ 4).
- What is the conditional probability for a sheep to have a positive reaction knowing that it is not ill?
P[ T | M ] = 1 − P[ T | M ] = 1 − 0_._ 9 = 0_._ 1_._
- What is the probability for a sheep not to be ill and have a positive reaction?
P[ T and M ] = P[ T | M ] P[ M ] = 0_._ 1 × 0_._ 7 = 0_._ 07_._
- What proportion of the sheep have a positive reaction?
Use the formula of total probabilities or compute it straight away, by distinguishing among those sheep reacting positively, those which are ill from those which are not.
P[ T ] = P[ T and M ] + P[ T and M ]
= P[ T | M ] P[ M ] + P[ T | M ] P[ M ]
= 0_._ 8 × 0_._ 3 + 0_._ 1 × 0_._ 7 = 0_._ 24 + 0_._ 07 = 0_._ 31_._
- What is the probability for a sheep to be ill, knowing that it has reacted posi- tively?
Use the Bayes formula or prove it again as follows.
P[ M | T ] =
P[ T and M ] P[ T ]
P[ T | M ] P[ M ]
P[ T | M ] P[ M ] + P[ T | M ] P[ M ]
0. 8 × 0. 3
0. 8 × 0. 3 + 0. 1 × 0. 7
- What is the probability for a sheep not to be ill, knowing it has reacted negatively?
Use the Bayes formula or prove it again as follows.
P[ M | T ] =
P[ T and M ] P[ T ]
P[ T | M ] P[ M ]
P[ T | M ] P[ M ] + P[ T | M ] P[ M ]
0. 9 × 0. 7
0. 9 × 0. 7 + 0. 2 × 0. 3
Exercise 1.2.2. There are three sorts of a given plant: early, normal, and late. It can also be either dwarf or tall. In a sample of plants grown from 1000 seeds, there are 600 dwarf, 200 late, 300 early dwarf, 250 normal tall, 100 late tall. Consider the plant grown from a seed taken at random.
- What is the probability that it is early? normal? late? dwarf? tall?
- A dwarf plant is observed. What is the probability that it is early? normal? late?
- A tall plant is observed. What is the probability that it is early? normal? late?
- A late plant is observed. What is the probability that it is dwarf? tall?
Exercise 1.2.3. In a batch of manufactured items, 5% are faulty. The items are checked, but the checking is not perfect. If the item is good, it is accepted with probability 0_._ 96 ; if it is faulty, it is rejected with probability 0_._ 98. An item is chosen at random, then checked.
- What is the probability that this item is rejected?
- What is the probability that it is good, knowing that it has been rejected?
- What is the probability that it is faulty, knowing that it has been accepted?
- What is the probability that there is an error in the checking (the item is good and rejected or bad and accepted)?
Exercise 1.2.4. Here are the percentages of the different blood types in France.
Group O A B AB Factor Rhesus + 37.0 38.1 6.2 2. Rhesus – 7.0 7.2 1.2 0.
- Determine the probability distribution of the four groups O, A, B, AB in the French population.
- Determine the probability distribution of the four groups among the persons with positive rhesus
- Determine the probability distribution of the four groups among the persons with negative rhesus
- If a person of group O is chosen at random, what is the probability that he/she has a negative rhesus? Same question for a person of group B.
- What is the probability for the surgery to succeed at least 3 times?
P[ X > 3] = P[ X = 3] + P[ X = 4] + P[ X = 5]
( 5 3
) 0_._ 93 0_._ 12 +
( 5 4
) 0_._ 94 0_._ 11 +
( 5 5
) 0_._ 95 0_._ 10
Exercise 1.3.2. When a hunter aims at a helpless rabbit, he has 1 chance out of 10 to hit it.
- Two hunters aim independently at the same rabbit. Find the probability that:
(a) neither of them hit; (b) only one of them hits; (c) both hunters hit.
- Four hunters aim independently at the same rabbit.
(a) What is the probability distribution of the number of shots suffered by the poor animal? Give the expectation and variance of that distribution. (b) What is the probability that the rabbit is hit at most twice? (c) What is the probability that the rabbit is hit at least twice?
- Ten hunters aim independently at the same rabbit.
(a) What is the probability for the rabbit not to be hit? (b) What is the probability that the rabbit becomes inedible (if it has received at least 5 shots).
Exercise 1.3.3. At an identification session, 6 witnesses are asked to identify a murderer among 4 suspects, including yourself.
- If each one of the 6 witnesses chooses at random, what are your chances:
(a) of not being pointed out? (b) of being pointed out exactly once? (c) of being pointed out twice or more?
- It turns out that 2 of the 6 witnesses have identified you as the murderer. Re- ferring to 1 (c), do you expect that the judge will think that this may be due to chance?
- What if 4 of the 6 witnesses have identified you?
1.4 Hypergeometric distribution
- In a set of N elements, among which m have been marked, n distinct elements are selected at random. The random variable X , equal to the number of marked elements among the selected n , follows the hypergeometric distribution with pa- rameters N, m, n.
- In the case where n ≤ m and n ≤ N − m , X may take all integer values between 0 and n.
- For any integer k between 0 and n , X takes value k with probability:
P[ X = k ] =
( m k
)( N − m n − k
) ( N n
- The expectation of X is nm/N.
Exercise 1.4.1. There are 18 girls and 11 boys in a certain group of students. A sample of 5 persons is chosen at random in that group. Let X be the random variable equal to the number of girls in that sample.
- What model do you propose for X?
The distribution of X is the hypergeometric distribution with parameters N = 29 (total number of persons), m = 18 (the “marked” individuals are the girls), and n = 5 (the size of the sample). The values are the integers between 0 and 5_. For any integer k_ = 0 , 1_... ,_ 5 :
P[ X = k ] =
( 18 k
)( (^11) 5 − k
) ( 29 5
- Give the expectation of X.
The expectation of X is 5 × 18 / 29 ' 3_._ 1_. It is the size of the sample, multiplied by the proportion of girls in the group._
- Find the probability of having only girls in the sample.
P[ X = 5] =
( 18 5
) ( 29 5
- Find the probability of having at least one girl in the sample.
The value of P[ X > 1] must be calculated. It could be done as P[ X = 1] + P[ X =
- If a random variable X follows the N ( μ, σ^2 ) distribution, then ( X − μ ) /
σ^2 follows the N (0 , 1) distribution. Thus:
P[ a 6 X 6 b ] = P
[ a − μ √ σ^2
X − μ √ σ^2
b − μ √ σ^2
]
= F
( b − μ √ σ^2
) − F
( a − μ √ σ^2
) ,
where F is the distribution function of the N (0 , 1).
- If X and Y are two independent random variables, with respective distributions N ( μx, σ^2 x ) and N ( μy, σ y^2 ), then X + Y follows the N ( μx + μy, σ^2 x + σ y^2 ) and X − Y follows the N ( μx − μy, σ^2 x + σ y^2 ).
Exercise 1.5.1. The height X of men in France is modeled by a normal distribution N (172 , 196) (unit: cm).
- What proportion of French men are less than 160 cm tall?
P[ X < 160] = P
[ X − 172 √ 196
] = F (− 0_._ 857) = 1 − F (0_._ 857) = 0_._ 1957 ,
where F denotes the distribution function of the N (0 , 1) distribution.
- What proportion of French men are more than two meters tall?
P[ X > 200] = P
[ X − 172 √ 196
] = 1 − F (2) = 0_._ 02275_._
- What proportion of French men are between 165 and 185 centimeters tall?
P[165 < X < 185] = P
[ 165 − 172 √ 196
X − 172
]
= F (0. 928) − F (− 0. 5) = 0. 8234 − 0. 3085 = 0. 5149.
- If ten thousand French men chosen at random were ranked by increasing size, how tall would be the 9000-th?
The question amounts to finding the size such that 90% of the French are smaller, i.e. the 90 -th quantile of the ninth decile. Let x be that size.
P[ X < x ] = P
[ X − 172 √ 196
x − 172 √ 196
] = 0_._ 9
Thus x √− 196172 is the value of the quantile function of the N (0 , 1) distribution for p = 0_._ 9 , that is 1_._ 2816_. Therefore:_
x = 172 + 1_._ 2816 ×
196 ' 190 cm.
- The height of French women is modeled by a normal distribution N (162 , 144) (in centimeters). What is the probability for a French man chosen at random to be taller than a French woman chosen at random?
Let X denote the size of the man and Y that of the woman, and suppose they are independent. Then X − Y follows the normal distribution N (10 , 340). The probability for X to be larger than Y is the probability for X − Y to be positive:
P [ X − Y > 0] = P
[ ( X − Y ) − 10 √ 340
] = 1 − F (− 0_._ 5423) = 0_._ 7062_._
Exercise 1.5.2. Let X be a random variable with N (0 , 1) distribution.
- Express with the distribution function of X , then compute using the table the following probabilities.
(a) P[ X > 1_._ 45] (b) P[− 1_._ 65 6 X 6 1_._ 34] (c) P[| X | < 2_._ 05]
- Find the value of u in the following cases.
(a) P[ X < u ] = 0_._ 63 (b) P[ X > u ] = 0_._ 63 (c) P[| X | < u ] = 0_._ 63
Exercise 1.5.3. Let X be a random variable with N (0 , 1) distribution. Let Y = 2 X − 3.
- What is the distribution of Y?
- Find P[ Y < −4].
- Find P[− 2 < Y < 3].
Exercise 1.5.4. Let X be a random variable with N (3 , 25) distribution.
- Express with the distribution function of the N (0 , 1) distribution, then compute using the table the following probabilities.
(a) P[ X < 6] (b) P[ X > −2] (c) P[− 1 6 X 6 1_._ 5]
- Find the value of u in the following cases.
- In that case, if X follows the B( n, p ) distribution, one computes the probability for X to be in the interval [ a, b ] by:
P[ a 6 X 6 b ] = P
(^) √ a^ −^ np np (1 − p )
X − np √ np (1 − p )
b − np √ np (1 − p )
' F
(^) √ b^ −^ np np (1 − p )
(^) − F
(^) √ a^ −^ np np (1 − p )
(^) ,
where F is the distribution function of the N (0 , 1).
Exercise 1.6.1. From past experience, it is known that a certain surgery has a 90% chance to succeed. This surgery is performed by a certain clinic 400 times each year. Let N be the number of successes next year. The normal approximation will be used for N.
- Find the expectation and variance of N.
The expectation is 400 × 0_._ 9 = 360 , the variance is 400 × 0_._ 9 × 0_._ 1 = 36_._
- Find the probability for the clinic to perform successfully the surgery at least 345 times.
P[ N > 345] = P
[ N − 360 √ 36
]
= 1 − F (− 2. 5) = F (2. 5) = 0. 9938.
- Find the probability that the surgery fails in that clinic more than 28 times in the year.
P[ N 6 372] = P
[ N − 360 √ 36
]
= F (2) = 0. 9772.
- The insurance accepts to cover a certain number of failed surgeries: that number has only a 1% chance to be exceeded. What number is it?
Let n be the number of failed surgeries that must be determined. The correspond- ing number of successes is 400 − n. Therefore P[ N 6 400 − n ] = 0_._ 01_. Now:_
P[ N 6 400 − n ] = P
[ N − 360 √ 36
400 − n − 360 √ 36
]
= F
( 40 − n √ 36
) = 0_._ 01_._
The number^40 √− 36 n is the quantile at 0_._ 01 of the N (0 , 1) , that is − 2_._ 3236_. Thus:_
40 − n √ 36
= − 2_._ 3263 =⇒ n = 40 + 2_._ 3263
The reasoning could also be applied to the number of failed surgeries R = 400− N. It follows the binomial distribution B(400 , 0_._ 1) , that can be approximated by the normal N (40 , 36). The desired number is such that P[ R > n ] = 0_._ 01_._
P[ R > n ] = P
[ R − 40 √ 36
n − 40 √ 36
]
= 1 − F
( n − 40 √ 36
)
= F
( 40 − n √ 36
) = 0_._ 01_._
Of course the result is the same.
Exercise 1.6.2. Among people old enough to receive an injection against the flu, 40% of them ask for it. In a population of 150000 persons old enough to receive the injection, let N be the number of those that will ask for it.
- What model would you propose for N?
- If 60500 syringes are prepared, what is the probability that these will not suffice?
- Find the number of syringes that should be prepared to ensure that there will be enough with 90% probability at least.
Exercise 1.6.3. A restaurant, serving only upon reservation, has 50 seats. The proba- bility that a someone with a reservation does not show up is 1 / 5. Let N be the number of meals served on a given day. The normal approximation will be used for N.
- If the chef accepts 50 reservations, what is the probability he will serve more than 45 meals?
- If he accepts 55 reservations, what is the probability he will find himself in an embarassing situation?
Exercise 1.6.4. Suppose there is probability 0_._ 1 of being controlled in the tramway. Mr A. makes 700 trips per year. The normal approximation will be used for the number of fraud checks.
- Find the probability that Mr. A will be controled between 60 and 80 times in the year.
2 Parametric estimation
2.1 Estimating a parameter
- For an unknown parameter, an estimator is a function of the data, taking values close to that parameter. It is unbiased if its expectation is equal to the parameter. It is convergent if the probability for it to take a value at distance up to ε from the parameter tends to 1 as the size of the sample tends to infinity.
- The empirical frequency of an event is an unbiased convergent estimator of the probability of that event.
- The empirical mean of a sample is an unbiased convergent estimator of the the- oretical expectation of the variables.
- The empirical variance of a sample is a convergent estimator of the theoretical variance of the variables. Un unbiased estimator is obtained by multiplying the empirical variance by n/ ( n −1), where n is the size of the sample.
Exercise 2.1.1. Consider the statistical sample (1 , 0 , 2 , 1 , 1 , 0 , 1 , 0 , 0).
- Find its empirical mean and variance.
x =
and s^2 x =
- Supposing that the data are realizations of a variable with an unknown dis- tribution, give unbiased estimates for the expectation and the variance of that distribution.
The empirical mean ( 2 / 3 ) is an unbiased estimate of the expectation. An unbiased estimate of the variance is obtained, multiplying s^2 x by 9 / 8 : this gives 1 / 2_._
- The data of the sample are modeled by a binomial distribution B(2 , p ). Use the empirical mean to propose an estimate for p.
The expectation of the B(2 , p ) distribution is 2 p. It is estimated by the empirical mean (here 2 / 3 ). Thus p can be estimated by:
2 / 3 2
- With the same model, use the empirical variance to propose another estimate for p.
The variance of the B(2 , p ) distribution is 2 p (1 − p ). It is estimated by 1 / 2_. The value of p can be estimated by solving the equation_ 2 p (1 − p ) = 1 / 2 , giving p = 1 / 2_._
- The data of the sample are now modeled by a Poisson distribution P( λ ), the expectation of which is λ. What estimate would you propose for λ?
The parameter λ can be estimated by the empirical mean, 2 / 3_._
Exercise 2.1.2. Consider the statistical sample (1 , 3 , 2 , 3 , 2 , 2 , 0 , 2 , 3 , 1).
- Supposing that the variables are realizations of a variable with unknown dis- tribution, give unbiased estimates for the expectation and the variance of that distribution.
- The data of that sample are modeled by a B(3 , p ) distribution. Use the empirical mean to propose an estimate for p.
Exercise 2.1.3. Consider the statistical sample (1_._ 2 , 0_._ 2 , 1_._ 6 , 1_._ 1 , 0_._ 9 , 0_._ 3 , 0_._ 7 , 0_._ 1 , 0_._ 4).
- The data of that sample are modeled by a uniform distribution on the interval [0 , θ ]. What estimate would you propose for θ?
- The data of the sample are now modeled by a normal distribution N ( μ, σ^2 ). What estimates would you propose for μ and σ^2?
2.2 Confidence intervals for a Gaussian sample
A Gaussian sample is a n -tuple ( X 1 ,... , Xn ) of independent random variables with normal distribution N ( μ, σ^2 ). The empirical mean and variance of the sample are given by:
X =
n
∑^ n
i =
Xi et S^2 =
( 1 n
∑^ n
i =
X i^2
) − X 2 ,
- If the theoretical variance σ^2 is known , a confidence interval at level 1 − α for μ is obtained by: (^) [
X − uα
σ^2 √ n
; X + uα
σ^2 √ n
] ,
where uα is the quantile of order 1 − α/ 2 for the normal distribution N (0 , 1).
- If the theoretical variance σ^2 is unknown , a confidence interval at level 1 − α for μ is obtained by: (^) [
X − tα
S^2
n − 1
; X + tα
S^2
n − 1
] ,
where tα is the quantile of order 1 − α/ 2 for the Student distribution with param- eter n − 1.
