WEBVTT 1 00:00:00.570 --> 00:00:01.860 Hello students, 2 00:00:01.860 --> 00:00:05.553 welcome to Biostatistics ER, Example 2 Chapter 7. 3 00:00:06.990 --> 00:00:08.040 In this example, 4 00:00:08.040 --> 00:00:13.040 we will learn how to perform hypothesis testing 5 00:00:13.410 --> 00:00:17.340 for two sample matched continuous outcome data. 6 00:00:17.340 --> 00:00:19.983 The problem is from our textbook, Problem 18. 7 00:00:21.270 --> 00:00:22.507 The problem states, 8 00:00:22.507 --> 00:00:24.540 "Suppose the results of the analysis 9 00:00:24.540 --> 00:00:27.000 in problem 15 through Problem 17 10 00:00:27.000 --> 00:00:30.930 are reported and criticized because the participants 11 00:00:30.930 --> 00:00:33.600 were not randomized to different diets, 12 00:00:33.600 --> 00:00:36.630 and that there may be other factors 13 00:00:36.630 --> 00:00:39.423 associated with changes in cholesterol. 14 00:00:40.710 --> 00:00:43.290 A third study is run to estimate the effect 15 00:00:43.290 --> 00:00:47.160 of the low-carbohydrate diet on cholesterol levels. 16 00:00:47.160 --> 00:00:48.210 In the third study, 17 00:00:48.210 --> 00:00:50.640 participants' cholesterol levels 18 00:00:50.640 --> 00:00:53.190 are measured before starting the program, 19 00:00:53.190 --> 00:00:56.790 and then again after 6 months on the program. 20 00:00:56.790 --> 00:00:58.473 The data are shown here. 21 00:00:59.550 --> 00:01:02.340 Is there a significant increase in cholesterol 22 00:01:02.340 --> 00:01:05.790 after 6 months on the low-carbohydrate diet? 23 00:01:05.790 --> 00:01:10.790 Run the appropriate test at a 5% level of significance." 24 00:01:11.190 --> 00:01:13.650 So here, the samples are matched 25 00:01:13.650 --> 00:01:16.770 because the cholesterol levels of the same participants 26 00:01:16.770 --> 00:01:21.603 were measured before and after 6 months on the program. 27 00:01:23.640 --> 00:01:26.070 As we know, cholesterol level is also measured 28 00:01:26.070 --> 00:01:27.513 on a continuous scale, 29 00:01:28.590 --> 00:01:30.810 and here we will use t-test 30 00:01:30.810 --> 00:01:34.200 because our sample size is less than 30. 31 00:01:34.200 --> 00:01:38.493 With that, we will move to Step 1, 32 00:01:40.260 --> 00:01:42.540 which is set up hypothesis 33 00:01:42.540 --> 00:01:45.480 and determine level of significance. 34 00:01:45.480 --> 00:01:49.230 The null hypothesis here is mu d equal to 0 35 00:01:49.230 --> 00:01:54.230 and the alternative hypothesis is mu d greater than 0, 36 00:01:54.240 --> 00:01:57.150 and the level of significance is 0.05 37 00:01:57.150 --> 00:01:58.563 as stated in the problem. 38 00:02:00.030 --> 00:02:01.470 Here, I have provided you 39 00:02:01.470 --> 00:02:03.390 with the appropriate test statistics, 40 00:02:03.390 --> 00:02:04.713 which is a t-test. 41 00:02:07.050 --> 00:02:09.520 In Step 3, we will set up the decision rule 42 00:02:10.680 --> 00:02:12.510 and we will reject. 43 00:02:12.510 --> 00:02:17.510 Now if our t calculated is greater than or equal to 2.015, 44 00:02:19.230 --> 00:02:24.230 and this value was from the t-table 45 00:02:26.550 --> 00:02:31.080 based on degrees of freedom being 5. 46 00:02:31.080 --> 00:02:34.680 Now we are going to calculate the test statistics. 47 00:02:34.680 --> 00:02:37.050 So here, I strongly encourage you 48 00:02:37.050 --> 00:02:39.240 to either use Excel or SPSS 49 00:02:39.240 --> 00:02:40.890 to calculate the average 50 00:02:40.890 --> 00:02:42.570 and the standard deviation, 51 00:02:42.570 --> 00:02:45.810 and I have used Excel to calculate the average 52 00:02:45.810 --> 00:02:47.520 and the standard deviation. 53 00:02:47.520 --> 00:02:49.620 So once those values are calculated, 54 00:02:49.620 --> 00:02:52.200 we have to insert them into our formula. 55 00:02:52.200 --> 00:02:54.690 and the numerator here is going to be 5, 56 00:02:54.690 --> 00:02:55.740 which is the average, 57 00:02:55.740 --> 00:02:59.040 and the denominator is going to be the standard deviation, 58 00:02:59.040 --> 00:03:02.880 11.8 divided by the square root of sample size, 59 00:03:02.880 --> 00:03:04.170 which is 6. 60 00:03:04.170 --> 00:03:06.573 And once we perform the algebra, 61 00:03:08.310 --> 00:03:10.743 our t calculated is 1.04. 62 00:03:12.090 --> 00:03:13.800 So now in Step 5, 63 00:03:13.800 --> 00:03:17.520 and we are going to reach our conclusion. 64 00:03:17.520 --> 00:03:19.800 And here we will fail to reject null 65 00:03:19.800 --> 00:03:24.750 because 1.04 is less than 2.015. 66 00:03:24.750 --> 00:03:28.170 So we do not have statistically significant evidence 67 00:03:28.170 --> 00:03:31.050 at the alpha of 0.05 to show 68 00:03:31.050 --> 00:03:34.650 that there is a significant difference in cholesterol 69 00:03:34.650 --> 00:03:36.723 after 6 months on the diet. 70 00:03:43.740 --> 00:03:46.350 Thank you for your time and attention, 71 00:03:46.350 --> 00:03:48.273 and I'll see you in the next video.