In: Math
1) Give an example of a research problem that you could use an independent –sample t test on.
2) Give an example of a research problem that you could use a paired-sample t test on.
REAL LIFE EXAMPLE(RESEARCH PROBLEM) FOR T TEST AND PAIRED T TEST
1)
RESEARCH PROBLEM
Let’s say you’re curious about whether New Yorkers and Kansans spend a different amount of money per month on movies. It’s impractical to ask every New Yorker and Kansan about their movie spending, so instead you ask a sample of each—maybe 300 New Yorkers and 300 Kansans—and the averages are $14 and $18. The t-test asks whether that difference is probably representative of a real difference between Kansans and New Yorkers generally or whether that is most likely a meaningless statistical fluke.
Technically, it asks the following: If there were in fact no difference between Kansans and New Yorkers generally, what are the chances that randomly selected groups from those populations would be as different as these randomly selected groups are? For example, if Kansans and New Yorkers as a whole actually spent the same amount of money on average, it’s very unlikely that 300 randomly selected Kansans each spend exactly $14 and 300 randomly selected New Yorkers each spend exactly $18. So if you’re sampling yielded those results, you would conclude that the difference in the sample groups is most likely representative of a meaningful difference between the populations as a whole.
2) paired sample t test
EXAMPLE: Drinking and Driving
Drunk driving is one of the main causes of car accidents. Interviews with drunk drivers who were involved in accidents and survived revealed that one of the main problems is that drivers do not realize that they are impaired, thinking “I only had 1-2 drinks … I am OK to drive.”
A sample of 20 drivers was chosen, and their reaction times in an obstacle course were measured before and after drinking two beers. The purpose of this study was to check whether drivers are impaired after drinking two beers. Here is a figure summarizing this study:
Since the measurements are paired, we can easily reduce the raw data to a set of differences and conduct a one-sample t-test.
Here are some of the results for this data:
Step 1: State the hypotheses
We define μd = the population mean difference in reaction times (Before – After).
As we mentioned, the null hypothesis is:
The null hypothesis claims that the differences in reaction times are centered at (or around) 0, indicating that drinking two beers has no real impact on reaction times. In other words, drivers are not impaired after drinking two beers.
Although we really want to know whether their reaction times are longer after the two beers, we will still focus on conducting two-sided hypothesis tests. We will be able to address whether the reaction times are longer after two beers when we look at the confidence interval.
Therefore, we will use the two-sided alternative:
Step 2: Obtain data, check conditions, and summarize data
Let’s first check whether we can safely proceed with the paired t-test, by checking the two conditions.
We can see from the histogram above that there is no evidence of violation of the normality assumption (on the contrary, the histogram looks quite normal).
Also note that the vast majority of the differences are negative (i.e., the total reaction times for most of the drivers are larger after the two beers), suggesting that the data provide evidence against the null hypothesis.
The question (which the p-value will answer) is whether these data provide strong enough evidence or not against the null hypothesis. We can safely proceed to calculate the test statistic (which in practice we leave to the software to calculate for us).
Test Statistic: We will use software to calculate the test statistic which is t = -2.58.
Step 3: Find the p-value of the test by using the test statistic as follows
As a special case of the one-sample t-test, the null distribution of the paired t-test statistic is a t distribution (with n – 1 degrees of freedom), which is the distribution under which the p-values are calculated.
We will let the software find the p-value for us, and in this case, gives us a p-value of 0.0183 (SAS) or 0.018 (SPSS).
The small p-value tells us that there is very little chance of getting data like those observed (or even more extreme) if the null hypothesis were true. More specifically, there is less than a 2% chance (0.018=1.8%) of obtaining a test statistic of -2.58 (or lower) or 2.58 (or higher), assuming that 2 beers have no impact on reaction times.
Step 4: Conclusion
In our example, the p-value is 0.018, indicating that the data provide enough evidence to reject Ho.
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