WEBVTT 1 00:00:01.410 --> 00:00:02.580 Hello students. 2 00:00:02.580 --> 00:00:05.763 Now we are going to solve problem one. 3 00:00:07.560 --> 00:00:11.070 So this problem is asking us, which measure, 4 00:00:11.070 --> 00:00:15.450 the mean or median is a better measure of a typical value 5 00:00:15.450 --> 00:00:17.550 and we need to justify it. 6 00:00:17.550 --> 00:00:18.930 So before I get started, 7 00:00:18.930 --> 00:00:20.850 I'm going to read the problem again. 8 00:00:20.850 --> 00:00:22.740 So the problem here states: 9 00:00:22.740 --> 00:00:26.610 A study is run to estimate the mean total cholesterol level 10 00:00:26.610 --> 00:00:29.460 in children 2 to 6 years of age. 11 00:00:29.460 --> 00:00:31.890 A sample of nine participants is selected 12 00:00:31.890 --> 00:00:35.193 and their total cholesterol levels are measured as follows: 13 00:00:36.952 --> 00:00:40.869 185, 225, 240, 196, 175, 180, 194, 147 and 223. 14 00:00:46.680 --> 00:00:49.140 Now to answer the question, 15 00:00:49.140 --> 00:00:52.770 we need to determine if we have any outliers 16 00:00:52.770 --> 00:00:56.670 and to identify if there are any outliers, 17 00:00:56.670 --> 00:00:59.820 we first need to calculate the Tukey Fences 18 00:00:59.820 --> 00:01:01.820 and basically the upper and lower limits 19 00:01:03.360 --> 00:01:04.410 of the Tukey Fences. 20 00:01:04.410 --> 00:01:06.540 So this is what we have to do here. 21 00:01:06.540 --> 00:01:09.423 We have to calculate the Tukey Fences, 22 00:01:11.190 --> 00:01:15.303 upper and lower limits. 23 00:01:16.380 --> 00:01:21.380 So there is a formula to calculate these limits and that is 24 00:01:24.420 --> 00:01:26.313 lower limit equal to Q. 25 00:01:42.780 --> 00:01:47.780 So it's Q1 minus 1.5, multiplied by Q3 minus Q1. 26 00:01:49.260 --> 00:01:53.673 Now Q3 minus Q1 has a name that is called IQR. 27 00:01:54.570 --> 00:01:59.040 And then for the upper limit, 28 00:01:59.040 --> 00:02:02.770 basically the formula is pretty much the same except 29 00:02:04.110 --> 00:02:05.980 we will replace 30 00:02:07.757 --> 00:02:11.130 Q1 with Q3 31 00:02:11.130 --> 00:02:13.080 and instead of a subtraction 32 00:02:13.080 --> 00:02:15.033 we are going to have an addition here. 33 00:02:17.100 --> 00:02:19.053 And I'm just gonna call this IQR. 34 00:02:20.340 --> 00:02:22.953 So now let's calculate the lower limit. 35 00:02:28.530 --> 00:02:29.640 So what is our Q1? 36 00:02:29.640 --> 00:02:31.440 We already calculated it before, 37 00:02:31.440 --> 00:02:36.140 so it is 1-8-0 or 180, minus 1.5. 38 00:02:39.030 --> 00:02:44.030 Our Q3 is 223, and our Q1 is 180. 39 00:02:45.570 --> 00:02:48.420 Again, we have already calculated these things. 40 00:02:48.420 --> 00:02:52.080 So what is our IQR? 41 00:02:52.080 --> 00:02:56.280 When we do the subtraction, our IQR is 43 42 00:02:56.280 --> 00:02:58.170 and when we do the multiplication, 43 00:02:58.170 --> 00:03:01.890 what we get is 180 minus 64.5. 44 00:03:04.950 --> 00:03:08.180 So that will be 115.5. 45 00:03:10.140 --> 00:03:13.563 Now for the upper limit, 46 00:03:14.880 --> 00:03:16.590 it's basically the same, 47 00:03:16.590 --> 00:03:20.640 except some of these values will be replaced. 48 00:03:20.640 --> 00:03:25.530 So our Q3 as we already know is 223. 49 00:03:25.530 --> 00:03:30.390 And we already know the second part of the calculation 50 00:03:30.390 --> 00:03:31.950 because we just performed it. 51 00:03:31.950 --> 00:03:33.810 So no reason to do it again. 52 00:03:33.810 --> 00:03:38.430 So it's going to be 223 plus 64.5. 53 00:03:38.430 --> 00:03:39.843 So when we add them up, 54 00:03:41.100 --> 00:03:42.120 what do we get? 55 00:03:42.120 --> 00:03:44.570 We get 287.5. 56 00:03:46.080 --> 00:03:48.420 Now as we look at our data set, 57 00:03:48.420 --> 00:03:49.830 what do we see? 58 00:03:49.830 --> 00:03:51.840 We do not see any value 59 00:03:51.840 --> 00:03:55.080 that is less than 115.5, 60 00:03:55.080 --> 00:04:00.080 or any value that is above 287.5. 61 00:04:00.240 --> 00:04:04.863 Therefore, we do not have any outliers. 62 00:04:07.680 --> 00:04:10.233 So because we do not have any outliers here, 63 00:04:15.450 --> 00:04:18.470 mean will be 64 00:04:21.240 --> 00:04:24.480 the better measure of a typical value. 65 00:04:24.480 --> 00:04:26.880 Now what will happen if we have an outlier? 66 00:04:26.880 --> 00:04:29.313 Well, we will discuss that in a later problem.