12 subjects with borderline high cholesterol levels were...

12 subjects with borderline-high cholesterol levels were recruited for a study that involved taking a nutrition education class. Cholesterol readings were taken before the class and months after the class. Subject 1 2 3 4 5 6 7 8 9 10 11 12 Pre class 285 279 250 235 255 290 310 260 275 240 235 230 Post class 265 266 245 240 230 255 235 250 250 215 245 235 Can you conclude that a significant amount of improvement took place? Use α = 0.005 = 0.5% Find the 95% confidence interval for μd (the average of all difference).

Transcript

00:01 All right, so here we have some data from 12 subjects who had borderline high cholesterol, and they were selected to take a class on nutrition, and their cholesterol levels were taken pre -class and post -class, and that's what these columns are.

00:17 And what we want to know is whether or not we can conclude that a significant amount of improvement took place.

00:25 So we're basically looking at is the difference, so the null hypothesis is that the difference, zero or even or even above zero, right? it would be above zero.

00:38 But what we're looking for is that was there improvement? meaning cholesterol went down.

00:44 This difference negative.

00:47 So what we're gonna do then is use, actually a notation, let me fix notation for us.

01:01 The notation that used is the use of d.

01:06 Alright, there we go.

01:07 So we're gonna do a haired samples t test and we have the alpha of 0 .005 for our level of significance and we're going to reject the null hypothesis if the alpha we find or assuming if the p value we find is less than the alpha so that the formula for this is going to be the average the sample the the average difference x bar sub d divided by the sample standard deviation of the difference divided by the square of the sample size.

01:46 So here the sample size is 12.

01:50 So i got part of that already.

01:53 Now the, so i use the spreadsheet to help me out here.

01:55 So i did post minus pre to find the difference to see, because that means if it's negative, it means it would have gone down.

02:03 So for this person, post cross was 266, pre -class is 225, so it's negative 20.

02:08 So you do that, on down the line, copy the formula.

02:11 Take the average or the mean of the data so the mean is negative 17 .75 with the sample standard deviation from this and i just used a spreadsheet function to generate that for us is 22 .94 and just you know i did all the calculations with the full precision that my spreadsheet gave us so if you round like i did hear your your answer might be slightly off but not by much and this gives us a t value of negative 2 .68 and the p value associated with that.

02:48 Again, you can use a table or a spreadsheet to look it up.

02:51 It is 0 .01, which is not less than its alpha -lux.

02:56 That means we fail to reject.

03:05 And that means that, and this is a pretty high level of significance too, 0 .005, that's a pretty high, or i guess low, a small alpha, i guess.

03:15 So in that sense, it's gonna be hard to reject because you have to find a really extreme case.

03:21 So we failed to reject the alpha 0 .005.

03:26 Now the second half of this question is to make a confidence interval and 95 % confidence interval for the difference of the means.

03:35 So the way we do that is x bar and sub d plus minus our t value, such that we have the alpha of the 2 and its degrees of freedom.

03:49 And oh, i didn't say this.

03:51 You need the degrees of freedom to get the p value for this, because for a t distribution, you have to note the degrees of freedom.

03:57 And for a paired samples, t test, it's n minus one.

04:01 So degrees of freedom.

04:01 In our case, is 12 minus 1, so it's 11...

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