Brief Summary
This video focuses on the one-sample t-test, explaining when and how to use it in hypothesis testing. Key points include the difference between Z tests and t-tests, defining null and alternative hypotheses, calculating degrees of freedom, setting a decision rule, calculating the test statistic, and interpreting results.
- One-sample t-tests are used when the population standard deviation is unknown.
- The process involves seven steps: defining hypotheses, stating alpha, calculating degrees of freedom, establishing a decision rule, calculating the test statistic, stating results, and drawing conclusions.
The video begins by contrasting Z tests with t-tests. Z tests are applicable when the population standard deviation is known, while t-tests are used when the population standard deviation is unknown, requiring an estimate and utilizing the T distribution.
An example is presented where a scientist investigates the effect of a medication on intelligence. The population mean IQ is known to be 100, and a sample of 30 participants who took the medication had a mean of 140 and a standard deviation of 20. The goal is to determine if this medication significantly affects intelligence with an alpha level of 0.05.
The null hypothesis (H0) is defined as the mean IQ being equal to 100, while the alternative hypothesis (H1) states that the mean IQ is not equal to 100. The chosen alpha level for the test is set at 0.05.
To compute degrees of freedom for the t-test, the formula n - 1 is used, leading to 29 degrees of freedom for the sample of 30 participants. This factor is crucial for determining the critical value in the t-test.
The decision rule is established based on the alpha level of 0.05 for a two-tailed test. The middle 95% of the distribution will fall between the critical values of negative 2.04 and positive 2.04. If the calculated t statistic falls outside this range, the null hypothesis will be rejected.
Using the sample mean (140), population mean (100), sample standard deviation (20), and sample size (30), the t statistic is calculated using the appropriate formula, resulting in a t value of 10.96.
With the calculated t value of 10.96 exceeding the critical value of 2.04, the null hypothesis is rejected. This indicates that the medication significantly affected intelligence, resulting in a sample mean that differs from the expected population mean of 100. The results can be formally stated as T = 10.96 with p < 0.05, indicating statistical significance.
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Transcript
Full transcript of “One Sample t-Test”. Click a timestamp to open the video at that moment. Need another video? Use the YouTube transcript generator.
0:00 and now the one sample t-test so far I've talked about Z tests and T test Z tests are commonly done when the population standard deviation is known T tests are commonly done when the population standard deviation is unknown we use the T distribution in these cases because the unknown population standard deviation is now a value that we must estimate we don't know us we have to put in an estimate for word that's why we have this new distribution instead of just using the Z distribution so here's a question we could do a t-test on in the population the average IQ is 180 mo scientist wants to test a new medication to see if it has either a positive or negative effect on intelligence or no effect at all a sample of 30 participants who have taken the
0:45 medication has a mean of 140 with a standard deviation of 20 did the medication effect intelligence so we're going to answer this question using these seven steps with an alpha level of 0.05 notice that we don't know the population standard deviation we only know that the population mean is 100 we do however know the sample standard deviation which in this case is 20 that's why we're doing a t-test instead of a Z test so step one is for us to define our null and alternative hypotheses then we will state the alpha calculate the degrees of freedom because this is a t-test then state the decision rule calculate the test statistic state the results and state the conclusion so let's get started with that first we're going to define our hypotheses in this
1:31 case h0 our null hypothesis is that the mean is equal to 100 because in the population the average IQ is 100 so that's what we would expect the mean of this sample to be h1 our alternative hypothesis is that the mean is not equal to 100 because we're testing to see if it's any different from the expected value of 100 next the Alpha level is just point zero five because I said use point zero five now we're going to calculate the degrees of freedom now connecting the degrees of freedom for a t-test is like this it's just n minus 1 same thing as it was before we have 30 participants minus 1 is 29 so this t-test has 29 degrees of freedom this has an effect on what credit value we use which is going to be coming up in just a bit step four is to state
2:17 the decision rule now we have an alpha of 0.05 so we want to find the middle 95% of where we would expect to the mean to be if it's outside of that in this two tailed test we're going to conclude that the mean we have is strange and likely is different from the population we're comparing it to so this is our T table remember that our alpha level is point zero five this is a two-tailed test and it has 29 degrees of freedom so you can see I have point zero five and 29 marked in red and now if we draw the area's the arrows over here this is a critical value we're going to use we're going to use two point zero four five two and that's how I found it using the alpha level and the degrees of freedom for the T test so that is our critical value we would expect most values we
3:05 calculate to fall within negative 2.0 four and positive 2.0 for if it falls outside of that range we're going to conclude that our sample is different from the expected population so now our decision rule is if T is less than negative 2.0 for or greater than positive 2.04 we're going to reject the null hypothesis so now let's actually calculate that T and it uses this equation right here where the sample mean is 140 the population mean is 100 the sample standard deviation is 20 and the sample size is 30 so I'm just going to plug those things into the equation and that's how I get a T of ten point nine six I'm going to go ahead but if you want to write that down you can pause the video right here and come back to it when you're done so next the next
3:53 step is for us to state our results now remember our decision rule was if the T was less than negative two point oh four or greater than two point oh four we're going to reject the null hypothesis and we found a T of ten point nine six that 10 point nine six is definitely greater than two point zero four so we're going to reject h0 we're going to reject the null hypothesis now our conclusion or what this means is that medication significantly affected intelligence basically it did something it changed it from the expected value of in this case it was 140 so it looks like it went up a lot we can't really say that but it is different from the 100 we thought it would be and then I have T equals 10 point 9 6 with P less than point zero 5 that's just the official format for stating the results if your
4:39 teacher wants to do that you can do that if you don't then you don't have to do that but that is how you would write your conclusion and that is a one sample t-test
