Paired t-test Assignment help
Paired sample t-test is a statistical technique that is used to compare two population means in the case of two samples that are correlated. Paired sample t-test is used in 'before-after' studies, or when the samples are the matched pairs, or when it is a case-control study.
Where d bar is the mean difference between two samples, s² is the sample variance, n is the sample size and t is a paired sample t-test with n-1 degrees of freedom. An alternate formula for paired sample t-test.
Testing of hypothesis or decision making: After calculating the parameter, we will compare the calculated value with the table value. If the calculated value is greater than the table value, then we will reject the null hypothesis for the paired sample t-test. If the calculated value is less than the table value, then we will accept the null hypothesis and say that there is no significant mean difference between the two paired samples. Set up hypothesis: We set up two hypotheses. The first is the null hypothesis, which assumes that the mean of two paired samples are equal. The second hypothesis will be an alternative hypothesis, which assumes that the means of two paired samples are not equal.
A paired t-test can be more powerful than a 2-sample t-test because the latter includes additional variation occurring from the independence of the observations. A paired t-test is not subject to this variation because the paired observations are dependent. Also, a paired t-test does not require both samples to have equal variance. Therefore, if you can logically address your research question with a paired design, it may be advantageous to do so, in conjunction with a paired t-test, to get more statistical power.
The Paired Samples t Test can only compare the means for two (and only two) related (paired) units on a continuous outcome that is normally distributed. The Paired Samples t Test is not appropriate for analyses involving the following: 1) unpaired data; 2) comparisons between more than two units/groups; 3) a continuous outcome that is not normally distributed; and 4) an ordinal/ranked outcome.
- To compare unpaired means between two groups on a continuous outcome that is normally distributed, choose the Independent Samples t Test.
- To compare unpaired means between more than two groups on a continuous outcome that is normally distributed, choose ANOVA.
- To compare paired means for continuous data that are not normally distributed, choose the nonparametric Wilcoxon Signed-Ranks Test.
- To compare paired means for ranked data, choose the nonparametric Wilcoxon Signed-Ranks Test.
The paired t-test also works well when the assumption of normality is violated, but only if the underlying distribution is symmetric, unimodal, and continuous. If the values are highly skewed, it might be appropriate to use a nonparametric procedure, such as a 1-sample sign test.
The dependent t-test (called the paired-samples t-test in SPSS Statistics) compares the means between two related groups on the same continuous, dependent variable. For example, you could use a dependent t-test to understand whether there was a difference in smokers' daily cigarette consumption before and after a 6 week hypnotherapy programme (i.e., your dependent variable would be "daily cigarette consumption", and your two related groups would be the cigarette consumption values "before" and "after" the hypnotherapy programme). If your dependent variable is dichotomous, you should instead use McNemar's test.
SPSS Statistics generates three tables in the Output Viewer under the title "T-Test", but you only need to look at two tables: the Paired Samples Statistics table and the Paired Samples Test table. In addition, you will need to interpret the boxplots that you created to check for outliers and the output from the Shapiro-Wilk test for normality, which you used to determine whether the distribution of the differences in the dependent variable between the two related groups were approximately normally distributed. This is explained in our enhanced guide. However, in this "quick start" guide, we focus on the two main tables you need to understand if your data has met all the necessary assumptions:
Assumption : There should be no significant outliers in the differences between the two related groups. An outlier is simply a single data point within your data that does not follow the usual pattern (e.g., in a study of 100 students' IQ scores, where the mean score was 108 with only a small variation between students, one student had a score of 156, which is very unusual, and may even put her in the top 1% of IQ scores globally). The problem with outliers is that they can have a negative effect on the paired t-test, distorting the differences between the two related groups (whether increasing or decreasing the scores on the dependent variable), which reduces the accuracy of your results. In addition, they can affect the statistical significance of the test. Fortunately, when using Stata to run a paired t-test on your data, you can easily detect possible outliers.
This output provides useful descriptive statistics for the two groups that you compared, including the mean and standard deviation, as well as actual results from the paired t-test. Looking at the Mean column, you can see that those people who used the nicotine patches had lower cigarette consumptions at the end of the experiment compared to those who received the placebo. You can see that there is a mean difference between the two trials of 0.1355 km (Mean) with a standard deviation of 0.09539 km (Std. Dev.), a standard error of the mean of 0.02133 km (Std. Err.), and 95% confidence intervals of 0.09085 to 0.18015 km (95% Conf. interval).
You are presented with an obtained t-value (t) of 6.3524, the degrees of freedom (degrees of freedom), which are 19, and the statistical significance (2-tailed p-value) of the paired t-test (Pr(|T| > |t|) under Ha: mean(diff) != 0), which is 0.0000. As the p-value is less than 0.05 (i.e., p < .05), it can be concluded that there is a statistically significant difference between our two variable scores (carb and carb_protein). In other words, the difference between the two run distances is not equal to zero.
The paired t–test assumes that the differences between pairs are normally distributed; you can use the histogram spreadsheet described on that page to check the normality. If the differences between pairs are severely non-normal, it would be better to use the Wilcoxon signed-rank test. I don't think the test is very sensitive to deviations from normality, so unless the deviation from normality is really obvious, you shouldn't worry about it.
The paired t–test does not assume that observations within each group are normal, only that the differences are normal. And it does not assume that the groups are homoscedastic. For the zinc concentration problem, if you do not recognize the paired structure, but mistakenly use the 2-sample t-test treating them as independent samples, you will not be able to reject the null hypothesis. This demonstrates the importance of distinguishing the two types of samples. Also, it is wise to design an experiment efficiently whenever possible.
Wiebe and Bortolotti (2002) examined color in the tail feathers of northern flickers. Some of the birds had one "odd" feather that was different in color or length from the rest of the tail feathers, presumably because it was regrown after being lost. They measured the yellowness of one odd feather on each of 16 birds and compared it with the yellowness of one typical feather from the same bird. There are two nominal variables, type of feather (typical or odd) and the individual bird, and one measurement variable, yellowness. Because these birds were from a hybrid zone between red-shafted flickers and yellow-shafted flickers, there was a lot of variation among birds in color, making a paired analysis more appropriate. The difference was significant (P=0.001), with the odd feathers significantly less yellow than the typical feathers (higher numbers are more yellow).