CONFIDENCE INTERVALS FOR FREQUENCIES AND FRACTIONS

© 2008

National Institute of Public Health, Oslo, Norway

The article describes and discusses the calculation of confidence intervals for frequencies and proportions using the Wald, Wilson, Clopper - Pearson methods, using the angular transformation and the Wald method with Agresti - Coull correction. The presented material provides general information about methods for calculating confidence intervals for frequencies and proportions and is intended to arouse the interest of journal readers not only in using confidence intervals when presenting the results of their own research, but also in reading specialized literature before starting work on future publications.

Keywords: confidence interval, frequency, proportion

One of the previous publications briefly mentioned the description of qualitative data and reported that their interval estimate is preferable to point estimate for describing the frequency of occurrence of the characteristic being studied in the population. Indeed, since research is conducted using sample data, the projection of the results onto the population must contain an element of sampling imprecision. The confidence interval is a measure of the accuracy of the parameter being estimated. It is interesting that some books on basic statistics for doctors completely ignore the topic of confidence intervals for frequencies. In this article we will look at several ways to calculate confidence intervals for frequencies, implying such sample characteristics as non-repetition and representativeness, as well as the independence of observations from each other. In this article, frequency is understood not as an absolute number showing how many times a particular value occurs in the aggregate, but as a relative value that determines the proportion of study participants in whom the studied characteristic occurs.

In biomedical research, 95% confidence intervals are most commonly used. This confidence interval is the area within which the true proportion falls 95% of the time. In other words, we can say with 95% reliability that the true value of the frequency of occurrence of a trait in the population will be within the 95% confidence interval.

Most statistics manuals for medical researchers report that frequency error is calculated using the formula

where p is the frequency of occurrence of the characteristic in the sample (value from 0 to 1). Most domestic scientific articles indicate the frequency of occurrence of a trait in a sample (p), as well as its error (s) in the form p ± s. It is more appropriate, however, to present a 95% confidence interval for the frequency of occurrence of a trait in the population, which will include values ​​from

before.

Some manuals recommend that for small samples, replace the value of 1.96 with the value of t for N – 1 degrees of freedom, where N is the number of observations in the sample. The t value is found from tables for the t-distribution, available in almost all statistics textbooks. The use of the t distribution for the Wald method does not provide visible advantages compared to other methods discussed below, and therefore is not recommended by some authors.

The method presented above for calculating confidence intervals for frequencies or proportions is named Wald in honor of Abraham Wald (1902–1950), since its widespread use began after the publication of Wald and Wolfowitz in 1939. However, the method itself was proposed by Pierre Simon Laplace (1749–1827) back in 1812.

The Wald method is very popular, but its application is associated with significant problems. The method is not recommended for small sample sizes, as well as in cases where the frequency of occurrence of a characteristic tends to 0 or 1 (0% or 100%) and is simply impossible for frequencies of 0 and 1. In addition, the approximation of the normal distribution, which is used when calculating the error , “does not work” in cases where n · p< 5 или n · (1 – p) < 5 . Более консервативные статистики считают, что n · p и n · (1 – p) должны быть не менее 10 . Более детальное рассмотрение метода Вальда показало, что полученные с его помощью доверительные интервалы в большинстве случаев слишком узки, то есть их применение ошибочно создает слишком оптимистичную картину, особенно при удалении частоты встречаемости признака от 0,5, или 50 % . К тому же при приближении частоты к 0 или 1 доверительный интревал может принимать отрицательные значения или превышать 1, что выглядит абсурдно для частот. Многие авторы совершенно справедливо не рекомендуют применять данный метод не только в уже упомянутых случаях, но и тогда, когда частота встречаемости признака менее 25 % или более 75 % . Таким образом, несмотря на простоту расчетов, метод Вальда может применяться лишь в очень ограниченном числе случаев. Зарубежные исследователи более категоричны в своих выводах и однозначно рекомендуют не применять этот метод для небольших выборок , а ведь именно с такими выборками часто приходится иметь дело исследователям-медикам.

Since the new variable is normally distributed, the lower and upper bounds of the 95% confidence interval for the variable φ will be φ-1.96 and φ+1.96left">

Instead of 1.96 for small samples, it is recommended to substitute the value t for N – 1 degrees of freedom. This method does not produce negative values ​​and allows more accurate estimates of confidence intervals for frequencies than the Wald method. In addition, it is described in many domestic reference books on medical statistics, which, however, has not led to its widespread use in medical research. Calculation of confidence intervals using angular transformation is not recommended for frequencies approaching 0 or 1.

This is where the description of methods for estimating confidence intervals in most books on the basics of statistics for medical researchers usually ends, and this problem is typical not only for domestic but also for foreign literature. Both methods are based on the central limit theorem, which implies a large sample.

Taking into account the shortcomings of estimating confidence intervals using the above methods, Clopper and Pearson proposed in 1934 a method for calculating the so-called exact confidence interval, given the binomial distribution of the trait being studied. This method is available in many online calculators, but the confidence intervals obtained this way are in most cases too wide. At the same time, this method is recommended for use in cases where a conservative assessment is necessary. The degree of conservativeness of the method increases as the sample size decreases, especially when N< 15 . описывает применение функции биномиального распределения для анализа качественных данных с использованием MS Excel, в том числе и для определения доверительных интервалов, однако расчет последних для частот в электронных таблицах не «затабулирован» в удобном для пользователя виде, а потому, вероятно, и не используется большинством исследователей.

According to many statisticians, the most optimal assessment of confidence intervals for frequencies is carried out by the Wilson method, proposed back in 1927, but practically not used in domestic biomedical research. This method not only allows one to estimate confidence intervals for both very small and very large frequencies, but is also applicable for a small number of observations. In general, the confidence interval according to Wilson’s formula has the form

where takes the value 1.96 when calculating the 95% confidence interval, N is the number of observations, and p is the frequency of occurrence of the characteristic in the sample. This method is available in online calculators, so its use is not problematic. and do not recommend using this method for n p< 4 или n · (1 – p) < 4 по причине слишком грубого приближения распределения р к нормальному в такой ситуации, однако зарубежные статистики считают метод Уилсона применимым и для малых выборок .

In addition to the Wilson method, the Wald method with Agresti–Coll correction is also believed to provide an optimal estimate of the confidence interval for frequencies. The Agresti-Coll correction is a replacement in the Wald formula of the frequency of occurrence of a characteristic in a sample (p) by p`, when calculating which 2 is added to the numerator and 4 is added to the denominator, that is, p` = (X + 2) / (N + 4), where X is the number of study participants who have the characteristic being studied, and N is the sample size. This modification produces results very similar to Wilson's formula, except when the event frequency approaches 0% or 100% and the sample is small. In addition to the above methods for calculating confidence intervals for frequencies, continuity corrections have been proposed for both the Wald and Wilson methods for small samples, but studies have shown that their use is inappropriate.

Let's consider the application of the above methods for calculating confidence intervals using two examples. In the first case, we study a large sample of 1,000 randomly selected study participants, of whom 450 have the trait being studied (this could be a risk factor, an outcome, or any other trait), representing a frequency of 0.45, or 45%. In the second case, the study is carried out using a small sample, say, only 20 people, and only 1 study participant (5%) has the studied trait. Confidence intervals using the Wald method, the Wald method with Agresti–Coll correction, and the Wilson method were calculated using an online calculator developed by Jeff Sauro (http://www. /wald. htm). Wilson's continuity-corrected confidence intervals were calculated using the calculator provided by Wassar Stats: Web Site for Statistical Computation (http://faculty.vassar.edu/lowry/prop1.html). Angular Fisher transformation calculations were performed manually using the critical t value for 19 and 999 degrees of freedom, respectively. The calculation results are presented in the table for both examples.

Confidence intervals calculated in six different ways for two examples described in the text

As can be seen from the table, for the first example the confidence interval calculated using the “generally accepted” Wald method enters the negative region, which cannot be the case for frequencies. Unfortunately, such incidents are not uncommon in Russian literature. The traditional way of presenting data in terms of frequency and its error partially masks this problem. For example, if the frequency of occurrence of a trait (in percentage) is presented as 2.1 ± 1.4, then this is not as “offensive to the eye” as 2.1% (95% CI: –0.7; 4.9), although and means the same thing. The Wald method with Agresti–Coll correction and calculation using angular transformation give a lower bound tending to zero. Wilson's continuity-corrected method and the "exact method" produce wider confidence intervals than Wilson's method. For the second example, all methods give approximately the same confidence intervals (differences appear only in thousandths), which is not surprising, since the frequency of occurrence of the event in this example is not much different from 50%, and the sample size is quite large.

For readers interested in this problem, we can recommend the works of R. G. Newcombe and Brown, Cai and Dasgupta, which provide the pros and cons of using 7 and 10 different methods for calculating confidence intervals, respectively. Among the domestic manuals, we recommend the book and, which, in addition to a detailed description of the theory, presents the methods of Wald and Wilson, as well as a method for calculating confidence intervals taking into account the binomial frequency distribution. In addition to free online calculators (http://www. /wald. htm and http://faculty. vassar. edu/lowry/prop1.html), confidence intervals for frequencies (and not only!) can be calculated using the CIA program ( Confidence Intervals Analysis), which can be downloaded from http://www. medschool. soton. ac. uk/cia/ .

The next article will look at univariate ways to compare qualitative data.

Bibliography

CONFIDENCE INTERVALS FOR PROPORTIONS

A. M. Grjibovski

National Institute of Public Health, Oslo, Norway

The article presents several methods for calculations confidence intervals for binomial proportions, namely, Wald, Wilson, arcsine, Agresti-Coull and exact Clopper-Pearson methods. The paper gives only a general introduction to the problem of confidence interval estimation of a binomial proportion and its aim is not only to stimulate the readers to use confidence intervals when presenting results of their own empirical research, but also to encourage them to consult statistics books prior to analyzing own data and preparing manuscripts.

Key words: confidence interval, proportion

Contact Information:

– Senior Advisor, National Institute of Public Health, Oslo, Norway

Confidence intervals ( English Confidence Intervals) one of the types of interval estimates used in statistics, which are calculated for a given significance level. They allow us to make the statement that the true value of an unknown statistical parameter of the population is within the obtained range of values ​​with a probability that is specified by the selected level of statistical significance.

Normal distribution

When the variance (σ 2) of the population of data is known, the z-score can be used to calculate confidence limits (confidence interval endpoints). Compared to using the t-distribution, using the z-score will allow you to construct not only a narrower confidence interval, but also more reliable estimates of the expected value and standard deviation (σ), since the z-score is based on a normal distribution.

Formula

To determine the boundary points of the confidence interval, provided that the standard deviation of the population of data is known, the following formula is used

Example

Assume that the sample size is 25 observations, the sample expected value is 15, and the population standard deviation is 8. For a significance level of α=5%, the Z-score is Z α/2 =1.96. In this case, the lower and upper limits of the confidence interval will be

Thus, we can say that with a 95% probability the mathematical expectation of the population will fall in the range from 11.864 to 18.136.

Methods for narrowing the confidence interval

Let us assume that the range is too wide for the purposes of our study. There are two ways to reduce the range of the confidence interval.

1. Reduce the level of statistical significance α.
2. Increase sample size.

Reducing the level of statistical significance to α=10%, we obtain a Z-score equal to Z α/2 =1.64. In this case, the lower and upper boundaries of the interval will be

And the confidence interval itself can be written in the form

In this case, we can make the assumption that with a 90% probability the mathematical expectation of the population will fall within the range .

If we want not to reduce the level of statistical significance α, then the only alternative is to increase the sample size. Increasing it to 144 observations, we obtain the following values ​​of confidence limits

The confidence interval itself will have the following form

Thus, narrowing the confidence interval without reducing the level of statistical significance is only possible by increasing the sample size. If increasing the sample size is not possible, then narrowing the confidence interval can be achieved solely by reducing the level of statistical significance.

Constructing a confidence interval for a distribution other than normal

If the standard deviation of the population is not known or the distribution is different from normal, the t-distribution is used to construct a confidence interval. This technique is more conservative, which is reflected in wider confidence intervals, compared to the technique based on the Z-score.

Formula

To calculate the lower and upper limits of the confidence interval based on the t-distribution, use the following formulas

The Student distribution or t-distribution depends on only one parameter - the number of degrees of freedom, which is equal to the number of individual values ​​of the attribute (the number of observations in the sample). The value of the Student's t-test for a given number of degrees of freedom (n) and the level of statistical significance α can be found in the reference tables.

Example

Assume that the sample size is 25 individual values, the sample expected value is 50, and the sample standard deviation is 28. It is necessary to construct a confidence interval for the level of statistical significance α=5%.

In our case, the number of degrees of freedom is 24 (25-1), therefore the corresponding table value of Student’s t-test for the level of statistical significance α=5% is 2.064. Therefore, the lower and upper limits of the confidence interval will be

And the interval itself can be written in the form

Thus, we can say that with a 95% probability the mathematical expectation of the population will be in the range .

Using the t distribution allows you to narrow the confidence interval either by reducing statistical significance or by increasing the sample size.

Reducing the statistical significance from 95% to 90% in the conditions of our example, we obtain the corresponding table value of the Student’s t-test of 1.711.

In this case, we can say that with a 90% probability the mathematical expectation of the population will be in the range .

If we do not want to reduce statistical significance, then the only alternative is to increase the sample size. Let's say that it is 64 individual observations, and not 25 as in the original condition of the example. The table value of the Student's t-test for 63 degrees of freedom (64-1) and the level of statistical significance α=5% is 1.998.

This allows us to say that with a 95% probability the mathematical expectation of the population will be in the range .

Large samples

Large samples are samples from a population of data in which the number of individual observations exceeds 100. Statistical studies have shown that larger samples tend to be normally distributed, even if the distribution of the population is not normal. In addition, for such samples, the use of a z-score and a t-distribution gives approximately the same results when constructing confidence intervals. Thus, for large samples, it is acceptable to use the z-score for the normal distribution instead of the t-distribution.

Let's sum it up

Confidence intervals.

The calculation of the confidence interval is based on the average error of the corresponding parameter. Confidence interval shows within what limits with probability (1-a) the true value of the estimated parameter lies. Here a is the significance level, (1-a) is also called confidence probability.

In the first chapter we showed that, for example, for the arithmetic mean, the true population mean in approximately 95% of cases lies within 2 standard errors of the mean. Thus, the boundaries of the 95% confidence interval for the mean will be separated from the sample mean by twice the mean error of the mean, i.e. we multiply the average error of the mean by a certain coefficient depending on the confidence level. For the average and difference of averages, the Student coefficient (critical value of the Student's test) is taken, for the share and difference of shares, the critical value of the z criterion. The product of the coefficient and the average error can be called the maximum error of a given parameter, i.e. the maximum that we can obtain when assessing it.

Confidence interval for arithmetic mean : .

Here is the sample mean;

Average error of the arithmetic mean;

s – sample standard deviation;

n

f = n-1 (Student's coefficient).

Confidence interval for differences of arithmetic means :

Here is the difference between sample means;

- average error of the difference between arithmetic means;

s 1 , s 2 – sample standard deviations;

n1,n2

The critical value of the Student's test for a given significance level a and the number of degrees of freedom f=n 1 +n 2-2 (Student's coefficient).

Confidence interval for shares :

.

Here d is the sample fraction;

– average fraction error;

n– sample size (group size);

Confidence interval for difference of shares :

Here is the difference in sample shares;

– average error of the difference between arithmetic means;

n1,n2– sample volumes (number of groups);

The critical value of the z criterion at a given significance level a ( , , ).

By calculating confidence intervals for the difference between indicators, we, firstly, directly see the possible values ​​of the effect, and not just its point estimate. Secondly, we can draw a conclusion about the acceptance or rejection of the null hypothesis and, thirdly, we can draw a conclusion about the power of the test.

When testing hypotheses using confidence intervals, you must adhere to the following rule:

If the 100(1-a) percent confidence interval of the difference in means does not contain zero, then the differences are statistically significant at significance level a; on the contrary, if this interval contains zero, then the differences are not statistically significant.

Indeed, if this interval contains zero, it means that the indicator being compared may be either greater or less in one of the groups compared to the other, i.e. the observed differences are due to chance.

The power of the test can be judged by the location of zero within the confidence interval. If zero is close to the lower or upper limit of the interval, then it is possible that with a larger number of groups being compared, the differences would reach statistical significance. If zero is close to the middle of the interval, then it means that both an increase and a decrease in the indicator in the experimental group are equally likely, and, probably, there really are no differences.

Examples:

To compare surgical mortality when using two different types of anesthesia: 61 people were operated on with the first type of anesthesia, 8 died, with the second type – 67 people, 10 died.

d 1 = 8/61 = 0.131; d2 = 10/67 = 0.149; d1-d2 = - 0.018.

The difference in lethality of the compared methods will be in the range (-0.018 - 0.122; -0.018 + 0.122) or (-0.14; 0.104) with a probability of 100(1-a) = 95%. The interval contains zero, i.e. the hypothesis of equal mortality with two different types of anesthesia cannot be rejected.

Thus, the mortality rate can and will decrease to 14% and increase to 10.4% with a probability of 95%, i.e. zero is approximately in the middle of the interval, so it can be argued that, most likely, these two methods really do not differ in lethality.

In the example discussed earlier, the average pressing time during the tapping test was compared in four groups of students who differed in exam scores. Let's calculate the confidence intervals for the average pressing time for students who passed the exam with 2 and 5 marks and the confidence interval for the difference between these averages.

Student's coefficients are found using Student's distribution tables (see appendix): for the first group: = t(0.05;48) = 2.011; for the second group: = t(0.05;61) = 2.000. Thus, confidence intervals for the first group: = (162.19-2.011*2.18; 162.19+2.011*2.18) = (157.8; 166.6), for the second group (156.55- 2,000*1.88 ; 156.55+2,000*1.88) = (152.8; 160.3). So, for those who passed the exam with a 2, the average pressing time ranges from 157.8 ms to 166.6 ms with a probability of 95%, for those who passed the exam with a 5 – from 152.8 ms to 160.3 ms with a probability of 95%.

You can also test the null hypothesis using confidence intervals for means, and not just for the difference in means. For example, as in our case, if the confidence intervals for the means overlap, then the null hypothesis cannot be rejected. To reject a hypothesis at a chosen significance level, the corresponding confidence intervals must not overlap.

Let's find the confidence interval for the difference in the average pressing time in the groups that passed the exam with grades 2 and 5. Difference of averages: 162.19 – 156.55 = 5.64. Student's coefficient: = t(0.05;49+62-2) = t(0.05;109) = 1.982. Group standard deviations will be equal to: ; . We calculate the average error of the difference between the means: . Confidence interval: =(5.64-1.982*2.87; 5.64+1.982*2.87) = (-0.044; 11.33).

So, the difference in the average pressing time in the groups that passed the exam with 2 and 5 will be in the range from -0.044 ms to 11.33 ms. This interval includes zero, i.e. The average pressing time for those who passed the exam well may either increase or decrease compared to those who passed the exam unsatisfactorily, i.e. the null hypothesis cannot be rejected. But zero is very close to the lower limit, and the pressing time is much more likely to decrease for those who passed well. Thus, we can conclude that there are still differences in the average time of pressing between those who passed 2 and 5, we just could not detect them given the change in the average time, the spread of the average time and the sample sizes.

The power of a test is the probability of rejecting an incorrect null hypothesis, i.e. find differences where they actually exist.

The power of the test is determined based on the level of significance, the magnitude of differences between groups, the spread of values ​​in groups and the size of samples.

For Student's t test and analysis of variance, sensitivity diagrams can be used.

The power of the criterion can be used to preliminarily determine the required number of groups.

The confidence interval shows within which limits the true value of the estimated parameter lies with a given probability.

Using confidence intervals, you can test statistical hypotheses and draw conclusions about the sensitivity of criteria.

LITERATURE.

Glanz S. – Chapter 6,7.

Rebrova O.Yu. – p.112-114, p.171-173, p.234-238.

Sidorenko E.V. – p.32-33.

Questions for self-testing of students.

1. What is the power of the criterion?

2. In what cases is it necessary to evaluate the power of criteria?

3. Methods for calculating power.

6. How to test a statistical hypothesis using a confidence interval?

7. What can be said about the power of the criterion when calculating the confidence interval?

Tasks.

Often the appraiser has to analyze the real estate market of the segment in which the property being assessed is located. If the market is developed, it can be difficult to analyze the entire set of presented objects, so a sample of objects is used for analysis. This sample does not always turn out to be homogeneous; sometimes it is necessary to clear it of extreme points - too high or too low market offers. For this purpose it is used confidence interval. The purpose of this study is to conduct a comparative analysis of two methods for calculating the confidence interval and select the optimal calculation option when working with different samples in the estimatica.pro system.

Confidence interval is an interval of attribute values ​​calculated on the basis of a sample, which with a known probability contains the estimated parameter of the general population.

The point of calculating a confidence interval is to construct such an interval based on sample data so that it can be stated with a given probability that the value of the estimated parameter is in this interval. In other words, the confidence interval contains the unknown value of the estimated value with a certain probability. The wider the interval, the higher the inaccuracy.

There are different methods for determining the confidence interval. In this article we will look at 2 methods:

• through the median and standard deviation;
• through the critical value of t-statistics (Student's coefficient).

Stages of comparative analysis of different methods for calculating CI:

1. form a data sample;

2. we process it using statistical methods: we calculate the average value, median, variance, etc.;

3. calculate the confidence interval in two ways;

4. analyze the cleaned samples and the resulting confidence intervals.

Stage 1. Data sampling

The sample was formed using the estimatica.pro system. The sample included 91 offers for the sale of 1-room apartments in the 3rd price zone with the “Khrushchev” type of layout.

Table 1. Initial sample

Fig.1. Initial sample

Stage 2. Processing the initial sample

Processing a sample using statistical methods requires calculating the following values:

1. Arithmetic mean

2. Median is a number characterizing the sample: exactly half of the sample elements are greater than the median, the other half are less than the median

(for a sample with an odd number of values)

3. Range - the difference between the maximum and minimum values ​​in the sample

4. Variance - used to more accurately estimate the variation of data

5. Sample standard deviation (hereinafter - SD) is the most common indicator of the dispersion of adjustment values ​​around the arithmetic mean.

6. Coefficient of variation - reflects the degree of scattering of adjustment values

7. oscillation coefficient - reflects the relative fluctuation of extreme price values ​​in the sample around the average

Table 2. Statistical indicators of the original sample

The coefficient of variation, which characterizes the homogeneity of the data, is 12.29%, but the coefficient of oscillation is too high. Thus, we can say that the original sample is not homogeneous, so let’s move on to calculating the confidence interval.

Stage 3. Confidence interval calculation

Method 1. Calculation using the median and standard deviation.

The confidence interval is determined as follows: minimum value - the standard deviation is subtracted from the median; maximum value - standard deviation is added to the median.

Thus, the confidence interval (47179 CU; 60689 CU)

Rice. 2. Values ​​falling within confidence interval 1.

Method 2. Constructing a confidence interval using the critical value of t-statistics (Student coefficient)

S.V. Gribovsky in his book “Mathematical Methods for Estimating Property Value” describes a method for calculating the confidence interval through the Student coefficient. When calculating using this method, the estimator must himself set the significance level ∝, which determines the probability with which the confidence interval will be constructed. Typically, significance levels of 0.1 are used; 0.05 and 0.01. They correspond to confidence probabilities of 0.9; 0.95 and 0.99. With this method, the true values ​​of the mathematical expectation and variance are assumed to be practically unknown (which is almost always true when solving practical estimation problems).

Confidence interval formula:

n - sample size;

The critical value of t-statistics (Student distribution) with a significance level ∝, the number of degrees of freedom n-1, which is determined from special statistical tables or using MS Excel (→"Statistical"→ STUDIST);

∝ - significance level, take ∝=0.01.

Rice. 2. Values ​​falling within confidence interval 2.

Stage 4. Analysis of different methods for calculating the confidence interval

Two methods of calculating the confidence interval - through the median and Student's coefficient - led to different values ​​of the intervals. Accordingly, we got two different cleaned samples.

Table 3. Statistics for three samples.

Based on the calculations performed, we can say that the confidence interval values ​​obtained by different methods intersect, so you can use any of the calculation methods at the discretion of the appraiser.

However, we believe that when working in the estimatica.pro system, it is advisable to choose a method for calculating the confidence interval depending on the degree of market development:

• if the market is undeveloped, use the calculation method using the median and standard deviation, since the number of retired objects in this case is small;
• if the market is developed, apply the calculation through the critical value of t-statistics (Student's coefficient), since it is possible to form a large initial sample.

In preparing the article the following were used:

1. Gribovsky S.V., Sivets S.A., Levykina I.A. Mathematical methods for assessing property value. Moscow, 2014

2. System data estimatica.pro

The confidence interval comes to us from the field of statistics. This is a certain range that serves to estimate an unknown parameter with a high degree of reliability. The easiest way to explain this is with an example.

Suppose you need to study some random variable, for example, the server's response speed to a client request. Each time a user types the address of a specific site, the server responds at different speeds. Thus, the response time under study is random. So, the confidence interval allows us to determine the boundaries of this parameter, and then we can say that with a 95% probability the server will be in the range we calculated.

Or you need to find out how many people know about the company’s trademark. When the confidence interval is calculated, it will be possible to say, for example, that with a 95% probability the share of consumers aware of this is in the range from 27% to 34%.

Closely related to this term is the value of confidence probability. It represents the probability that the desired parameter is included in the confidence interval. How large our desired range will be depends on this value. The larger the value it takes, the narrower the confidence interval becomes, and vice versa. Typically it is set to 90%, 95% or 99%. The value 95% is the most popular.

This indicator is also influenced by the dispersion of observations and its definition is based on the assumption that the characteristic under study obeys. This statement is also known as Gauss’s Law. According to him, a distribution of all probabilities of a continuous random variable that can be described by a probability density is called normal. If the assumption of a normal distribution is incorrect, the estimate may be incorrect.

First, let's figure out how to calculate the confidence interval for There are two possible cases here. Dispersion (the degree of spread of a random variable) may or may not be known. If it is known, then our confidence interval is calculated using the following formula:

xsr - t*σ / (sqrt(n))<= α <= хср + t*σ / (sqrt(n)), где

α - sign,

t - parameter from the Laplace distribution table,

σ is the square root of the variance.

If the variance is unknown, then it can be calculated if we know all the values ​​of the desired feature. The following formula is used for this:

σ2 = х2ср - (хср)2, where

х2ср - average value of the squares of the studied characteristic,

(хср)2 is the square of this characteristic.

The formula by which the confidence interval is calculated in this case changes slightly:

xsr - t*s / (sqrt(n))<= α <= хср + t*s / (sqrt(n)), где

xsr - sample average,

α - sign,

t is a parameter that is found using the Student distribution table t = t(ɣ;n-1),

sqrt(n) - square root of the total sample size,

s is the square root of the variance.

Consider this example. Suppose that based on the results of 7 measurements, the studied characteristic was determined to be equal to 30 and the sample variance to be equal to 36. It is necessary to find, with a probability of 99%, a confidence interval that contains the true value of the measured parameter.

First, let's determine what t is equal to: t = t (0.99; 7-1) = 3.71. Using the above formula, we get:

xsr - t*s / (sqrt(n))<= α <= хср + t*s / (sqrt(n))

30 - 3.71*36 / (sqrt(7))<= α <= 30 + 3.71*36 / (sqrt(7))

21.587 <= α <= 38.413

The confidence interval for the variance is calculated both in the case of a known mean and when there is no data on the mathematical expectation, and only the value of the point unbiased estimate of the variance is known. We will not give formulas for calculating it here, since they are quite complex and, if desired, can always be found on the Internet.

Let us only note that it is convenient to determine the confidence interval using Excel or a network service, which is called that way.