Anyone know how to find the d-value for a paired samples t-test? I want the formula, thanks!!! Urgently
SPSS paired t-test d-value calculation statement:
How do you write the formula for the paired t-test?
The formula for the paired t-test is as follows:
The single overall t-test is a test of whether the difference between a sample mean and a known overall mean is significant. When the overall distribution is normal, such as the overall standard deviation is unknown and the sample size is less than 30, then the departure statistic of the sample mean from the overall mean is t-distributed.
A paired design is one in which test units are first paired two by two according to the requirements for pairing, and then the two paired test units are randomly assigned to the two treatment groups.
The requirement of pairing is that the initial conditions of the two experimental units paired together are the same as far as possible, and the initial conditions of the experimental units are allowed to differ between pairs, and each pair is a replication of the experimental treatment. There are two ways of pairing: self-pairing and homologous pairing.
Self-pairing: refers to the same test unit in two different time respectively before and after the two treatments, with its before and after the two observations of its own control comparison or the same test unit of different parts of the observations or different methods of the observations of its own control comparison. Such as the observation of a disease before and after the treatment of clinical examination results of changes; observation of two different methods of agricultural products in the determination of toxic substances or drug residues changes in the results.
Homologous pairing: refers to the source of the same, the same nature of the two individuals into a pair, such as livestock species, litter, gender, age, weight of the same two test animals into a pair, and then randomly implement the pair of two individuals with different treatments paired with the design of the experimental data of the general form.
Applicable conditions: an overall mean is known; a sample mean and the sample standard deviation can be obtained; samples from the normal or nearly normal overall
What is the t-test formula?
Test formula: t=(x-μ0)/S/√n.
The t-test, also known as the studentt test, is mainly used for normal distributions with small sample size (e.g., n<30), and an unknown overall standard deviation σ. The t-test uses the theory of the t-distribution to infer the probability of a difference occurring, and thus compares the difference between two averages to determine whether the difference is significant. It is ranked alongside the f-test and the chi-square test. t-test was invented by Gost to observe the quality of winemaking and published in Biometrika in 1908.
The most common uses of the t-test
1, One-sample mean test (One-sample-test) is used to test whether the mean of a single sample with unknown overall variance, normal data, or approximately normal is equal to the known overall mean.
2, two independent samples mean test (Independenttwo-samplet-test) is used to test the two pairs of independent normal data or nearly normal samples of the mean is equal, here can be discussed according to the overall variance of the overall classification is equal.
3, paired samples mean test (Dependentt-testforpairedsamples) is used to test a pair of paired samples of the mean of the difference is equal to a certain value.
4, regression coefficient of significance test (t-testforregressioncoefficientsignificance) is used to test the regression model of the explanatory variables of the explained variables whether there is a significant effect.
Paired test problem. Know some statistics to tell me very simple
In short, t-test and u-test is the statistic t,u hypothesis test, both are common hypothesis testing methods. When the sample content n is large, the sample mean conforms to the normal distribution, so it can be analyzed with the u test. When the sample content n is small, if the observation value x meets the normal distribution, then use the t test (so then the sample mean meets the t distribution), when x is an unknown distribution should be used when the rank sum test.
First, the sample mean and the overall mean t-test
The sample mean and the overall mean comparison of the t-test is actually inferred from the sample from the overall mean µ with a known overall mean µ0 (often for the theoretical value or standard value)
There is no difference. If, on the basis of a large number of surveys, it is known that the mean pulse rate of healthy adult males is 72 beats per minute, a doctor in a mountainous area randomly sampled 25 healthy males, and found that the mean pulse rate was 74.2 beats per minute, with a standard deviation of 6.0 beats per minute, asked whether it can be assumed that the adult males of the mountainous area have a higher mean pulse rate than that of the general adult males.
The inequality of the above two means may be due to sampling error, but also may really be the effect of environmental differences, for this reason, the t-test can be used to determine the test process is as follows:
1. Establish the hypothesis
H0: µ=µ0=72 beats/min, H0: µ>µ0, the level of the test is unilateral 0.05.
2. Calculate the statistical Quantity
The t-test for comparing the sample mean with the overall mean is performed with the t-value as the absolute value of the difference between the sample mean and the overall mean divided by the quotient of the standard error, where the standard error is the standard deviation divided by the quotient of the arithmetic square root of the sample content.
3. Determine the probability, make a judgment
With the degree of freedom v (sample content n minus 1) to check the t-boundary value table, 0.025<P<0.05, reject H0, accept H1, it can be considered that the mountainous region of adult males of the average number of pulse is higher than that of the average adult male.
It should be noted that when the sample content n is larger, the u test can be used instead of the t test.
Second, the t-test of the paired design
The paired design is a special design, which can control the influence of non-experimental factors on the results, and there are self-paired and non-self-paired. The t-test of paired design data is actually a comparison of the paired difference with the overall mean of “0”, i.e., to infer whether the overall mean of the difference is “0”. Therefore, the test process is similar to the t-test comparing the sample mean with the overall mean, i.e.
1. Establishment of hypotheses
H0: µd=0, i.e. the overall mean of the difference is “0”, H1: µd>0 or µd<0, i.e. the overall mean of the difference is not “0”, H1: µd>0 or µd<0, i.e. the overall mean of the difference is not “0”. “
2. Calculation of statistics
The t-value of the paired design t-test is the absolute value of the difference between the mean of the difference and 0 divided by the quotient of the standard error of the difference, which is the quotient of the standard deviation of the difference divided by the arithmetic square root of the sample content.
3. Determine the probability, make a judgment
To the degree of freedom v (the number of pairs minus 1) to check the t-boundary value table, if P & lt; 0.05, then reject H0, accept H1, if P & gt; = 0.05, it can not yet be rejected H0.
Third, the group design of the two samples of the means of the comparison of the t-test
The group design of the two sample mean comparisons of the t The t-test is also known as a group comparison or completely randomized design of the t-test, the purpose of which is to infer whether the two samples represented by the overall mean is equal. The test process is not very different from the above two t-tests, only the expression of the hypothesis and the formula for calculating the t-value is different.
The t-test for comparing the means of two samples is generally based on the assumptions that H0: µ1=µ2, i.e., the overall means of the two samples are equal, and H1: µ1> µ2 or µ1< µ2, i.e., the overall means of the two samples are unequal, with the test level of 0.05.
The t-statistic is calculated by dividing the absolute value of the difference in the mean of two samples by the standard error of the difference in the mean of two samples, and the t-value is calculated by dividing the standard error of the mean of two samples by the standard error of the difference in the mean of two samples. The standard error of the difference between the two sample means.
It should be noted that when the sample content of n is large (such as greater than 100 when) can be used to replace the t-test u test, the formula for calculating the value of u than the t-value of the formula is much simpler.
Fourth, the application of the t test conditions and precautions
The t test for the comparison of the means of two small samples have the following conditions:
(1) the two samples from the overall are in line with the normal distribution,
(2) the two samples from the overall variance of the Qi.
Therefore, in the two small samples before the t-test for comparison of means, to use the variance chi-square test to infer the two samples on behalf of the overall variance is equal, the variance chi-square test using the F-test, the principle of which is to see the larger samples of the variance of the smaller samples of the variance of the quotient is close to “1”. If close to “1”, then the two samples can be considered to represent the overall variance of the overall chi-square. To determine whether the two samples from the overall distribution of normal, available normality test method.
If the two samples from the overall variance is not uniform, also does not meet the normal distribution, in line with the lognormal distribution of information available for the t-test of its geometric mean, and other information can be analyzed with the t-test or rank-sum test.