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T-Tests

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Source: Table created by Katherine Williams

It is likely that we do not know a population standard deviation. So this tutorial on t-tests covers tests that are very common. Because it is common for us not to know the population standard deviation.

Instead we use the sample standard deviation s in order to estimate sigma. If our sample size is small, using the normal distribution underestimates the proportion of the extreme values. And so instead of using the normal distribution, we use this student's t-distribution.

And before we talk about that, just a quick reminder, for t-tests the standard error is s divided by the square root of n, our sample size. This is very much like our z-test, except because we don't know sigma we use s here. The variation in s is what causes us to need to shift off of the normal distribution and onto student's t-distribution.

The student's t-distribution is very similar to the normal distribution. And for a large sample size, the t-distribution approaches normal distribution. When you have a small sample size, there's a lot more variation in s. And that's why it's much more important to use t-distributions for those smaller sample size. With the smaller sample size, the t-distribution does not diminish towards the tails as fast as the normal distribution does. So that's why it's better for the smaller sample sizes than the normal.

A t-test is a hypothesis test for a population mean when sigma, the population standard deviation, is not known. The steps for taking a t-test are listed here. First, you need to formulate the null and alternative hypotheses and choose the significance level.

Next, you need to check that the conditions of the hypothesis test are met for the random sample you use. Third, you need to use s to calculate a test statistic and compare to a critical value or find the p value and compare that to the significance level. Then we need to use a t-table. We have to decide whether reject or not the null hypothesis and draw conclusion.

In this example, we're looking at a school of 400 students. In that school, the spelling quiz scores for 50 students were randomly selected for analysis. The sample mean was 7.5 questions correct with a standard deviation of 0.5.

We want to test the hypothesis that the school performed worse than the state average of 7.7 questions correct. So first, our null hypothesis is that the mean of that sample and of the school-- sorry, the mean of the school is in fact 7.7. Our alternative hypothesis is that the school mean is something lower than 7.7.

And we're going to choose a significance level of 5%, so our alpha is 0.05. Now, the conditions are met. The students were selected randomly, and the other ones are met as well.

Next, we're going to calculate a test statistic. So we have that calculation here. We have the 7.5 minus the 7.7, the sample mean minus the state mean. And then divided by 0.5, divided by the square root of 50, n, our sample size.

When you enter this new calculator, you end up with negative 2.83. Now, in order to find the p value, I entered this negative 2.83 and the degrees of freedom into a calculator. And I used an online calculator, you can also use a table.

And I find a p value of 0.03. Now, based on this, we're able to reject the null hypothesis. It does not meet the significance level. So we're able to reject the null hypothesis.