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A t-test is any statistical hypothesis test in which the test statistic has a Student's t distribution if the null hypothesis is true. It is applied when the population is assumed to be normally distributed but the sample sizes are small enough that the statistic on which inference is based is not normally distributed because it relies on an uncertain estimate of standard deviation rather than on a precisely known value.
## History

The t statistic was introduced in 1908 by William Sealy Gosset, a statistician working for the Guinness brewery in Dublin, Ireland ("Student" was his pen name). Gosset had been hired due to Claude Guinness's innovative policy of recruiting the best graduates from Oxford and Cambridge to apply biochemistry and statistics to Guinness' industrial processes. Gosset devised the t-test as a way to cheaply monitor the quality of beer. He published the test in Biometrika in 1908, but was forced to use a pen name by his employer, who regarded the fact that they were using statistics as a trade secret. In fact, Gosset's identity was known to fellow statisticians.## Use

Among the most frequently used t tests are:## Assumptions

### Determining type

For novices, the most difficult issue is often whether the samples are independent or dependent. Independent samples typically consist of two groups with no relationship. Dependent samples typically consist of a matched sample (or a "paired" sample) or one group that has been tested twice (repeated measures). ## Calculations

### Independent one-sample t-test

### Slope of a regression line

### Independent two-sample t-test

#### Equal sample sizes, equal variance

This test is only used when both:
#### Unequal sample sizes, equal variance

This test is used only when it can be assumed that the two distributions have the same variance. (When this assumption is violated, see below.) The t statistic to test whether the means are different can be calculated as follows:#### Unequal sample sizes, unequal variance

This test is used only when the two sample sizes are unequal and the variance is assumed to be different. See also Welch's t test. The t statistic to test whether the means are different can be calculated as follows:### Dependent t-test

This test is used when the samples are dependent; that is, when there is only one sample that has been tested twice (repeated measures) or when there are two samples that have been matched or "paired".

### Example

## Alternatives to the t test

Recall that the t test can be used to test the equality of the means of two normal populations with unknown, but equal, variance.## Implementations

Most spreadsheet programs and statistics packages, such as Microsoft Excel, DAP, gretl, R and PSPP include implementations of Student's t-test.
## Notes

## References

*## External links

### Online calculators

Today, the t-test is more generally applied to the confidence that can be placed in judgments made from small samples.

- A test of whether the mean of a normally distributed population has a value specified in a null hypothesis.
- A test of the null hypothesis that the means of two normally distributed populations are equal. Given two data sets, each characterized by its mean, standard deviation and number of data points, we can use some kind of t test to determine whether the means are distinct, provided that the underlying distributions can be assumed to be normal. All such tests are usually called Student's t tests, though strictly speaking that name should only be used if the variances of the two populations are also assumed to be equal; the form of the test used when this assumption is dropped is sometimes called Welch's t test. There are different versions of the t test depending on whether the two samples are
- unpaired, independent of each other (e.g., individuals randomly assigned into two groups, measured after an intervention and compared with the other group), or
- paired, so that each member of one sample has a unique relationship with a particular member of the other sample (e.g., the same people measured before and after an intervention).

- If the calculated p-value is below the threshold chosen for statistical significance (usually the 0.10, the 0.05, or 0.01 level), then the null hypothesis which usually states that the two groups do not differ is rejected in favor of an alternative hypothesis, which typically states that the groups do differ.

- A test of whether the slope of a regression line differs significantly from 0.

Once a t value is determined, a p-value can be found using a table of values from Student's t-distribution.

- Normal distribution of data, tested by using a normality test, such as Shapiro-Wilk and Kolmogorov-Smirnov test.
- Equality of variances, tested by using either the F test, the more robust Levene's test, Bartlett's test, or the Brown-Forsythe test.
- Samples may be independent or dependent, depending on the hypothesis and the type of samples:
- Independent samples are usually two randomly selected groups
- Dependent samples are either two groups matched on some variable (for example, age) or are the same people being tested twice (called repeated measures)

Since all calculations are done subject to the null hypothesis, it may be very difficult to come up with a reasonable null hypothesis that accounts for equal means in the presence of unequal variances. In the usual case, the null hypothesis is that the different treatments have no effect — this makes unequal variances untenable. In this case, one should forgo the ease of using this variant afforded by the statistical packages. See also Behrens–Fisher problem.

One scenario in which it would be plausible to have equal means but unequal variances is when the 'samples' represent repeated measurements of a single quantity, taken using two different methods. If systematic error is negligible (e.g. due to appropriate calibration) the effective population means for the two measurement methods are equal, but they may still have different levels of precision and hence different variances.

Dependent t-tests are also used for matched-paired samples, where two groups are matched on a particular variable. For example, if we examined the heights of men and women in a relationship, the two groups are matched on relationship status. This would call for a dependent t-test because it is a paired sample (one man paired with one woman). Alternatively, we might recruit 100 men and 100 women, with no relationship between any particular man and any particular woman; in this case we would use an independent samples test.

Another example of a matched sample would be to take two groups of students, match each student in one group with a student in the other group based on an achievement test result, then examine how much each student reads. An example pair might be two students that score 90 and 91 or two students that scored 45 and 40 on the same test. The hypothesis would be that students that did well on the test may or may not read more. Alternatively, we might recruit students with low scores and students with high scores in two groups and assess their reading amounts independently.

An example of a repeated measures t-test would be if one group were pre- and post-tested. (This example occurs in education quite frequently.) If a teacher wanted to examine the effect of a new set of textbooks on student achievement, (s)he could test the class at the beginning of the year (pretest) and at the end of the year (posttest). A dependent t-test would be used, treating the pretest and posttest as matched variables (matched by student).

In testing the null hypothesis that the population mean is equal to a specified value μ_{0}, one uses the statistic

- $t\; =\; frac\{overline\{x\}\; -\; mu\_0\}\{s\; /\; sqrt\{n\}\},$

where s is the sample standard deviation of the sample. n is the sample size. The degrees of freedom used in this test is n − 1.

Suppose one is fitting the model

- $Y\_i\; =\; alpha\; +\; beta\; x\_i\; +\; varepsilon\_i,$

where x_{i}, i = 1, ..., n are known, α and β are unknown, and ε_{i} are independent normally distributed random errors with expected value 0 and unknown variance σ^{2}, and Y_{i}, i = 1, ..., n are observed. It is desired to test the null hypothesis that the slope β is equal to some specified value β_{0} (often taken to be 0, in which case the hypothesis is that x and y are unrelated).

Let

- $$

Then

- $frac\{(widehatbeta\; -\; beta\_0)sqrt\{n-2\}\}\{\; sqrt\{text\{SSE\}sum\_\{i=1\}^n\; left(x\_i\; -\; overline\{x\}right)^2\}\; \}$

has a t-distribution with n − 2 degrees of freedom if the null hypothesis is true.

- *the two sample sizes (that is, the n or number of participants of each group) are equal;

- *it can be assumed that the two distributions have the same variance.

The t statistic to test whether the means are different can be calculated as follows:

- $t\; =\; frac\{bar\; \{X\}\_1\; -\; bar\{X\}\_2\}\{S\_\{X\_1X\_2\}\; cdot\; sqrt\{frac\{2\}\{n\}\}\}$

- $S\_\{X\_1X\_2\}\; =\; sqrt\{frac\{S\_\{X1\}^2+S\_\{X2\}^2\}\{2\}\}\; .$

Here $S\_\{X\_1X\_2\}$ is the grand standard deviation (or pooled standard deviation), 1 = group one, 2 = group two. The denominator of t is the standard error of the difference between two means. For significance testing, the degrees of freedom for this test is 2n − 2 where n = # of participants in each group.

- $t\; =\; frac\{bar\; \{X\}\_1\; -\; bar\{X\}\_2\}\{S\_\{X\_1X\_2\}\; cdot\; sqrt\{frac\{1\}\{n\_1\}+frac\{1\}\{n\_2\}\}\}$

where

- $S\_\{X\_1X\_2\}\; =\; sqrt\{frac\{(n\_1-1)S\_\{X1\}^2+(n\_2-1)S\_\{X2\}^2\}\{n\_1+n\_2-2\}\}.$

Note that the formulae above are generalizations for the case where both samples have equal sizes (substitute n_{1} and n_{2} for n and you'll see).

$S\_\{X\_1X\_2\}$ is the unbiased estimator of the variance of the two samples, n = number of participants, 1 = group one, 2 = group two. n − 1 is the number of degrees of freedom for either group, and the total sample size minus two (that is, n_{1} + n_{2} − 2) is the total number of degrees of freedom, which is used in significance testing.

The statistical significance level associated with the t value calculated in this way is the probability that, under the null hypothesis of equal means, the absolute value of t could be that large or larger just by chance—in other words, it's a two-tailed test, testing whether the means are different when, if they are, either one may be the larger (see Press et al., 1999, p. 616).

- $t\; =\; \{overline\{X\}\_1\; -\; overline\{X\}\_2\; over\; s\_\{overline\{X\}\_1\; -\; overline\{X\}\_2\}\}$

- $s\_\{overline\{X\}\_1\; -\; overline\{X\}\_2\}\; =\; sqrt\{\{s\_1^2\; over\; \{n\}\_1\}\; +\; \{s\_2^2\; over\; \{n\}\_2\}\}$

Where s^{2} is the unbiased estimator of the variance of the two samples, n = number of participants, 1 = group one, 2 = group two. Note that in this case, $s\_\{overline\{X\}\_1\; -\; overline\{X\}\_2\}^2$ is not a pooled variance. For use in significance testing, the distribution of the test statistic is approximated as being an ordinary Student's t distribution with the degrees of freedom calculated using

- $mathrm\{D.F.\}\; =\; frac\{(s\_1^2/n\_1\; +\; s\_2^2/n\_2)^2\}\{(s\_1^2/n\_1)^2/(n\_1-1)\; +\; (s\_2^2/n\_2)^2/(n\_2-1)\}.$

This is called the Welch-Satterthwaite equation. Note that the true distribution of the test statistic actually depends (slightly) on the two unknown variances: see Behrens–Fisher problem.

This test can be used as either a one-tailed or two-tailed test.

- $t\; =\; frac\{overline\{X\}\_D\; -\; mu\_0\}\{\{s\_D\}/sqrt\{N\}\}.$

For this equation, the differences between all pairs must be calculated. The pairs are either one person's pre-test and post-test scores or between pairs of persons matched into meaningful groups (for instance drawn from the same family or age group: see table). The average (X_{D}) and standard deviation (s_{D}) of those differences are used in the equation. The constant μ_{0} is non-zero if you want to test whether the average of the difference is significantly different than μ_{0}. The degree of freedom used is N − 1.

Example of repeated measures | |||

Number | Name | Test 1 | Test 2 |
---|---|---|---|

1 | Mike | 35% | 67% |

2 | Melanie | 50% | 46% |

3 | Melissa | 90% | 86% |

4 | Mitchell | 78% | 91% |

Example of matched pairs | |||

Pair | Name | Age | Test |
---|---|---|---|

1 | Jon | 35 | 250 |

1 | Jane | 36 | 340 |

2 | Jimmy | 22 | 460 |

2 | Jessy | 21 | 200 |

A random sample of screws have weights

- 30.02, 29.99, 30.11, 29.97, 30.01, 29.99

Calculate a 95% confidence interval for the population's mean weight.

Assume the population is distributed as N(μ, σ^{2}).

The samples' mean weight is 30.015 with standard deviation of 0.0497. With the mean and the first five weights it is possible to calculate the sixth weight. Consequently there are five degrees of freedom.

We can lookup in the table that for a confidence range of 95% and five degrees of freedom, the value is 2.571.

i.e.

- $$

- $29.9630.07.,\; math>$

- To relax the normality assumption, a non-parametric alternative to the t test can be used, and the usual choices are:
- for independent samples, the Mann-Whitney U test
- for related samples, either the binomial test or the Wilcoxon signed-rank test
- To test the equality of the means of more than two normal populations, an analysis of variance can be performed
- To test the equality of the means of two normal populations with known variance, a Z-test can be performed

*

- GraphPad's Paired/Unpaired/Welch T-Test Calculator
- OpenEpi software program

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Last updated on Monday October 06, 2008 at 09:31:48 PDT (GMT -0700)

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