Estimated of COVID-19 Sampling Mean in Burkina Faso

Our objective in the development of this document, was to establish an estimate of the average of different variables in particular, the case of daily contamination of covid-19, the case of recovery and finally the case of lethality of COVID-19 in Burkina to give an idea of the real rate of contamination in Burkina Faso. To achieve this objective we used the tools of inferential statistics.


Introduction
The COVID-19 pendulum began in China in December 2019 and since humanity has been fighting against this disease. This emerging disease, little known around the world, has been declared a public health emergency of international concern (USPPI) by WHO. In light of the country's trade with China and the rest of the world, the spread of this pandemic was predictable. Burkina Faso registered its first case of COVID-19 on March 9, 2020. Since that day, it has been registering cases daily.
Coronaviruses, also called crown viruses,are transmitted from human to human by air, by coughing or sneezing and by close contact (touching or shaking hands). They can also be transmitted by contact with an object and / or a surface contaminated by the virus (door handle, stair railing, elevator buttons, etc.).
In view of the extent of COVID's disease, it is difficult for a scientist to sit around and do nothing.This in this dynamic that our small research team has decided to make our contribution. We offer through certain properties of mathematical statistics an estimate of the average of the following parameters: Number of daily contamination, number of healing and number of cotidient deaths This modeling propose in our document is an empirical study. We have observed the evolution of the disease and the reaction of the population to the disease. For 63 days we collected data on the disease through local media. We believe that behind the figures announced lies the behavior of the people of Burkina. Working on 63 days of data is acceptable. We could have waited 90 days to observe again, but the urgency of the situation will not allow it.

Sample as Creation of a Random Tuple
If we consider the samples as the realization of a random variable with n dimensions (X 1 , X 2 , ..., X n ), (Y 1 , Y 2 , ..., Y n ) and (Z 1 , Z 2 , ..., Z n ). These multidimensional variables is defined on the set of all possible samples of size n that can be extracted from the mother population (Burkina Faso). They then have all their own probability distributions, linked to that of X for X i , Y for Z i and Z for Z i . All the quantities that can be calculated in the sub-population sample here become realizations of random variables, defined, like the tuple, on the set of all samples of size n that we can extract from the population.
x i , is the realization, in the sample considered, of empirical random mean variable defined byX = 1 n n i=1 X ī y = 1 n n i=1 y i , is the realization, in the sample considered, of empirical random mean variable defined byȲ = 1 n n i=1 Y ī z = 1 n n i=1 z i , is the realization, in the sample considered,of empirical random mean variable defined byZ = 1 is the realization, in the sample considered, of the random variable empirical variance defined by (y i −ȳ) 2 is the realization, in the sample considered, of the random variable empirical variance defined by (z i −z) 2 is the realization, in the sample considered, of the random variable empirical variance defined by Given a risk α = 0, 05, the centered-reduced normal distribution table gives a number u α/2 = 1, 645 such that :

Confidence Interval Estimate of the Variance in the Number of Contamination
Cases µ x is unknown. We take S 2 n−1,x as an estimator of σ 2 . SiX Given a risk α = 0, 05, the chi-square table at 63 degrees of freedom gives z 1 and z 2 such as : Since the interval [z 1 , z 2 ] is not unique, it depends on the risks α 1 and α 2 admitted outside the interval.

Estimated Average Contamination
X follows a normal law, σ x is unknown. In this case, we consider the variable with s n−1,x = 63 62 × s n,x = 9.95 . which follows a Student law with 63 degrees of freedom. Given a risk α = 0.05, Student's table at 63 degrees of freedom gives t α/2 = 1.669 such as :

Classical Estimation (From a Sample of Size n)
Estimation of the different means : It comes that after observation and analysis of the different diagrams: The largest number (49 cases) of gestation was recorded on the 38th day out of the 63 test days with a median of 7. The peak (41 cases) of COVID-19 comtamination with burkina faso is reached on the 58th day out of the 63 days of testing with a median of 9. On the 30th day the Bukina recorded its largest case of death (4 deaths). We also have a median of 1 case of death over the 63 days of testing.  The following table summarizes the point values of some characteristic parameters of the random values X, Y and Z We observe in the following graph on the X axis we have the number of daily contamination of COVID-19, the Y axis represents the cases of recovery and finally the Z axis represents the daily lethality of COVID-19 in Burkina.
The properties of sampling an average gives:

Confidence Interval Estimation
Let X, Y and are a random variable defined on the mother population, of mean µ x , µ y and µ z , of variance σ 2 x , σ 2 y and σ 2 z . We have 63 days of data, we assume that the different means are the means of another sampling to allow us to use the laws of inferential statistics. We also have different situations that presents itself to us. Either the laws of X, Y and Z are known or they are not. Which brings us to the following hypotheses.
3.2.1 Confidence Interval of an Average (so µ is Unknown) Given a risk α = 0.05, the centered-reduced normal distribution table gives a number u α/2 = 1.645 such that : T 2 (X 1 , X 2 , ..., X n ) = 14.09 We can therefore say that in Burkina Faso, the average daily contamination is between 10.02 and 14.09 with an error rate of 5%. In short if we consider that σ x is known and X follows a normal law; the number of contamination for the 63 days varies between 631.26 and 887.67. The total number of official contamination is from 760 the 63rd day. The 95% confidence interval is well verified.
In the same dynamic if we want to have an idea after 90 days how many cases we can have in the same conditions we will have an interval of [901.8;1,268.1]. The maximum number of cases that can be expected with this method is 1.269.

Average Healing Estimate
If Y N µ y , σ y , so ∀i ∈ {1, ..., n}, Y i N µ y , σ y andȲ N µ y , σ y √ n . Given a risk α = 0.05, the centered-reduced normal distribution table gives a number u α/2 = 1.645 such that : The average of daily healing is between 7.31 and 11.52 with a 95% confidence level.
If we now consider that σ x is unknown and that X always follows a normal distribution.

Estimated Lethality Mean
Z follows a normal law, σ z is unknown. In this case, we consider the variable T z =Z − µ z S n−1,z √ n with s n−1,z = 63 62 × s n,z = 0.96. which follows a Student law with 63 degrees of freedom. Given a risk α = 0.05, Student's table at 44 degrees of freedom gives t α/2 = 1.669 such as : T 2 (Z 1 , Z 2 , ..., Z n ) = 1.22 we have (1 − 0.05) × 100 odds out of 100 that the interval [0.58, 0.99] covers the average µ z .

Third Case
X follows an unknown law, n is large (n = 63).
The hypotheses of the central limit theorem are then satisfied and we deduce thatX N µ x , σ x √ 63 .
• If σ is known, we are brought back to the first case.
• If σ is unknown, we should make the normality assumption for the law of X in order to be able to use the Student variable. However, as n is large, we would approach the student law by a reduced centered normal law and the results are identical to those of the first case (where we estimate σ by s n−1 ). We will therefore content ourselves with estimating σ by s n−1 without the constraining assumption of normality for X.

0.95 Confidence Interval Estimate of the Variance in the Number of Contamination Cases
µ x is unknown. We take S 2 n−1,x as an estimator of σ 2 . SiX N (µ x , σ x ) then Given a risk α = 0.05, the chi-square table at 63 degree of freedom gives z 1 and z 2 such as : Since the interval [z 1 , z 2 ] is not unique, it depends on the risks α 1 and α 2 admitted outside the interval.
In fact z 1 is such that for n > 30 the chi-2 law can be approximated by the normal law N 63, by the same process we get from where z 2 = 40.69. If we take symmetrical risks α 1 = α 2 = α 2 , we have (1 − 0.05) × 100 chance out of 100 that the confidence interval at risk that the confidence interval because the X i are independent and of the same law N (µ x , σ x ). Given a risk α, the chi-square

0.95 Confidence Interval Estimate of Variance in Number of Healing Cases
µ Y is unknown. We take S 2 n−1,y as an estimator of σ 2 . SiȲ N µ y , σ y then Given a risk α = 0.05, the chi-square table at 63 degree of freedom gives z 1 and z 2 such as : Since the interval [z 1 , z 2 ] is not unique, it depends on the risks α 1 and α 2 admitted outside the interval.
In fact z 1 is such that z 2 is such that from where z 2 = 40.69. If we take symmetrical risks α 1 = α 2 = α 2 , we have (1 − 0.05) × 100 chance out of 100 that the confidence interval at risk that the confidence interval because the Y i are independent and of the same law N µ y , σ y .
Given a risk α, the chi-square

0.95 Confidence Interval Estimate of Variance in Number of Deaths
µ Y is unknown. As resolved in cases of contamination or healing we get : If we take symmetrical risks α 1 = α 2 = α 2 , we have (1 − 0.05) × 100 chance out of 100 that the confidence interval at risk that the confidence interval because the X i are independent and of the same law N (µ, σ). Given a risk α, the chi-square We observe in the following graph on the X axis we have the number of daily contamination of COVID-19, the Y axis represents the cases of recovery and finally the Z axis represents the daily lethality of COVID-19 in Burkina.

Statistics and Data Analysis
In the following table we offer a summary of the different estimates that we were able to model.
• I 1 tolerance interval with 95% confidence starting from the fact that X follows a normal law and known sigma.
• I 2 tolerance interval with 95% confidence starting from the fact that X follows a normal law and unknown sigma.
• I 63 1 is the tolerance interval with 95% confidence starting from the fact that X follows a normal law and known sigma after 63 days.
• I 90 1 is the tolerance interval with 95% confidence starting from the fact that X follows a normal law and known sigma after 90 days.  Vol. 12, No. 4; • I 63 2 is the tolerance interval with 95% confidence starting from the fact that X follows a normal law and unknown sigma after 63 days.
• I 90 2 is the tolerance interval with 95% confidence starting from the fact that X follows a normal law and unknown sigma after 90 days.
• I σ is the variance tolerance interval with a 95% confidence level.
• I 150 1 is the tolerance interval with 95% confidence starting from the fact that X follows a normal law and known sigma after 150 days.
• I 150 2 is the tolerance interval with 95% confidence starting from the fact that X follows a normal law and unknown sigma after 150 days.

Ancillary Analysis
The second case gives us a much wider interval. In this estimation context we think that it is the most suitable interval.
If we consider the information in table 1, the number of cases of contamination will be between [896; 1,274], the number of cases of recovery will be between [651;1,044], the number of deaths will be between [52;90] after three months .
The number of cases of contamination will be between [1,494,2,123], the number of cases of recovery will be between [1,086;1,739], the number of deaths will be between [87;149] after three months . Report any other analyses performed, including subgroup analyses and adjusted analyses, indicating those that were prespecified and those that were exploratory (though not necessarily in the level of detail of primary analyses). Consider putting the detailed results of these analyses on the supplemental online archive. Discuss the implications, if any, of the ancillary analyses for statistical error rates.

Discussion
This study is based on a 5% error assumption. In clear terms, we have a 5% chance of being wrong. We also consider that the random variables follow a law. This forecasts is 2123 cases of contamination, 1739 cases of cure and 148 cases of death in 5 months if no palliative measure is taken. This study is an empirical modeling, these figures will remain true as long as the response to COVID-19 from the government and people of Burkina Faso does not evolve.
The maximum number of cures (1728) is lower than the number of new cases (2123). We believe that the Burkinabè State should optimize the system in place, so that the number of cures exceeds the number of new cases. It is in this sense that the disease can be overcome.
The rate of 90 dead over 3 months and 149 deaths after 5 months of covid-19 in Burkina. It is less than that of European countries. But it is not acceptable to let people die. The strategy implemented in Burkina needs to be improved. By popularizing the barrier gesture protocols. Undoubtebly taking a large-scale test. This will make it possible to detect and isolate vulnerable or affected people more quickly.

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