The method allows you to test the hypothesis that the average values of two general populations from which the compared ones are extracted dependent selections differ from each other. The assumption of dependence most often means that the characteristic is measured on the same sample twice, for example, before the intervention and after it. In the general case, each representative of one sample is assigned a representative from another sample (they are combined in pairs) so that the two data series are positively correlated with each other. Weaker types of sample dependence: sample 1 - husbands, sample 2 - their wives; sample 1 - one-year-old children, sample 2 is made up of twins of children in sample 1, etc.
Testable statistical hypothesis, as in the previous case, H 0: M 1 = M 2(mean values in samples 1 and 2 are equal). If it is rejected, the alternative hypothesis is accepted that M 1 more less) M 2.
Initial assumptions for statistical testing:
Each representative of one sample (from one general population) is associated with a representative of another sample (from another general population);
The data from the two samples are positively correlated (form pairs);
The distribution of the studied characteristic in both samples corresponds to the normal law.
Source data structure: there are two values of the studied feature for each object (for each pair).
Restrictions: the distribution of the characteristic in both samples should not differ significantly from normal; the data of two measurements corresponding to one and the other sample are positively correlated.
Alternatives: Wilcoxon T test, if the distribution for at least one sample differs significantly from normal; t-Student test for independent samples - if the data for two samples do not correlate positively.
Formula for the empirical value of the Student's t test reflects the fact that the unit of analysis for differences is difference (shift) attribute values for each pair of observations. Accordingly, for each of the N pairs of attribute values, the difference is first calculated d i = x 1 i - x 2 i.
where M d is the average difference of values; σ d - standard deviation differences.
Calculation example:
Let’s assume that during testing the effectiveness of the training, each of the 8 members of the group was asked the question “How often does your opinion coincide with the opinion of the group?” - twice, before and after the training. A 10-point scale was used for responses: 1 - never, 5 - half the time, 10 - always. The hypothesis was tested that as a result of the training, the self-esteem of conformity (the desire to be like others in the group) of the participants would increase (α = 0.05). Let's create a table for intermediate calculations (Table 3).
Table 3
The arithmetic mean for the difference M d = (-6)/8 = -0.75. Subtract this value from each d (the penultimate column of the table).
The formula for standard deviation differs only in that d appears in it instead of X. Let's substitute everything required values, we get:
σ d = = 0.886.
Step 1. Calculate the empirical value of the criterion using formula (3): average difference Md= -0.75; standard deviation σ d = 0,886; t e = 2,39; df = 7.
Step 2. Using the table of critical values of the t-Student criterion, we determine the p-level of significance. For df = 7 the empirical value is between the critical values for R= 0.05 and R - 0.01. Hence, R< 0,05.
| df | R | ||
| 0,05 | 0,01 | 0,001 | |
| 2,365 | 3,499 | 5,408 |
Step 3. We make a statistical decision and formulate a conclusion. The statistical hypothesis of equality of average values is rejected. Conclusion: the indicator of self-assessment of participants’ conformity after the training increased statistically significantly (at significance level p< 0,05).
Parametric methods include comparison of variances of two samples according to the criterion F-Fisher. Sometimes this method leads to valuable meaningful conclusions, and in the case of comparing means for independent samples, comparison of variances is mandatory procedure.
To calculate F em you need to find the ratio of the variances of the two samples, and so that the larger variance is in the numerator, and the smaller one is in the denominator.
Comparison of Variances. The method allows you to test the hypothesis that the variances of the two general populations from which the compared samples are drawn differ from each other. Tested statistical hypothesis H 0: σ 1 2 = σ 2 2 (the variance in sample 1 is equal to the variance in sample 2). If it is rejected, the alternative hypothesis is accepted that one variance is greater than the other.
Initial assumptions: two samples are drawn randomly from different populations with a normal distribution of the characteristic being studied.
Source data structure: the characteristic being studied is measured in objects (subjects), each of which belongs to one of the two samples being compared.
Restrictions: the distributions of the trait in both samples do not differ significantly from normal.
Alternative method: Levene's test, the use of which does not require checking the assumption of normality (used in the SPSS program).
Formula for the empirical value of the Fisher's F test:
(4)
where σ 1 2 — large dispersion, and σ 2 2 - smaller dispersion. Since it is not known in advance which dispersion is greater, then to determine the p-level it is used Table of critical values for non-directional alternatives. If F e > F Kp for the corresponding number of degrees of freedom, then R< 0,05 и статистическую гипотезу о равенстве дисперсий можно отклонить (для α = 0,05).
Calculation example:
The children were given regular arithmetic problems, after which one randomly selected half of the students were told that they had failed the test, and the rest were told the opposite. Each child was then asked how many seconds it would take them to solve a similar problem. The experimenter calculated the difference between the time the child called and the result of the completed task (in seconds). It was expected that the message of failure would cause some inadequacy in the child's self-esteem. The hypothesis tested (at the α = 0.005 level) was that the variance of the aggregate self-esteem does not depend on reports of success or failure (H 0: σ 1 2 = σ 2 2).
The following data was obtained:
Step 1. Calculate the empirical value of the criterion and the number of degrees of freedom using formulas (4):
Step 2. According to the table of critical values of the Fisher f-criterion for non-directional alternatives we find the critical value for df number= 11; df know= 11. However, there is a critical value only for df number= 10 and df know = 12. A larger number of degrees of freedom cannot be taken, so we take the critical value for df number= 10: For R= 0,05 F Kp = 3.526; For R= 0,01 F Kp = 5,418.
Step 3. Acceptance statistical solution and meaningful conclusion. Since the empirical value exceeds the critical value for R= 0.01 (and even more so for p = 0.05), then in this case p< 0,01 и принимается альтернативная гипо-теза: дисперсия в группе 1 превышает дисперсию в группе 2 (R< 0.01). Consequently, after a message about failure, the inadequacy of self-esteem is higher than after a message about success.
where f is the degree of freedom, which is defined as
Example . Two groups of students were trained using two different methods. At the end of the training, they were given a test throughout the course. It is necessary to assess how significant the differences in acquired knowledge are. The test results are presented in Table 4.
Table 4
Let's calculate the sample mean, variance and standard deviation:
Let's determine the value of t p using the formula t p = 0.45
Using Table 1 (see appendix) we find the critical value t k for the significance level p = 0.01
Conclusion: since the calculated value of the criterion is less than the critical value of 0.45<2,88 гипотеза Но подтверждается и существенных различий в методиках обучения нет на уровне значимости 0,01.
Algorithm for calculating Student's t-test for dependent samples of measurements
1. Determine the calculated value of the t-test using the formula
, Where 
3. Determine the critical value of the t-test according to Table 1 of the Appendix.
4. Compare the calculated and critical value of the t-test. If the calculated value is greater than or equal to the critical value, then the hypothesis of equality of average values in two samples of changes is rejected (Ho). In all other cases it is accepted at a given significance level.
U- criterionManna- Whitney
Purpose of the criterion
The criterion is intended to assess differences between two non-parametric samples in terms of the level of any quantitatively measured characteristic. It allows you to identify differences between small samples when n< 30.
Description of criterion
This method determines whether the area of overlapping values between two series is small enough. The smaller this area, the more likely it is that the differences are significant. The empirical value of the U criterion reflects how large the area of agreement between the rows is. Therefore, the smaller U is, the more likely it is that the differences are significant.
Hypotheses
BUT: The level of the trait in group 2 is not lower than the level of the trait in group 1.
HI: The level of the trait in group 2 is lower than the level of the trait in group 1.
Algorithm for calculating the Mann-Whitney criterion (u)
Transfer all test subjects’ data to individual cards.
Mark the cards of the subjects in sample 1 with one color, say red, and all the cards from sample 2 with another color, for example, blue.
Arrange all the cards in a single row according to the degree of increase in the attribute, regardless of which sample they belong to, as if we were working with one large sample.
where n 1 is the number of subjects in sample 1;
n 2 – number of subjects in sample 2,
T x – the larger of the two rant amounts;
n x – the number of subjects in the group with a larger sum of ranks.
9. Determine the critical values of U according to table 2 (see appendix).
If U em.> U cr0.05, then the hypothesis But is accepted. If U emp.≤ U cr, then it is rejected. How less than value U, the higher the reliability of the differences.
Example. Compare the effectiveness of two teaching methods in two groups. The test results are presented in Table 5.
Table 5
Let's transfer all the data to another table, highlighting the data of the second group with an underline, and make a ranking of the overall sample (see the ranking algorithm in the guidelines for task 3).
|
Values | ||||||||||||||||||
Let's find the sum of the ranks of two samples and choose the larger one: T x = 113
Let's calculate the empirical value of the criterion using formula 2: U p = 30.
Using table 2 in the appendix, we determine the critical value of the criterion at a significance level of p = 0.05: U k = 19.
Conclusion: since the calculated value of the criterionUis greater than critical at the significance level p = 0.05 and 30 > 19, then the hypothesis about the equality of means is accepted and the differences in teaching methods are insignificant.
The method allows you to test the hypothesis that the average values of two general populations from which the compared ones are extracted dependent samples differ from each other. The assumption of dependence most often means that the trait is measured on the same sample twice, for example, before the intervention and after it. In the general case, each representative of one sample is assigned a representative from another sample (they are combined in pairs) so that the two data series are positively correlated with each other. Weaker types of sample dependence: sample 1 - husbands, sample 2 - their wives; sample 1 - one-year-old children, sample 2 is made up of twins of children in sample 1, etc.
Testable statistical hypothesis, as in the previous case, H 0: M 1 = M 2(the average values in samples 1 and 2 are equal). If it is rejected, the alternative hypothesis is accepted that M 1 more less) M 2.
Initial assumptions for statistical testing:
□ each representative of one sample (from one general population) is associated with a representative of another sample (from another general population);
□ data from two samples are positively correlated (form pairs);
□ the distribution of the studied characteristic in both samples corresponds to the normal law.
Source data structure: there are two values of the studied feature for each object (for each pair).
Restrictions: the distribution of the characteristic in both samples should not differ significantly from normal; the data of the two measurements corresponding to both samples are positively correlated.
Alternatives: Wilcoxon T-test, if the distribution for at least one sample differs significantly from normal; t-Student test for independent samples - if the data for the two samples are not positively correlated.
Formula for the empirical value of the Student's t test reflects the fact that the unit of analysis for differences is difference (shift) characteristic values for each pair of observations. Accordingly, for each of the N pairs of attribute values, the difference is first calculated d i = x 1 i - x 2 i.
(3) where M d – average difference of values; σ d – standard deviation of differences.
Calculation example:
Suppose, during testing the effectiveness of the training, each of the 8 group members was asked the question “How often do your opinions coincide with the opinions of the group?” - twice, before and after the training. A 10-point scale was used for responses: 1 - never, 5 - half the time, 10 - always. The hypothesis was tested that as a result of the training, the self-esteem of conformity (the desire to be like others in the group) of the participants would increase (α = 0.05). Let's create a table for intermediate calculations (Table 3).
Table 3
The arithmetic mean for the difference M d = (-6)/8= -0.75. Subtract this value from each d (the penultimate column of the table).
The formula for the standard deviation differs only in that d appears in it instead of X. We substitute all the necessary values, and we get
σ d = = 0.886.
Step 1. Calculate the empirical value of the criterion using formula (3): average difference Md= -0.75; standard deviation σ d = 0,886; t e = 2,39; df = 7.
Step 2. Using the table of critical values of the t-Student criterion, we determine the p-level of significance. For df = 7, the empirical value is between the critical values for p = 0.05 and p - 0.01. Therefore, p< 0,05.
| df | R | ||
| 0,05 | 0,01 | 0,001 | |
| 2,365 | 3,499 | 5,408 |
Step 3. We make a statistical decision and formulate a conclusion. The statistical hypothesis of equality of means is rejected. Conclusion: the indicator of self-assessment of participants’ conformity after the training increased statistically significantly (at significance level p< 0,05).
Parametric methods include comparison of variances of two samples according to the criterion F-Fisher. Sometimes this method leads to valuable meaningful conclusions, and in the case of comparing means for independent samples, comparing variances is mandatory procedure.
To calculate F em you need to find the ratio of the variances of the two samples, and so that the larger variance is in the numerator, and the smaller one is in the denominator.
Comparison of Variances. The method allows you to test the hypothesis that the variances of the two populations from which the compared samples are drawn differ from each other. Tested statistical hypothesis H 0: σ 1 2 = σ 2 2 (the variance in sample 1 is equal to the variance in sample 2). If it is rejected, the alternative hypothesis is accepted that one variance is greater than the other.
Initial assumptions: two samples are drawn randomly from different populations with a normal distribution of the trait being studied.
Source data structure: the characteristic being studied is measured in objects (subjects), each of which belongs to one of the two samples being compared.
Restrictions: the distributions of the trait in both samples do not differ significantly from normal.
Alternative method: Levene's test, the use of which does not require checking the assumption of normality (used in the SPSS program).
Formula for the empirical value of the Fisher's F test:
(4)
where σ 1 2 - large dispersion, and σ 2 2 - smaller dispersion. Since it is not known in advance which dispersion is greater, then to determine the p-level it is used Table of critical values for non-directional alternatives. If F e > F Kp for the corresponding number of degrees of freedom, then R < 0,05 и статистическую гипотезу о равенстве дисперсий можно отклонить (для α = 0,05).
Calculation example:
The children were given regular arithmetic problems, after which one randomly selected half of the students were told that they had failed the test, and the rest were told the opposite. Each child was then asked how many seconds it would take them to solve a similar problem. The experimenter calculated the difference between the time the child called and the result of the completed task (in seconds). It was expected that the message of failure would cause some inadequacy in the child's self-esteem. The hypothesis being tested (at the α = 0.005 level) was that the variance of the aggregate self-esteem does not depend on reports of success or failure (H 0: σ 1 2 = σ 2 2).
The following data was obtained:
Step 1. Calculate the empirical value of the criterion and the number of degrees of freedom using formulas (4):
Step 2. According to the table of critical values of the Fisher f-criterion for undirected alternatives we find the critical value for df number = 11; df know= 11. However, there is a critical value only for df number= 10 and df know = 12. It is impossible to take a larger number of degrees of freedom, so we take the critical value for df number= 10: For R = 0,05 F Kp = 3.526; For R = 0,01 F Kp = 5,418.
Step 3. Making a statistical decision and meaningful conclusion. Since the empirical value exceeds the critical value for R= 0.01 (and even more so for p = 0.05), then in this case p< 0,01 и принимается альтернативная гипотеза: дисперсия в группе 1 превышает дисперсию в группе 2 (R< 0.01). Consequently, after a message about failure, the inadequacy of self-esteem is higher than after a message about success.
/ practical statistics / reference materials / student t-test values
Meaningt -Student's t-test at significance levels of 0.10, 0.05 and 0.01
ν – degrees of freedom of variation
Standard Student's t-test values
|
Number of degrees of freedom |
Significance levels |
Number of degrees of freedom |
Significance levels |
||||||
Table XI
Standard Fisher test values used to assess the significance of differences between two samples
|
Degrees of freedom |
Significance level |
Degrees of freedom |
Significance level |
||||
Student's t-test
Student's t-test- a general name for a class of methods for statistical testing of hypotheses (statistical tests) based on the Student distribution. The most common uses of the t-test involve testing the equality of means in two samples.
t-statistics are usually built according to the following general principle: the numerator is a random variable with zero mathematical expectation (if the null hypothesis is satisfied), and the denominator is the sample standard deviation of this random variable, obtained as Square root from the unconfounded variance estimate.
Story
This criterion was developed by William Gossett to evaluate the quality of beer at the Guinness company. In connection with obligations to the company regarding non-disclosure of trade secrets (Guinness management considered the use of statistical apparatus in its work as such), Gosset’s article was published in 1908 in the journal Biometrics under the pseudonym “Student”.
Data requirements
To apply this criterion, it is necessary that the original data have a normal distribution. In the case of applying a two-sample test for independent samples, it is also necessary to comply with the condition of equality of variances. There are, however, alternatives to the Student's t test for situations with unequal variances.
The requirement of normal distribution of data is necessary for an accurate t (\displaystyle t) -test. However, even with other data distributions, it is possible to use t (\displaystyle t) -statistics. In many cases, this statistic asymptotically has a standard normal distribution - N (0, 1) (\displaystyle N(0,1)) , so quantiles of this distribution can be used. However, even in this case, often quantiles are used not of the standard normal distribution, but of the corresponding Student distribution, as in the exact t (\displaystyle t) test. They are asymptotically equivalent, but in small samples the confidence intervals of the Student distribution are wider and more reliable.
One-sample t-test
Used to test the null hypothesis H 0: E (X) = m (\displaystyle H_(0):E(X)=m) about the equality of the mathematical expectation E (X) (\displaystyle E(X)) to some known value m ( \displaystyle m) .
Obviously, if the null hypothesis is satisfied, E (X ¯) = m (\displaystyle E((\overline (X)))=m) . Taking into account the assumed independence of observations, V (X ¯) = σ 2 / n (\displaystyle V((\overline (X)))=\sigma ^(2)/n) . Using an unbiased variance estimate s X 2 = ∑ t = 1 n (X t − X ¯) 2 / (n − 1) (\displaystyle s_(X)^(2)=\sum _(t=1)^(n )(X_(t)-(\overline (X)))^(2)/(n-1)) we obtain the following t-statistics:
t = X ¯ − m s X / n (\displaystyle t=(\frac ((\overline (X))-m)(s_(X)/(\sqrt (n)))))
Under the null hypothesis, the distribution of this statistic is t (n − 1) (\displaystyle t(n-1)) . Therefore, if the statistics value is exceeded absolute value the critical value of a given distribution (at a given significance level), the null hypothesis is rejected.
Two-sample t-test for independent samples
Let there be two independent samples of volumes n 1, n 2 (\displaystyle n_(1)~,~n_(2)) of normally distributed random variables X 1, X 2 (\displaystyle X_(1),~X_(2)). It is necessary to test the null hypothesis of equality of mathematical expectations of these random variables H 0: M 1 = M 2 (\displaystyle H_(0):~M_(1)=M_(2)) using sample data.
Consider the difference between sample means Δ = X ¯ 1 − X ¯ 2 (\displaystyle \Delta =(\overline (X))_(1)-(\overline (X))_(2)) . Obviously, if the null hypothesis is true E (Δ) = M 1 − M 2 = 0 (\displaystyle E(\Delta)=M_(1)-M_(2)=0) . The variance of this difference is equal, based on the independence of the samples: V (Δ) = σ 1 2 n 1 + σ 2 2 n 2 (\displaystyle V(\Delta)=(\frac (\sigma _(1)^(2))( n_(1)))+(\frac (\sigma _(2)^(2))(n_(2)))) . Then using the unbiased variance estimate s 2 = ∑ t = 1 n (X t − X ¯) 2 n − 1 (\displaystyle s^(2)=(\frac (\sum _(t=1)^(n)( X_(t)-(\overline (X)))^(2))(n-1))) we obtain an unbiased estimate of the variance of the difference between sample means: s Δ 2 = s 1 2 n 1 + s 2 2 n 2 (\ displaystyle s_(\Delta )^(2)=(\frac (s_(1)^(2))(n_(1)))+(\frac (s_(2)^(2))(n_(2) ))) . Therefore, the t-statistic for testing the null hypothesis is
T = X ¯ 1 − X ¯ 2 s 1 2 n 1 + s 2 2 n 2 (\displaystyle t=(\frac ((\overline (X))_(1)-(\overline (X))_( 2))(\sqrt ((\frac (s_(1)^(2))(n_(1)))+(\frac (s_(2)^(2))(n_(2))))) ))
If the null hypothesis is true, this statistic has a distribution t (d f) (\displaystyle t(df)), where d f = (s 1 2 / n 1 + s 2 2 / n 2) 2 (s 1 2 / n 1) 2 / (n 1 − 1) + (s 2 2 / n 2) 2 / (n 2 − 1) (\displaystyle df=(\frac ((s_(1)^(2)/n_(1)+s_(2 )^(2)/n_(2))^(2))((s_(1)^(2)/n_(1))^(2)/(n_(1)-1)+(s_(2 )^(2)/n_(2))^(2)/(n_(2)-1))))
Case of equal variance
If the variances of the samples are assumed to be equal, then
V (Δ) = σ 2 (1 n 1 + 1 n 2) (\displaystyle V(\Delta)=\sigma ^(2)\left((\frac (1)(n_(1)))+(\ frac (1)(n_(2)))\right))
Then the t-statistic is:
T = X ¯ 1 − X ¯ 2 s X 1 n 1 + 1 n 2 , s X = (n 1 − 1) s 1 2 + (n 2 − 1) s 2 2 n 1 + n 2 − 2 (\ displaystyle t=(\frac ((\overline (X))_(1)-(\overline (X))_(2))(s_(X)(\sqrt ((\frac (1)(n_(1 )))+(\frac (1)(n_(2))))))~,~~s_(X)=(\sqrt (\frac ((n_(1)-1)s_(1)^ (2)+(n_(2)-1)s_(2)^(2))(n_(1)+n_(2)-2))))
This statistic has distribution t (n 1 + n 2 − 2) (\displaystyle t(n_(1)+n_(2)-2))
Two-sample t-test for dependent samples
To calculate the empirical value of the t (\displaystyle t) -criterion in the situation of testing a hypothesis about differences between two dependent samples (for example, two samples of the same test with a time interval), the following formula is used:
T = M d s d / n (\displaystyle t=(\frac (M_(d))(s_(d)/(\sqrt (n)))))
where M d (\displaystyle M_(d)) is the average difference of values, s d (\displaystyle s_(d)) is the standard deviation of the differences, and n is the number of observations
This statistic has a distribution t (n − 1) (\displaystyle t(n-1)) .
Testing a Linear Constraint on Linear Regression Parameters
The t-test can also test an arbitrary (single) linear constraint on the parameters linear regression, estimated by the usual method least squares. Let it be necessary to test the hypothesis H 0: c T b = a (\displaystyle H_(0):c^(T)b=a) . Obviously, if the null hypothesis is satisfied, E (c T b ^ − a) = c T E (b ^) − a = 0 (\displaystyle E(c^(T)(\hat (b))-a)=c^( T)E((\hat (b)))-a=0) . Here we use the property of unbiased least squares estimates of the model parameters E (b ^) = b (\displaystyle E((\hat (b)))=b) . In addition, V (c T b ^ − a) = c T V (b ^) c = σ 2 c T (X T X) − 1 c (\displaystyle V(c^(T)(\hat (b))-a )=c^(T)V((\hat (b)))c=\sigma ^(2)c^(T)(X^(T)X)^(-1)c) . Using instead of the unknown variance its unbiased estimate s 2 = E S S / (n − k) (\displaystyle s^(2)=ESS/(n-k)) we obtain the following t-statistics:
T = c T b ^ − a s c T (X T X) − 1 c (\displaystyle t=(\frac (c^(T)(\hat (b))-a)(s(\sqrt (c^(T) (X^(T)X)^(-1)c)))))
This statistic, when the null hypothesis is satisfied, has a distribution t (n − k) (\displaystyle t(n-k)) , so if the value of the statistic is higher than the critical value, then the null hypothesis of a linear constraint is rejected.
Testing hypotheses about the linear regression coefficient
A special case of a linear constraint is testing the hypothesis that the regression coefficient b j (\displaystyle b_(j)) is equal to a certain value a (\displaystyle a) . In this case, the corresponding t-statistic is:
T = b ^ j − a s b ^ j (\displaystyle t=(\frac ((\hat (b))_(j)-a)(s_((\hat (b))_(j)))))
where s b ^ j (\displaystyle s_((\hat (b))_(j))) is the standard error of the coefficient estimate - the square root of the corresponding diagonal element of the covariance matrix of the coefficient estimates.
If the null hypothesis is true, the distribution of this statistic is t (n − k) (\displaystyle t(n-k)) . If the absolute value of the statistic is higher than the critical value, then the difference between the coefficient and a (\displaystyle a) is statistically significant (non-random), otherwise it is insignificant (random, that is, the true coefficient is probably equal to or very close to the estimated value of a (\ display style a))
Comment
A one-sample test for mathematical expectations can be reduced to testing a linear constraint on the linear regression parameters. In a one-sample test, this is a "regression" on a constant. Therefore, s 2 (\displaystyle s^(2)) of regression is a sample estimate of the variance of the random variable being studied, the matrix X T X (\displaystyle X^(T)X) is equal to n (\displaystyle n) , and the estimate of the “coefficient” of the model is equal to sample mean. From here we obtain the expression for the t-statistic given above for the general case.
Similarly, it can be shown that a two-sample test with equal sample variances also reduces to testing linear constraints. In a two-sample test, this is a "regression" on a constant and a dummy variable identifying the subsample depending on the value (0 or 1): y = a + b D (\displaystyle y=a+bD) . The hypothesis about the equality of the mathematical expectations of the samples can be formulated as a hypothesis about the equality of the coefficient b of this model to zero. It can be shown that the appropriate t-statistic for testing this hypothesis is equal to the t-statistic given for the two-sample test.
It can also be reduced to checking the linear constraint in the case of different dispersions. In this case, the model error variance takes two values. From this you can also obtain a t-statistic similar to that given for the two-sample test.
Nonparametric analogues
An analogue of the two-sample test for independent samples is the Mann-Whitney U test. For the situation with dependent samples, the analogues are the sign test and the Wilcoxon T-test
Literature
Student. The probable error of a mean. // Biometrika. 1908. No. 6 (1). P. 1-25.
Links
On the criteria for testing hypotheses about the homogeneity of means on the website of the Novosibirsk State Technical University
Throughout the example, we will use fictitious information so that the reader can make the necessary transformations on his own.
So, let’s say, in the course of research, we studied the effect of drug A on the content of substance B (in mmol/g) in tissue C and the concentration of substance D in the blood (in mmol/l) in patients divided according to some criterion E into 3 groups of equal volume (n = 10). The results of such a fictitious study are shown in the table:
| Content of substance B, mmol/g |
Substance D, mmol/l |
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|---|---|---|---|---|---|---|---|
|
increase in concentration |
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We would like to warn you that we consider samples of size 10 for ease of data presentation and calculations; in practice, such a sample size is usually not enough to form a statistical conclusion.
As an example, consider the data in the 1st column of the table.
Descriptive Statistics
Sample mean
The arithmetic mean, often simply called the "mean", is obtained by adding all the values and dividing that sum by the number of values in the set. This can be shown using an algebraic formula. A set of n observations of a variable x can be represented as x 1 , x 2 , x 3 , ..., x n
The formula for determining the arithmetic mean of observations (pronounced “X with a line”):
= (X 1 + X 2 + ... + X n) / n
= (12 + 13 + 14 + 15 + 14 + 13 + 13 + 10 + 11 + 16) / 10 = 13,1;
Sample variance
One way to measure the dispersion of data is to determine the degree to which each observation deviates from the arithmetic mean. Obviously, the greater the deviation, the greater the variability, variability of observations. However, we cannot use the average of these deviations as a measure of dispersion, because positive deviations compensate for negative deviations (their sum is zero). To solve this problem, we square each deviation and find the average of the squared deviations; this quantity is called variation, or dispersion. Let's take n observations x 1, x 2, x 3, ..., x n, average which is equal to. Calculating the variance this, usually referred to ass2,these observations:
The sample variance of this indicator is s 2 = 3.2.
Standard deviation
Standard (mean square) deviation is the positive square root of the variance. Using n observations as an example, it looks like this:
We can think of standard deviation as a kind of average deviation of observations from the mean. It is calculated in the same units (dimensions) as the original data.
s = sqrt (s 2) = sqrt (3,2) = 1.79.
The coefficient of variation
If you divide the standard deviation by the arithmetic mean and express the result as a percentage, you get the coefficient of variation.
CV = (1.79 / 13.1) * 100% = 13.7
Sample mean error
1.79/sqrt(10) = 0.57;
Student's t coefficient (one-sample t-test)
Used to test the hypothesis that the average value differs from a certain value known value m
The number of degrees of freedom is calculated as f=n-1.
In this case confidence interval for the average it lies between the boundaries of 11.87 and 14.39.
For level confidence probability 95% m=11.87 or m=14.39, that is= |13.1-11.82| = |13.1-14.38| = 1.28
Accordingly, in this case, for the number of degrees of freedom f = 10 - 1 = 9 and the 95% confidence level t = 2.26.
Dialog Basic Statistics and Tables
In the module Basic statistics and tables let's choose Descriptive Statistics.
A dialog box will open Descriptive Statistics.
In field Variables let's choose Group 1.
Pressing OK, we obtain tables of results with descriptive statistics of the selected variables.
A dialog box will open One-sample t-test.
Suppose we know that the average content of substance B in tissue C is 11.
The table of results with descriptive statistics and Student's t-test is as follows:
We had to reject the hypothesis that the average content of substance B in tissue C is 11.
Since the calculated value of the criterion is greater than the tabulated value (2.26), the null hypothesis is rejected at the selected significance level, and the differences between the sample and the known value are considered statistically significant. Thus, the conclusion about the existence of differences made using the Student's test is confirmed using this method.
One of the most famous statistical tools is the Student's t test. It is used to measure the statistical significance of various pairwise quantities. Microsoft Excel has a special function for calculating this indicator. Let's learn how to calculate the Student's t-test in Excel.
But first, let’s find out what the Student’s t-test is in general. This indicator is used to check the equality of the average values of two samples. That is, it determines the significance of the differences between two groups of data. At the same time, a whole set of methods is used to determine this criterion. The indicator can be calculated taking into account one-sided or two-sided distribution.
Calculation of an indicator in Excel
Now let's move directly to the question of how to calculate this indicator in Excel. It can be done through the function STUDENT TEST. In 2007 and earlier versions of Excel, it was called TTEST. However, it was left in later versions for compatibility purposes, but in them it is still recommended to use a more modern one - STUDENT TEST. This function can be used in three ways, which will be discussed in detail below.
Method 1: Function Wizard
The easiest way to calculate this indicator is through the Function Wizard.
The calculation is performed, and the result is displayed on the screen in a pre-selected cell.
Method 2: Working with the Formulas tab
Function STUDENT TEST can also be called by going to the tab "Formulas" using a special button on the ribbon.
Method 3: Manual Entry
Formula STUDENT TEST can also be entered manually into any cell on the worksheet or into the function row. Its syntactic form looks like this:
STUDENT TEST(Array1,Array2,Tails,Type)
What each of the arguments means was considered when analyzing the first method. These values should be substituted into this function.
After the data has been entered, press the button Enter to display the result on the screen.
As you can see, calculating the Student's test in Excel is very simple and quick. The main thing is that the user who performs the calculations must understand what he is and what input data is responsible for what. The program performs the direct calculation itself.