WEBVTT mathematics/statistics/son
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Hi and welcome to www.educator.com.
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We are going to talk about confidence interval and hypothesis testing for the difference of two paired means.
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We have been talking about independent samples so far, one example, two independent samples.
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We are going to talk about paired samples.
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We are going to look at the difference between independent samples and paired samples.
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We are also going to try and clarify the difference between independent sample
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and independent variables because paired samples still use independent variables.
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We are going to talk about two types of t-tests.
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One that we covered or also called hypothesis testing and one that we covered so far with independent samples.
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The new one that was cover with paired samples.
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We are going to introduce some notation for paired samples, go through the steps of hypothesis testing
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for paired samples and adjust or add on to the rules of SDOD that we already have looked at.
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Finally we are going to go over the formulas that go with the steps of hypothesis testing for independent as well as paired samples.
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We are going to briefly cover confidence interval for paired samples.
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Here is the goal of hypothesis testing.
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Remember, with one sample our goal was to reject the null when we get a sample
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that significantly different from the hypothesized population.
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When we talk about two-tailed hypotheses we are really saying the
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hypothesized population might be significantly higher or significantly lower.
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Either way, we do not care.
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The sample is too low or too high, it is too extreme in some way.
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If that is the case, we reject the null.
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In two samples, what we do is we reject the null when we get samples that
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are significantly different from each other in some way.
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Either one is significantly lower than the other or the other is significantly lower than the one.
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It does not matter.
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Our null hypothesis becomes this idea that x - y either = 0 because they are the same
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and the alternative is that it does not equal 0 because they are different from each other.
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If they are the same that is considered the null hypotheses and when they are different that considered the alternative hypotheses.
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Remember another way you could write this is by adding y to each side and then you get x=y.
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X = y they are the same.
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In that way you know that you are covering the entire space of all the differences and the end of the day
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we can figure out whether they are the same or we do not think that the they are the same.
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Let us talk about independent samples versus paired samples because from here on out,
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we are totally going to be dealing with paired samples.
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It would help to know what those are.
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Independent samples, the scores are derived separately from each other.
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For instance they came from separate people, separate schools, separate dishes.
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The samples are independent from each other.
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My getting of the sample had nothing to do with my getting of this other sample.
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In dependent, another word for paired, in dependent or paired samples the scores are linked in some way.
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For instance, they are linked by the same person so my score on the math test and my score on the english test are linked because they both come from me.
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Maybe we are one married couple, we ask one spouse how many children would you like to have
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and you ask the other spouse how many children would you like to have?
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In that way, although they come from different people these scores are linked because they come from the same married couple.
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Another thing might be a pre and post tests of the class.
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Maybe a statistics class might do a pre and post test.
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Maybe 10 different statistics classes from all over the United States picked to do a pre and post test.
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Those tests are linked because the same class did the first test and the second test.
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10 different classes did the pairs.
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It is not just a hodgepodge of pretests scores and a hodgepodge of posttest scores, it is more like a neat line
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where the pretests scores for this guy, but for this class is lined up with the pretests scores for that class.
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They are all lined up next to each other.
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We know these definitions, let us see if we can pick them out.
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Which of these is which?
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The test scores from Professor x’s class versus test scores from professor y class.
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Will these be independent samples because they just come from different classes?
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They are not each score is not linked in any particular way.
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River samples from 8 feet deep versus 16 feet deep.
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This also does not really seem like paired samples unless they went through
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some procedure to make sure it is the same spot in the river.
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That is probably an independent sample.
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Male heights versus female heights, they just a jumble of heights over here and a jumble of heights over here.
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They are not like match to each other.
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They are independent samples.
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Left hand span versus right hand span will in this case basically these two spans came from the same person.
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It is not a hodgepodge like left hand right hand from person 1, left hand right hand for person2 or person 3.
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I would say this is a paired sample.
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Productive vocabulary of two-year-old infant often raised by bilingual parents versus monolingual parents.
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It is a bunch of scores here and a bunch of scores here.
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They are not lined up in any way.
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I would say independent.
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Productive vocabulary of identical twins, twin 1, twin 2.
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Here we see paired samples.
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Scores on an eye gaze by autistic individual and age matched controls.
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Autistic individuals often have trouble with eye gaze and in order to know that you
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would have to match them with people who are the same age who are not autistic.
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Here we have autistic individual lined up with somebody who is their same age is not autistic.
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They are these nice even pairs and each pair has eye gaze scores.
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I would say these are paired samples.
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Hopefully that give you a better idea of some examples of paired samples.
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What about independence samples versus independent variables?
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What you will also see is IV.
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In multi sample statistics like 2, 3, 4 samples we are often trying to find some
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predictive relationship between the IV and the DV.
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The independent variable and the dependent variable.
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Usually this is often called the test or the score.
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The independent variable is seen as the predictor and the dependent variable
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is the thing that is been predicted the outcome.
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We might be interested in the IV of parent language and you might have two levels of bilingual and monolingual.
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You might be interested in how that impacts the DV of children’s vocabulary.
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Here we have these two groups, bilingual and monolingual.
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We have these scorers from children and these are independent samples because
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although we have two groups these scores are not linked to each other in any particular way.
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They are just a hodgepodge of scores here and a hodgepodge of scores here.
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On the other hand, if our IV is something like age of twin.
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We have slightly older like a couple of minutes or seconds, and younger.
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We want to know is that has an impact on vocabulary.
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We will have a bunch of scorers for older twins versus younger twins, but these scores are not just in a jumble.
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They are linked to each other because these are twins.
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They are identical.
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This is the picture you could draw and the IV tells you how you determine these groups. The paired parts tells you whether these groups scores are linked to some scores
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in the other group for some reason or another.
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Here they are linked but here they are not linked.
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In all t tests, we are calling them hypothesis testing.
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We are going to have other hypothesis tests but so far we are using t test.
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T tests always have some sort of categorical IV so that you can create different groups
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and in t-tests it is always technically two groups, two means, paired means.
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The DV is always continuous.
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The reason that the dependent variable or the scores always continuous is because you need to calculate means in order to do a t test.
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We are comparing means too and looking at standard error and you can compute mean
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and standard error for categorical variables.
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If you have a categorical variables such as you know, yes or no, you cannot quite compute a mean for that.
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Or if you have a categorical variable like red or yellow, you cannot compute a standard error for that.
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If you did have a categorical DV and a categorical IV, you would use what it is called the logistic test.
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We are actually not going to cover that.
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That does not usually get covered in descriptive and inferential statistics. T
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Usually you have to graduate level work or higher level statistics courses.
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There are two types of t test given all of this.
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Remember all t tests have this.
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These are all t tests.
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Both of these t tests are going to use categorical IV and continuous DV.
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The first kind of t test is what we have been talking about so far, independent samples t tests.
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The second type is what we are going to cover today called paired or dependent samples.
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Both of these have categorical IV and continuous DV.
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Let us have some notations for paired samples.
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Just like before, with two sample independent sample t test, for one example,
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you might call it x so that its individual members are x sub 1, x sub 2, x sub 3.
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Remember each sample is a set of numbers.
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It is not just one number but a set of numbers.
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Second sample, you might call y.
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I did not have to pick x and y though.
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I could pick other letters.
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Y could just mean another sample.
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You could have picked w or p or n.
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We usually try to reserve n, t, f, d, k for other things in statistics, but it is mostly by culture more than we have to do it by rules.
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Here is the third thing you need to know for paired samples.
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With paired samples remember x sub 1 and y sub 1 are somehow linked to each other.
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They either come from the same person or the same married couple or
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they are a set of twins or it is an autistic person and age matched control.
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All these reasons why these are linked to each other in some way.
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And because of that you can actually subtract these scores from each other and get a set of different scores.
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That is what we call d.
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D is x sub 1 – y sub 1.
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What is the difference between these two scores?
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What is the difference between these two scores and what is the difference between these two scores?
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These are paired differences.
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Let us think about this.
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If the mean of x is denoted as x bar and the mean of y is denoted as y bar, what do you think the mean of d might be?
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I guess d bar and that is what it is.
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If you got the mean of this entire set that would be d bar.
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Once you have d bar, you could imagine having a sampling distribution made of d bars.
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It is not x bars anymore, sampling distribution of the mean is the sampling distribution of the mean of a whole bunch of differences.
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That is a new idea here.
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Imagine getting a sample of d, calculating the mean d bar and placing it somewhere here.
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You will get a sampling distribution of d bars.
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That is what we are going to learn about next.
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These are means of a bunch of linked differences.
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When we go through the steps of hypothesis testing for paired samples it is going
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to be very similar to hypothesis testing for independent samples with just a few tweaks.
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First you need to stay to hypothesis and often our null hypothesis is that the two groups of scores, the two samples x and y are the same.
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Usually that is the null hypothesis.
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You put the significance level, how weird does our sample has to be for us to reject that null hypothesis.
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We set a decision stage and we draw here the SDOD d bar.
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We identify the critical limits and rejection regions and we find the critical test statistic.
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From here on out I am going to assume that you are almost never going to
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be given the actual standard deviation of the population.
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From here on out I am usually going to be using t instead of z.
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Then we use the actual sample differences and SDOD in order to compute the mean differences.
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We are not dealing with just the means, we are dealing with mean differences, test statistics, and p value.
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We compare the sample to the population and we decide whether to reject the null or not.
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Things are very similar so far.
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It is going to make us figure out what SDOD is all about.
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The rules of SDOD we are now adding on to sampling distribution of
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the differences between means that we talked about before you.
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We are going to add onto that.
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The SDOM for x and y are normal then the SDOD is normal too.
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That is the same here.
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The mean for the null hypotheses now looks like this.
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Remember the SDOD with the bar, the mean here is no longer called the μ sub x bar - y bar because it is actually x bar - y bar.
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A whole bunch of them and then you find the mean of them.
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That is called d bar.
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That is the new notation for the differences of paired samples.
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Here the μ of d bar for the null hypotheses equal 0.
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Remember for independent samples = that for μ sub x bar - y bar that = 0.
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It is very similar.
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For standard error for independent samples when various is not homogenous, which is largely the case,
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what we would use is s sub x bar - y bar.
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Instead here for paired samples, we would use s sub d bar.
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Here what we would do is take the square root of the variance of
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the standard error from x and the standard error variance of y bar and add that together.
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If you wanted to write that out more fully, that would be s sub x² the variance of x / n sub x + variance of y / n sub y.
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That is what you would do if life was easy and you have independent samples.
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That is what we know so far.
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What about for paired samples?
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For paired samples you have to think about the world differently.
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You have to think first we are getting a whole bunch of differences then we are finding the standard error of those differences.
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Here is that we are going to do.
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Here we would find standard error of those differences by looking at
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the standard deviation of the differences ÷ how many differences we have.
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This is a little crazy, but when I show you it, it will be much more easy to understand.
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I think a lot of people have trouble understanding what is difference between this and this?
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I cannot keep track all these differences.
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We have to draw SDOD.
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You have to remember it is made up of a whole bunch of d bars.
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He is made up of a whole bunch of these.
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You have to imagine pulling out samples, finding the differences,
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averaging those differences together, then plotting it here.
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Each single sample it has a standard deviation made up of differences.
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Once you plot a whole bunch of these d bars on here, this is going to have a standard deviation and that is called standard error.
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Here we have μ sub d bar and this standard error is standard error sub d bar.
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Standard deviations of d bar whereas this is just for one sample.
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This guy is for entire sampling distribution.
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Let us talk about the different formulas that go with the steps of hypothesis testing.
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Hopefully we can drive home the difference between SDOD from before and SDOD now, we will call it SDOD bar.
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For independent samples, first we had to write down a null hypothesis and alternative hypothesis.
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Often a null hypothesis was that the μ sub x bar - y bar = 0 or μ sub x bar - y bar does not equal 0 as the alternative.
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In paired samples our hypothesis looks very similar except now we are not dealing with x bar - y bars but we are dealing with difference bars.
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The average of differences.
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The mean differences.
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This is the differences of means.
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This is mean of differences.
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We will get into the other one.
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μ sub d bar does not =0.
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This so far it seems like okay.
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Here difference of means and d bar is the mean of a whole bunch of differences.
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We get a whole bunch of differences first, then we find the mean of it.
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Here we find the means first and we find the difference between the means.
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This part is actually the same.
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It is α =.05 usually two tailed.
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Step 2, we got that.
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Significant level, we get it.
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Step 3 is where we draw the SDOD here.
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Here we draw the SDOD bar.
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Thankfully you could draw it in similar ways, but conceptually they are talking about different things.
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Here how we got it was we pulled a bunch of x.
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We got the mean then we pulled a bunch of y then we got the mean and subtracted those means and plotted that here.
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We did that millions and millions of time with a whole bunch of that.
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We got the entire sampling distribution of differences of means.
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Here what we did was we pull the sample of x and y.
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We got a bunch of the differences and then we average those differences and then we plot it back.
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Here this is the sampling distribution of the mean of differences.
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Where the mean go in the order is really important.
00:26:39.100 --> 00:26:46.800
Here we get μ sub x bar - y bar, but here we get μ sub d bar.
00:26:46.800 --> 00:26:55.400
In order to find the degrees of freedom for the differences here what we did was
00:26:55.400 --> 00:27:00.500
we found the degrees of freedom for x and add it to it the degrees of freedom for y.
00:27:00.500 --> 00:27:05.900
We are going to do something else in order to find the degrees of freedom for
00:27:05.900 --> 00:27:17.300
the difference we are going to count how many differences we average together and subtract 1.
00:27:17.300 --> 00:27:24.000
This is how many n sub d – 1.
00:27:24.000 --> 00:27:30.100
Finally we need to know the standard error of the sucker.
00:27:30.100 --> 00:27:38.900
The standard error of differences here we called it s sub x bar - y bar and that
00:27:38.900 --> 00:27:50.000
was the standard error of x, the variance of x bar + the variance of y bar.
00:27:50.000 --> 00:27:56.200
The variance of these two things added together then take the square root.
00:27:56.200 --> 00:28:07.800
This refers to this distribution with the spread of this distribution.
00:28:07.800 --> 00:28:15.800
This difference here is actually going to be called s sub d bar and that is
00:28:15.800 --> 00:28:27.500
going to be standard deviation of your sample of differences ÷ √n of those differences.
00:28:27.500 --> 00:28:38.900
Last thing, I am leaving off step 5 because step 5 is explanatory.
00:28:38.900 --> 00:28:43.300
Step 4, now we have to find the sample t.
00:28:43.300 --> 00:28:51.700
Our sample is really two independent samples.
00:28:51.700 --> 00:28:54.600
We have a sample of x and a sample of y.
00:28:54.600 --> 00:29:00.700
Because of that we need to find the difference between those two means.
00:29:00.700 --> 00:29:06.900
We find the mean of this group first, the mean of this group and we subtract.
00:29:06.900 --> 00:29:15.900
We find the means first then we subtract - the μ sub x bar - y bar.
00:29:15.900 --> 00:29:21.500
I want you to contrast this with this new sample t.
00:29:21.500 --> 00:29:26.500
Here we get a bunch of x and y, we have two samples.
00:29:26.500 --> 00:29:32.700
We find the differences first then we average.
00:29:32.700 --> 00:29:36.500
Here we find the average first and find a different.
00:29:36.500 --> 00:29:42.000
Here we find the differences then we find the average.
00:29:42.000 --> 00:29:44.600
That is going to be d bar.
00:29:44.600 --> 00:29:50.600
D bar – μ sub d bar.
00:29:50.600 --> 00:29:56.300
This is getting a little bit cramped.
00:29:56.300 --> 00:30:03.600
We divide all of that by the standard error of the difference and you could substitute that in.
00:30:03.600 --> 00:30:11.100
Divide all that by the standard error of the differences.
00:30:11.100 --> 00:30:20.200
You see how here it really matters when you take the differences.
00:30:20.200 --> 00:30:24.000
Here you find the differences first and then you just deal with the differences.
00:30:24.000 --> 00:30:30.100
Here, you have to keep finding means first then you find the differences between those means.
00:30:30.100 --> 00:30:40.900
Let us talk about the confidence interval for these paired samples.
00:30:40.900 --> 00:30:48.900
The confidence intervals are going to be very similar to the confidence intervals that you saw before with independent samples.
00:30:48.900 --> 00:30:51.500
I am just covering it very briefly.
00:30:51.500 --> 00:30:54.200
Let us think about independent samples.
00:30:54.200 --> 00:31:17.500
In this case, the confidence interval was just going to be the difference of means and + or - t × the standard error.
00:31:17.500 --> 00:31:24.300
You need to put in the appropriate standard error and use the appropriate degrees of freedom as well.
00:31:24.300 --> 00:31:37.000
In confidence intervals for paired samples it is going to look very similar except instead of having the differences of means
00:31:37.000 --> 00:31:45.800
you are going to put in the mean difference d bar + or - t × the standard error.
00:31:45.800 --> 00:31:53.900
Remember standard error here is going to mean s sub x bar - y bar.
00:31:53.900 --> 00:31:57.900
The standard error here is going to be s sub d bar.
00:31:57.900 --> 00:32:08.200
In order to find degrees of freedom you have to take the degrees of freedom for x and add that to the degrees of freedom for y.
00:32:08.200 --> 00:32:15.500
In order to find degrees of freedom you have to find the degrees of freedom for d
00:32:15.500 --> 00:32:25.600
your sample of differences and that equals how many differences you have -1.
00:32:25.600 --> 00:32:33.100
Let us talk about examples.
00:32:33.100 --> 00:32:38.300
There is a download available for you and says this data set includes the highway
00:32:38.300 --> 00:32:41.900
and city gas mileage for random sample of 8 cars.
00:32:41.900 --> 00:32:44.800
Assume gas mileage is normally distributed.
00:32:44.800 --> 00:32:50.200
It says that because we could see your sample is quite small so we do not have
00:32:50.200 --> 00:32:54.600
a reason to assume that normal distribution of the SDOM.
00:32:54.600 --> 00:33:04.300
Construct and interpret the confidence interval and also conduct an appropriate t test to check your confidence interval interpretation.
00:33:04.300 --> 00:33:09.600
Here I have my example and going to example 1.
00:33:09.600 --> 00:33:23.700
Here we have 8 models of cars, their highway miles per gallon, as well as their city miles per gallon.
00:33:23.700 --> 00:33:30.300
You can see that there is a reason to consider these things as linked.
00:33:30.300 --> 00:33:33.400
They are linked because they come from the same model car.
00:33:33.400 --> 00:33:38.200
Let us construct the confidence interval.
00:33:38.200 --> 00:33:48.300
Remember in confidence interval what we are going to do is use our sample in order to predict something about our population.
00:33:48.300 --> 00:33:56.500
Here we will use our sample differences to say something about the real difference between these two populations.
00:33:56.500 --> 00:34:02.600
Here is the big step of difference when you work with paired samples.
00:34:02.600 --> 00:34:08.400
You have to first find the paired differences so the set of d.
00:34:08.400 --> 00:34:14.100
That is going to be one of these will take highway - the city.
00:34:14.100 --> 00:34:21.700
That x1 – y1, x2 – y2, x sub 3 – y sub 3.
00:34:21.700 --> 00:34:27.200
Here are all our differences and we can now find the average differences.
00:34:27.200 --> 00:34:31.400
We can find the standard deviation of these differences and all the stuff.
00:34:31.400 --> 00:34:49.700
Let us find confidence interval and this helps me to say what I need is my d bar + or - t × the standard error.
00:34:49.700 --> 00:34:57.800
In order to find my t but in order to do that I need to find my degrees of freedom.
00:34:57.800 --> 00:35:07.100
My degrees of freedom is just going to be the degrees of freedom of the d.
00:35:07.100 --> 00:35:10.400
How many differences I have -1.
00:35:10.400 --> 00:35:21.000
That is count how many differences they should have the same number of differences as cars -1 =7.
00:35:21.000 --> 00:35:26.100
Once I have that, I could find my t.
00:35:26.100 --> 00:35:30.000
I also need to find d bar.
00:35:30.000 --> 00:35:34.200
Let us find t.
00:35:34.200 --> 00:35:46.000
I need to find t and t inverse and I probably am going to assume a 95% confidence interval.
00:35:46.000 --> 00:35:57.300
My two tailed probability is .05 and my degrees of freedom is down here and so that will be 2.36.
00:35:57.300 --> 00:36:04.800
Those are my outer boundaries and let us also find d bar, the average.
00:36:04.800 --> 00:36:11.900
I almost have everything I need.
00:36:11.900 --> 00:36:14.100
I just need standard error.
00:36:14.100 --> 00:36:27.300
Standard error here is going to be s sub d ÷ the square root of how many differences I have.
00:36:27.300 --> 00:36:45.800
That is going to be the standard deviation of my differences ÷ the square root of 8 because I have 8 differences.
00:36:45.800 --> 00:36:49.600
Once I have that, then I can find the confidence interval.
00:36:49.600 --> 00:37:23.700
The upper boundary will be the d bar + t × standard error and the lower boundary is the same thing, except that this - t × standard error.
00:37:23.700 --> 00:37:29.200
My upper boundary is that 10.6.
00:37:29.200 --> 00:37:33.300
My lower boundary is that 7.6.
00:37:33.300 --> 00:37:43.800
To interpret my confidence interval I would say the real difference between highway miles per gallon
00:37:43.800 --> 00:37:54.200
and city miles per gallon I have 95% confidence that the real difference in the population is between 10.6 and 7.6.
00:37:54.200 --> 00:37:59.700
Notice that 0 is not included in here in this confidence interval.
00:37:59.700 --> 00:38:08.100
It would be 0 if highway and city miles per gallon could be equal to each other by chance.
00:38:08.100 --> 00:38:15.600
There is less than 5% chance of them being equal to each other.
00:38:15.600 --> 00:38:24.200
Because of that, I would guess that we would also reject the null because it does not include 0.
00:38:24.200 --> 00:38:31.800
Let us do hypothesis testing to see if we do really reject the null because it does not include 0
00:38:31.800 --> 00:38:34.000
I would predict that we would reject the null.
00:38:34.000 --> 00:38:37.100
Let us go straight into hypothesis testing here.
00:38:37.100 --> 00:38:39.800
First things first.
00:38:39.800 --> 00:38:52.600
The step 1, the null hypothesis this should be that the μ of d bar.
00:38:52.600 --> 00:39:04.300
Here let us do hypothesis testing.
00:39:04.300 --> 00:39:16.300
The first step is μ sub d bar is equal to 0.
00:39:16.300 --> 00:39:25.900
Highway and city gas mileage are the same but the alternative is that one of them is different from the other.
00:39:25.900 --> 00:39:29.100
That they are different from each other in some way.
00:39:29.100 --> 00:39:31.200
It is significantly stand out.
00:39:31.200 --> 00:39:33.400
This difference stands out.
00:39:33.400 --> 00:39:37.800
That would be that μ sub d bar does not equal 0.
00:39:37.800 --> 00:39:52.100
Step 2, my significance level, the false alarm rate is very low .05 and two tailed.
00:39:52.100 --> 00:39:56.700
Let us set our decision stage.
00:39:56.700 --> 00:40:17.700
I need to draw an SDOD bar and here I put my μ as 0 because the μ sub d bar will be 0.
00:40:17.700 --> 00:40:21.600
Let us also find the standard error here.
00:40:21.600 --> 00:40:34.600
The standard error here is going to be s sub d bar and that is really the standard deviation of the d / √n sub d.
00:40:34.600 --> 00:40:39.300
That I could compute here.
00:40:39.300 --> 00:40:49.500
Actually, we already computed that because we have the standard deviation of the d bars / the square root of how many d I have.
00:40:49.500 --> 00:40:55.600
That is .64.
00:40:55.600 --> 00:40:58.600
What is my degrees of freedom?
00:40:58.600 --> 00:41:06.500
That is 7 because that is how many differences I have -1.
00:41:06.500 --> 00:41:15.800
Based on that I can find my t and my t is going to be + or - 2.36.
00:41:15.800 --> 00:41:23.100
Let us deal with our sample.
00:41:23.100 --> 00:41:41.900
When we talk about the sample t, what we really mean is what the x bar of our sample differences that would be d bar.
00:41:41.900 --> 00:41:45.000
I would just put x bar sub d because it is a simpler way of doing it.
00:41:45.000 --> 00:41:52.500
- the μ which is 0 / the standard error which is .64.
00:41:52.500 --> 00:42:07.200
I could just put this here so I can skip directly to step 4 and I will compute my sample t.
00:42:07.200 --> 00:42:12.700
I should say this is my critical t so that I do not get confused.
00:42:12.700 --> 00:42:26.600
My sample t is going to be d bar - μ / standard error.
00:42:26.600 --> 00:42:44.000
That is d bar - μ which is 0 ÷ standard error = 14.3.
00:42:44.000 --> 00:42:52.800
I can also find the p value and I'm guessing my p value is probably be tiny.
00:42:52.800 --> 00:42:57.200
Here 14.3 is really small.
00:42:57.200 --> 00:43:05.700
My p value is going to be t dist because I want my probability.
00:43:05.700 --> 00:43:12.800
I put in my t, my degrees of freedom which is 7, and I have a two-tailed hypotheses.
00:43:12.800 --> 00:43:30.600
That is going to be 2 × 10⁻⁶.
00:43:30.600 --> 00:43:49.700
Imagine .000002 given this tiny p value much smaller than .05 we should say at step 5 reject the null.
00:43:49.700 --> 00:43:56.600
We had predicted that we would reject the null because the CI, the confidence interval did include 0.
00:43:56.600 --> 00:44:01.200
Good job confidence interval and hypothesis testing working together.
00:44:01.200 --> 00:44:10.300
Example 2, see the download again, this data set shows the average salary earned by first-year college graduates.
00:44:10.300 --> 00:44:21.200
Graduated at the bottom or top 15% of their class for random sample of 10 colleges ranked in the top 100 public colleges in the US.
00:44:21.200 --> 00:44:27.600
Is there a significant difference in earnings that is unlikely to have occurred by chance alone?
00:44:27.600 --> 00:44:34.400
We want to know is there a difference between these top 15% folks and the bottom 15% folks.
00:44:34.400 --> 00:44:38.100
They are linked to having graduated from the same college.
00:44:38.100 --> 00:44:46.800
We would not necessarily want to compare people from the top 15% of one college that might be really good to one
00:44:46.800 --> 00:44:53.000
to the bottom percentage of people from a college that might be not as great.
00:44:53.000 --> 00:44:59.400
We would really want from the same college does not matter if you are in the top 15 or bottom 15%.
00:44:59.400 --> 00:45:18.400
If you go to example 2, you will see these randomly selected colleges and the earnings in dollars per year, salary per year for the bottom 15%, as well as the top 15%.
00:45:18.400 --> 00:45:24.300
Because it is a paired sample what we want to do is start off with d or set up d.
00:45:24.300 --> 00:45:29.200
What is the difference between bottom and top?
00:45:29.200 --> 00:45:38.000
We are going to get probably a whole bunch of negative numbers assuming that top earners earn more than bottom.
00:45:38.000 --> 00:45:40.400
Indeed we do, we have a bunch of negative numbers.
00:45:40.400 --> 00:45:45.500
If you wanted to turn these negatives into positives, you just have to remember
00:45:45.500 --> 00:45:50.600
which one you decided as x and which one you decided to be y.
00:45:50.600 --> 00:45:58.800
I will call this one x and I will call this one y.
00:45:58.800 --> 00:46:03.700
It will help me remember which one I subtracted from which.
00:46:03.700 --> 00:46:11.000
I am going to reverse all of these and it is just going to give me the positive versions of this.
00:46:11.000 --> 00:46:13.000
Here is my d.
00:46:13.000 --> 00:46:17.400
Let us go ahead and start with hypothesis testing.
00:46:17.400 --> 00:46:22.700
This part I will do by hand.
00:46:22.700 --> 00:46:36.300
Step 1, the null hypothesis says something that the top 15% folks and the bottom 15% folks are the same.
00:46:36.300 --> 00:46:39.900
Their difference is going to be 0.
00:46:39.900 --> 00:46:47.400
The μ sub d bar should be 0 but the alternative is that they are different.
00:46:47.400 --> 00:46:50.800
We are neutral as to how they are different.
00:46:50.800 --> 00:46:53.500
We do not know whether one earns more than the other.
00:46:53.500 --> 00:46:57.700
Whether they are top earns more than bottom or bottom earns more than the top.
00:46:57.700 --> 00:47:06.700
We can use our common sense to predict that the top ranking folks might earn more, but right now we are neutral.
00:47:06.700 --> 00:47:14.400
Step 2, is our α level .05 or significance level.
00:47:14.400 --> 00:47:16.800
Let us say two details.
00:47:16.800 --> 00:47:30.500
Step 3, drawing the SDOD, the mean differences and here we will put 0.
00:47:30.500 --> 00:47:37.300
And let us figure out the standard error.
00:47:37.300 --> 00:47:50.700
The standard error here would be s sub d bar and that would be the standard deviation of d / √(n ) sub d.
00:47:50.700 --> 00:48:01.500
We also want to figure out the degrees of freedom so that is going to be n sub b -1 and we also want to find out the t.
00:48:01.500 --> 00:48:04.200
These are all things you can do in Excel.
00:48:04.200 --> 00:48:23.100
Step 3, standard error is going to be s sub d bar and that will be s sub d ÷ √n sub d.
00:48:23.100 --> 00:48:43.500
That will be the standard deviation of our sample of d ÷ the square root of how many there are and there is 10.
00:48:43.500 --> 00:48:46.400
Here is our standard error.
00:48:46.400 --> 00:48:50.600
What is our degrees of freedom?
00:48:50.600 --> 00:48:54.800
That is going to be 10-1 =9.
00:48:54.800 --> 00:48:59.200
What is our critical t?
00:48:59.200 --> 00:49:04.500
We know it is a critical t because we are still in step 3 the decision stage.
00:49:04.500 --> 00:49:06.900
We are just setting up our boundaries.
00:49:06.900 --> 00:49:13.100
That is going to be t inverse because we already know the probability .05 two-tailed,
00:49:13.100 --> 00:49:18.800
degrees of freedom being 9 and we get 2.26.
00:49:18.800 --> 00:49:26.100
It is + or -2.26 those are our boundaries of t.
00:49:26.100 --> 00:49:33.400
Step 4, this will say what is our sample t?
00:49:33.400 --> 00:49:40.900
And that is going to be our d bar – μ / standard error/
00:49:40.900 --> 00:49:54.500
I will write step 4 here and so I need to find t which is d bar – μ/ standard error.
00:49:54.500 --> 00:50:00.200
I need to find the bar for sure and standard error.
00:50:00.200 --> 00:50:15.700
My d bar is the average of all my differences and that is about $12,000 - $13,000 a year.
00:50:15.700 --> 00:50:18.100
That is just right after college.
00:50:18.100 --> 00:50:32.900
I need to find the d bar - 0 ÷ the standard error to give me my sample t.
00:50:32.900 --> 00:50:40.700
That is the difference between sample t and critical t.
00:50:40.700 --> 00:50:56.500
8.05 is actually the average of the differences.
00:50:56.500 --> 00:51:05.100
The top 15% are on average earning $13,000 more than the bottom 15%.
00:51:05.100 --> 00:51:15.300
The sample t gives us how far that differences from 0 in terms of standard error.
00:51:15.300 --> 00:51:20.300
We know that is way more extreme than 2.26.
00:51:20.300 --> 00:51:22.800
Let us find the p value.
00:51:22.800 --> 00:51:27.400
We put it in t dist because we want to know the probability.
00:51:27.400 --> 00:51:34.500
Put in our t, degrees of freedom, and we have a two-tailed hypotheses.
00:51:34.500 --> 00:51:40.600
That would be 2 × 10⁻⁵.
00:51:40.600 --> 00:51:53.400
Our p value = 2 × 10⁻⁵ which is a very tiny, tiny number, much smaller than the α.
00:51:53.400 --> 00:51:58.000
We would reject the null hypotheses.
00:51:58.000 --> 00:52:04.800
Is there a significant difference in earnings that is unlikely to have occurred by chance alone?
00:52:04.800 --> 00:52:11.000
There is always going to be a difference in earnings between these two groups of people, the top 15 and the bottom 15%.
00:52:11.000 --> 00:52:14.700
Is this difference greater than would be expected by chance?
00:52:14.700 --> 00:52:20.900
Yes it is because we are rejecting the model that they are equal to each other.
00:52:20.900 --> 00:52:30.800
Example 3, in fitting hearing aids to individuals, researchers wanted to examine whether
00:52:30.800 --> 00:52:35.800
there was a difference between hearing words in silence or in the presence of background noise.
00:52:35.800 --> 00:52:40.500
Two equally difficult wordless are randomly presented to each person.
00:52:40.500 --> 00:52:45.800
One less than silence and the other with white noise in a random order for each person.
00:52:45.800 --> 00:52:51.500
This means that some people get silence than noise, other people get noise and silence.
00:52:51.500 --> 00:52:57.900
Are the hearing aid equally effective in silence or with background noise?
00:52:57.900 --> 00:53:05.600
First conduct the t test assuming that these are independent samples then conduct the t test assuming that these are paired samples.
00:53:05.600 --> 00:53:07.700
Which is more powerful?
00:53:07.700 --> 00:53:12.300
The independent sample t-test or paired samples t test?
00:53:12.300 --> 00:53:16.200
We need to figure out what it means by more powerful.
00:53:16.200 --> 00:53:25.300
I need some scratch paper here because the problem was so long I am just going to divide the space in half.
00:53:25.300 --> 00:53:30.800
This top part I am going to use for assuming independent samples.
00:53:30.800 --> 00:53:43.800
They are not actually independent samples, but I want you to see the difference between doing them as independent sample and doing them as paired samples doing this hypothesis testing as paired samples.
00:53:43.800 --> 00:53:50.800
Step 1, the hypothesis, the null hypothesis is that if I get these sample
00:53:50.800 --> 00:53:59.900
and they are independent this difference of means on average is going to be 0.
00:53:59.900 --> 00:54:04.000
The μ sub x bar - y bar is going to = 0.
00:54:04.000 --> 00:54:12.300
The alternative hypothesis is that the μ sub x bar - y bar does not equal to 0.
00:54:12.300 --> 00:54:20.700
Here I am going to put α =.05, two-tailed.
00:54:20.700 --> 00:54:27.800
I am going to draw myself an SDOD.
00:54:27.800 --> 00:54:33.500
Just to let you know it is the differences of means.
00:54:33.500 --> 00:54:43.200
Here we know that this is going to be 0 and we probably should find out the standard error.
00:54:43.200 --> 00:55:02.800
The standard error of this difference of means is going to be the square roots of the variance of x bar + the variance of y bar.
00:55:02.800 --> 00:55:14.900
I am going to write this out to be s sub x²/ n sub x + s sub y² /n sub y.
00:55:14.900 --> 00:55:21.100
The variance of x and x bar, the variance of x /n, the variance of y/n.
00:55:21.100 --> 00:55:39.900
We will probably need to find the degrees of freedom and that is going to be n sub x – 1 + n sub y -1.
00:55:39.900 --> 00:55:45.600
Finally we will probably need to know the critical t but I will put that up here.
00:55:45.600 --> 00:55:52.900
Let us look at this data, go to example 3.
00:55:52.900 --> 00:55:56.400
Click on example 3 and take a look at this data.
00:55:56.400 --> 00:55:59.300
Let us assume independent samples.
00:55:59.300 --> 00:56:08.300
Here we are going to assume that this silence is just one group of scores and
00:56:08.300 --> 00:56:12.300
this background noise is another group of scores and they are not paired.
00:56:12.300 --> 00:56:14.300
They are actually paired.
00:56:14.300 --> 00:56:20.000
This belongs to subject one, these 2 belongs to subject 3, this belongs to subject 5.
00:56:20.000 --> 00:56:24.100
Here is the list order, it is A, B.
00:56:24.100 --> 00:56:28.100
We get A list first then list B and here is the noise order.
00:56:28.100 --> 00:56:30.300
They get it silent first then noisy.
00:56:30.300 --> 00:56:32.700
This guy gets noisy first then silent.
00:56:32.700 --> 00:56:46.200
All these orders are randomly assigned and the noise orders are randomly assigned as well.
00:56:46.200 --> 00:56:52.300
For this exercise, we are going to assume we do not have any of this stuff.
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We are going to assume this is gone and that this just a bunch of scores from one group of subjects
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that listen to a list of words in silence and another group of subjects that listen to list of words in background noise.
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We do the independent samples t test and we start with step 3.
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We know we need to find the standard error, which is going to be the square root of the variance of x ÷ n(x) + the variance of y / n sub y.
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All that added together and a square root.
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We need to find the variance of x.
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We need to find n sub x.
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We also need to find the variance of y and n sub y before we can find standard error.
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Variance is pretty easy.
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We will just call silence x and the count of this is 24.
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The count for y is going to be the same, but what is the variance of y?
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The variance of y slightly different.
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In order to find this guy, the standard error, we are going to put in square root of the variance of x ÷ 24 + the variance of y ÷ 24.
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We get a standard error of 2.26 and standard error gives it just in terms of number of words accurately heard.
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We also need to find the degrees of freedom.
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In order to find degrees of freedom, we need the degrees of freedom for x + degrees of freedom for y.
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The degrees of freedom for x is just going to be 24 - 1 and the degrees of freedom for y is also going to be 24 – 1.
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The new degrees of freedom is 23 + 23 = 46.
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Once we have that we can find our critical t.
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Our critical t, we know that α is .05 so we are going to put in t in and
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put in our two-tailed probability and the degrees of freedom 46.
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We get a critical t of + or -2.01.
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Our critical t is + or -2.01.
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I will just leave that stuff on the Excel file.
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Given all this now let us deal with the sample.
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When we find the sample t what we are doing is finding the difference in means and then find the difference
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between that difference and our expected difference 0 and divide all of that by standard error to find how many standard errors away we are.
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Here I will put step 4, sample t.
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In order to find sample t we need to find x bar - y bar - μ and all of that ÷ standard error.
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Thankfully, we have a bunch of those things available to us quite easily.
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We have x bar, we can get y bar, we can get standard error.
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Let us find x bar, the average number of words heard accurately in silence and that is about 33 words.
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The average number of words heard correctly with background noise, and that is 29 words.
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Is the difference of about 4 words big enough to be statistically different?
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We would take this - this and we know μ = 0 so I am going to ignore that / standard error found up here.
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That would give us 1.75.
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1.75 that is more extreme than + or -2.01.
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1.75 we will actually say do not reject.
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We should find the p value too.
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This p value should be greater than .05.
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We will put in t dist then our sample t, degrees of freedom which is 46 and we want a two tailed and we get .09.
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.09 is greater than .05.
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Step 5, fail to reject.
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Now that we have all that, we want to know is it more sensitive?
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Can we detect the statistical difference better if we used paired examples?
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Let us start.
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Here we would say p =.09 and 5 is failed to reject.
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It is not outside of our rejection zone, it is inside our fail to reject zone.
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Let us talk about the null hypotheses here.
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What we are going to do is find the differences first then the mean of those differences.
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We are saying if they are indeed not that different from each other that mean different should be 0.
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The alternative is that the mean difference is not equal to 0.
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Once again α = .05 two tailed and now we will draw our SDOD bar which means
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it is a standard sampling distribution of means mean made of differences.
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Here we want to put 0.
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We probably also want to figure out standard error somewhere along the line,
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which is going to be s sub d bar which is s sub d ÷ √n sub d.
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We probably also want to find the degrees of freedom, which is going to be n sub d -1.
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We probably also want to find the critical t.
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Let us find out that.
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Here I will start my paired samples section.
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I will also start with step 3.
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Let me move all of these over here.
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Let us start here with step 3
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Let us find standard error and that is going to be s sub d not d bar ÷ √n sub d.
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We can find s sub d very easily and we could also find n sub d.
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First we need to create a column of d.
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I will find the standard deviation of the d but I realized that I do not have any d.
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The d look something like this silence - background noise.
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This is how many more words, they are able to hear accurately in silence and background noise.
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Here we see that some people hear a lot of words better in silence.
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Some people here words better with a little bit of background noise.
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Some people are exactly the same.
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We could find a standard deviation of all these differences.
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We could also find the mean of them
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The n of them will be the same as 24 because there are 24 people that came from.
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There is 24 differences.
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We could find out standard error.
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Standard deviation of d ÷ √24.
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That is standard error, notice that is quite different from finding a standard error of independent samples.
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Let us find degrees of freedom for d and that is going to be n sub d -1 and that is 24 -1.
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Our critical t should be t inverse .05 two tailed=23 and we get 2.07.
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So far it seems that our standard for how extreme it has to be is more far out.
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That makes sense because the degrees of freedom is smaller than 46.
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+ or -2.07.
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Let us talk about our sample.
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In order to find our sample t, we want to find the average of difference subtract from the hypothesized μ
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and divide all of that by standard error to find out how many standard errors away our sample mean difference is.
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We also want to find p value.
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Here is step 4, our sample t would be d bar - μ ÷ standard error.
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What is d bar and how would we find it?
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Just use the average function and the average of our d like this d bar.
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We can do d bar -0 / standard error= 2.97.
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That is more extreme than 2.06.
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Let us figure out why.
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We might look at standard error, the standard error is much smaller and the steps are smaller.
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How many steps we need to take to get all the way out to this d bar?
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There is more of them than these the bigger steps.
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These are almost twice as big.
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These bigger steps, there is few of them that you need.
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That is what the sample t I get is how many of these standard errors,
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how many of these steps does it take to get all the way out to d bar or x bar – y bar?
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We need almost 3 steps out.
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What is our p value?
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Our p value should be less than .05 that is going to be t dist.
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Here is our t value I will put in our degrees of freedom and two tailed and its .007.
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That certainly less than .05.
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Step 5, here we reject whereas here we fail to reject.
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Since there is this difference and we detected it with this one but not with this one,
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we would say that this is the more sensitive test given that there is something to detect out there.
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This is the difference if it does exist.
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This one is a little coarser, there is a couple of reasons for that.
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One of the reasons is because the standard error are usually larger than the standard error of differences.
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Another issue is that x bar - y bar, the difference here if we look at x bar - y bar this difference is roughly around the same.
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This difference is the same as this difference.
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It is not that bad but it is that you are dividing by a smaller standard error here then you are here.
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Here, the standard error is quite large.
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The steps are quite large.
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Here, the standard errors are small.
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The steps are quite small.
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It is because you are taking out some of the variation caused by having some people
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just being able to hear a lot of words accurately all the time with noise.
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Some people are very good at hearing anyway.
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They might have over a low number of scores but with d bar you do not care about those individual differences.
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You end up accounting for those by subtracting them out.
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Here this is a more sensitive test.
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Here we get p=.006 and we reject.
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Which test is more sensitive?
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Which test is able to detect the difference, if there is a difference?
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Paired samples.
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That principles are little more complicated to collect that data but it is worth it because it is a more sensitive test.
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