Hi, guys, this is a Jonathan Lambert. With the mathematics development and support service at the National College of Ireland and this short video is going to detail how to undertake a single sample t-test and in particular, a two-tailed version of the single sample t-test another video after this particular video will deal with the one tailed alternative to this particular test. But for this particular video, We’ll do a two-tailed test. Okay, so we have a particular scenario. This scenario has been taken from a classic text. It’s the burns and Levine and business statistics text. Its chapter 9 and this is exercise nine point five five from that particular test. What we’re going to do is we’re gonna walk through the single sample. Hair t-test. I suppose a process for this particular scenario. Okay, so what we have in an article? It was claimed that’s the typical supermarket trip takes a mean of 22 minutes. Okay, and that suppose that in an effort to test this claim you select a sample of 50 shoppers at a local supermarket and you find that the mean shopping time for the sample of 50 shoppers is twenty five point three six minutes with an Associated standard deviation of seven point two four minutes and the question is using the zero point one. Oh, level of significance. Is there evidence that the mean shopping time at the local supermarket is different from the claimed value of twenty two minutes? Okay, so for every hypothesis test that we perform, there’s five stages. A stage one is where we define our hypothesis, and the hypothesis is a statement about the population parameter. Okay, so, in this case we’re talking about average trip times to the supermarket, so the hypothesis is going to be a statement about the population mean or the average population time. Okay, now when it comes to a hypothesis, there’s two. I suppose positions that we can take. There’s the null position zero and there’s the alternative position. Hey, J. Okay, it’s important to note that the alternative is always what we’re trying to prove or the alternative is the position that err evidence suggests. Okay, and in this particular question, we’re being asked. Is there evidence that the mean shopping time at the local supermarket is different from the claimed value of 22 minutes, so the alternative is what we want to try to prove or is where we’d like our evidence to take us to, and that’s that the mean shopping time is different to 22 minutes so that the main shopping time is not equal to 22 minutes, in which case the null hypothesis must be that the main shopping time is equal to 22 minutes. Okay, and what we do from a hypothesis perspective is we assumed a null position to be true. Okay, until we capture evidence to suggest that this is not the case. Okay, So that’s stage one dawn of our hypothesis test, and we’ve defined our hypothesis. Okay, don’t forget that the alternative is always what you want to try to prove in this case we’ve been asked. Is there evidence that the mean shopping time at the local supermarket is different from the claim value of 22 minutes, okay, and the second stage is where we define the significance of our test or the significance level of our test, so our disc significance value and that’s given in this particular in this particular scenario, it says, using the 0.1 Oh, level of significance. So what we know is the Alpha must be equal to 0.1 okay. I suppose more importantly, the significance is the probability of was incorrectly rejecting the null hypothesis. Okay, so if we reject the null hypothesis, what we’re really saying if our significance is that Alpha is equal to 0.1 all what we’re saying is that if we do reject a null hypothesis, we’ll only reject it incorrectly 10% at a time, which means that if we do reject a null hypothesis that we’re 90% confident that we’ve made the right decision, but let’s just keep in mind that we might be, but we will only be wrong 10% of the time. Okay, No, we also need to keep in mind that this is going to be a two-tailed test because the alternative position for us to prove that D that the null position that the mean shopping time is 22 minutes for us to prove that incorrect. All we need to show. Is that the mean shopping time is less than 22 minutes or at the mean. Shopping time is greater than 22 minutes. So this is a two-tailed test. Okay, and the next stage is our test statistic tree. Okay, is our test statistic where we calculate how far our evidence is away from? Our hypothesized a center of our distribution? Now we’re going to do a t-test on this. A single sample t-test, so our test statistic is. T is equal to X bar minus Mu over S divided by the square root of N. Okay, so to calculate the test statistic, there’s one two three four values that we require, okay. I suppose what we should do first of all is we should calculate what the sample distribution are. The sample values are that’s given in in this scenario. Okay, so we suppose it’s suppose. In an effort to test this claim, you select a sample of 50 shoppers, So what we know is that N is equal to 50 okay, and we noticed that the mean shopping time for the sample of 50 shoppers is 25 point three six million minutes. So the mean shopping time X-bar is twenty five point three, six minutes. Okay, would an associate standard deviation of seven point two four minutes, so our standard deviation. S is equal to seven point two four minutes. Okay, so we have most values. We have the sample size N. We have the sample mean x-bar. And we have the sample standard deviation for this particular scenario where s the question is? What is Mu well? Mew is the value of our null hypothesis, So Mu in this case is going to be equal to 22 so our test statistic. T is going to be equal to X bar, which is 25 points trees. X minus Mu, which is 22 divided by S, which is 7.2 for which needs to be divided by the square root of N, Which is which is 50 Okay now. I’m going to do this on a calculator. Okay, so first of all. I’m going to calculate what the numerator is. Its twenty five point. Three six minus. Twenty two gives us a value of three point three six. I’m going to divide that by 7.24 okay, and which gives us a value of zero point four six. And when we when we divide by a fraction, it’s equivalent to inverting. I’m multiplying by by the inverted fraction, so I’m going to multiply this by the square root of 50 so we want to multiply this by this square root of 50 to give us a test statistic of approximately three point two eight. Okay, so what we know? Is that in standard units? Maybe the evidence is a distance of three point. Two eight standard standard units away from our hypothesized mean value now. The question is, is this far enough away from 22 to be significantly Different to it? Okay, so to figure this out what we need to do? Is we need to calculate? I suppose we need to calculate our critical values or critical values for our distribution. Okay, so we’re assuming a t-distribution. So our distribution is going to be bell-shaped. It’s centered on zero. It’s a two-tailed test. Okay, so we have two tails and we take our significance level, and we split our significance level in half. We pull Alpha over two in this tail, and we put Alpha over two in the left tail, okay, and well, Alpha over 2 is the same as 0.05 okay, So our critical values in this case are the critical values associated with 0.05 of an area Okay in the right hand tail of our distribution or 0.05 of an area in the left hand tail of our distribution. Now we have set of tables that we use. That will allow us to calculate what critical value has 0.05 of an area to the to the right-hand side. Okay, and the tables? Tell us that, okay if we go for. T distribution tables. I just grabbed a set of our T distribution tables. Okay, we have already created these particular tables for the T distribution. Our T distribution tables. Look, something like this, okay, And this is a table of the students. T distribution for critical values for right hand tail areas. Okay, so what we’re interested in is we’re interested in that. There should be 0.05 of an area in the right tail, so there should be 0.05 in the right tail. So this is this column here, which is 0.05 and then what we do is we look up our degrees of freedom for our test, which is our sample size minus 1 so we’re going to come down this column to 49 Our sample size is 50 minus 1 gives us 49 so the question is what critical value and with 49 degrees of freedom. Okay has 0.05 of the area to the right hand side. Okay, now on our tables here. 49 Is here’s degrees of freedom. And the column is p is equal to 0.05 If we triangulate in gives us a test statistic of sorry. A critical value of 1 point 6 7 7 so its 1 point 6 7 7 That’s what we get when we look up there. T distribution. So actually, this value here is equal to 1 point 6 7 7 and this value over here is equal to minus 1 point 6 7 7 to the Symmetric properties of the distribution. OK, so now we have our critical values. So what we’re saying, is this? Is that a fair test statistic? OK, falls into any of these critical regions. Okay, we reject a null hypothesis in favor of the alternative. Okay, and clearly what we can see is. 3.28 Is in the right hands tail area. Let me just maybe do that a little bit bigger. Okay, so our our distribution looks something like this. It’s centered on zero a critical value here for a right hand tail that has 0.05 in It is one point six seven seven, okay. Our test statistic is three point two eight, so three point two eight, which is our T value three point two eight falls in the rejection region of our distribution. Okay, so what we’re gonna say here is? This is that the probability of observing a test statistic as extreme as one point six seven seven. Okay, is at least, or is that most 0.05 okay, so what we know? Is that observing? This particular test statistic is very, very improbable. Yeah, if the null hypothesis is actually true, which makes it that it would be very probable to observe this test statistic if the null hypothesis was actually incorrect, so the next stage of our test is to make our decision. So step five is our decision. Okay, that’s where we compare. Let me just do this. We compare our. T value to a critical value. Okay, and what we have here is clearly clearly our T statistic is bigger than our critical value. What, I mean by that is three point. Two five is bigger than one point six, seven, seven. And as such as such we reject, we reject h0 in favor of hey, J. At the ten percent significance significance level. Okay, tensing, ten percent significance level. Wait, oh, really, what we’re saying is? This is that there’s evidence to suggest that the mean waiting time, so we’re rejecting. Okay, so there’s evidence to suggest that the mean waiting time is not twenty two minutes and actually from our evidence or we’re suggesting, is that the mean waiting time is significantly greater than twenty two minutes and falls into a right-tail. But the test was a two-tailed test If we had a found that the mean waiting time is less than twenty two minutes and into the right tail that would have also been evidence for us to reject the null hypothesis. Okay, guys. This was Jonathan Lambert with the mathematics development and support service at the National College of Ireland. And I hope this video was helpful.