WEBVTT 1 00:00:03.360 --> 00:00:05.610 Hello students and welcome to 2 00:00:06.720 --> 00:00:08.400 Biostat ER, example seven. 3 00:00:08.400 --> 00:00:11.010 In this example, we are going to learn how to calculate 4 00:00:11.010 --> 00:00:15.060 the 95% confidence interval for relative risk. 5 00:00:15.060 --> 00:00:17.820 We are using the same problem we used before. 6 00:00:17.820 --> 00:00:21.150 Nevertheless, I will read the problem to you. 7 00:00:21.150 --> 00:00:23.707 We used this problem for example six. 8 00:00:23.707 --> 00:00:25.237 "A clinical trial is conducted 9 00:00:25.237 --> 00:00:28.087 "to compare an experimental medication to placebo 10 00:00:28.087 --> 00:00:30.607 "to reduce the symptoms of asthma. 11 00:00:30.607 --> 00:00:32.887 "200 participants are enrolled to the study 12 00:00:32.887 --> 00:00:34.207 "and randomized to receive 13 00:00:34.207 --> 00:00:37.053 "either the experimental medication or the placebo. 14 00:00:38.017 --> 00:00:41.587 "The primary outcome is self-reported reduction of symptoms. 15 00:00:41.587 --> 00:00:43.417 "Among 100 participants who received 16 00:00:43.417 --> 00:00:46.537 "the experimental medication, 38 report reduction 17 00:00:46.537 --> 00:00:50.527 "in symptoms as compared to 21 participants of 100 18 00:00:50.527 --> 00:00:51.747 "assigned to placebo." 19 00:00:54.120 --> 00:00:56.370 First here, I have presented the information 20 00:00:56.370 --> 00:00:58.563 in a tabular format. 21 00:00:59.400 --> 00:01:03.660 And before we get started, in terms of solving the problem, 22 00:01:03.660 --> 00:01:06.660 I want to draw your attention to our textbook Chapter Six, 23 00:01:06.660 --> 00:01:11.660 page 114 and here it states clearly why we need to use 24 00:01:12.720 --> 00:01:15.930 the natural log and subsequently the anti-log 25 00:01:15.930 --> 00:01:17.823 to determine the relative risk. 26 00:01:19.620 --> 00:01:21.750 I have copied and pasted the information here 27 00:01:21.750 --> 00:01:25.770 from our textbook and I am not reading it 28 00:01:25.770 --> 00:01:26.940 in the interest of time, 29 00:01:26.940 --> 00:01:29.160 but I'm hoping that you will read it. 30 00:01:29.160 --> 00:01:32.490 Also, I will like to mention that for this 31 00:01:32.490 --> 00:01:35.350 Biostats ER example, the solution is typed 32 00:01:37.170 --> 00:01:38.910 because as I mentioned previously, 33 00:01:38.910 --> 00:01:43.200 once I heard from a student that all of my Biostats ER examples 34 00:01:43.200 --> 00:01:44.970 are solved by hand. 35 00:01:44.970 --> 00:01:48.330 Hence, I have decided to type some of the solutions 36 00:01:48.330 --> 00:01:52.320 and this is one of those that I typed. 37 00:01:52.320 --> 00:01:56.133 Nevertheless, I will go over the solution step by step. 38 00:01:57.060 --> 00:01:59.550 So first here, as you can see, this is the formula 39 00:01:59.550 --> 00:02:02.190 we are going to use to solve the problem. 40 00:02:02.190 --> 00:02:05.460 The Ln here stands for the natural log. 41 00:02:05.460 --> 00:02:08.610 The first thing we have to do is calculate the RR hat 42 00:02:08.610 --> 00:02:12.540 and the formula for that is p1 divided by p2 43 00:02:12.540 --> 00:02:17.540 And to determine p1, we need to divide x1 44 00:02:17.910 --> 00:02:21.300 divided by n1 and then to determine p2, 45 00:02:21.300 --> 00:02:24.780 we have to divide x2 by n2. 46 00:02:24.780 --> 00:02:29.780 So once we do these divisions, we get 1.81 47 00:02:29.910 --> 00:02:32.100 and that will be inserted here, 48 00:02:32.100 --> 00:02:37.100 and we will subsequently obtain the natural log of 1.81 49 00:02:37.320 --> 00:02:39.633 and for which we have to use our calculator. 50 00:02:40.530 --> 00:02:44.670 Now we will insert the Z value here, which is going 51 00:02:44.670 --> 00:02:48.360 to be 1.96, and then under the square root 52 00:02:48.360 --> 00:02:51.153 we are going to insert the values. 53 00:02:53.640 --> 00:02:58.640 For n1 minus x1, the value is 62 and x1 is 38, 54 00:02:59.640 --> 00:03:01.383 and n1 is 100. 55 00:03:02.280 --> 00:03:05.670 Then here we have for n2 minus x2, 56 00:03:05.670 --> 00:03:08.190 the value is going to be 79 57 00:03:08.190 --> 00:03:12.090 and for x2, the value is going to be 21 58 00:03:12.090 --> 00:03:14.823 and then we are going to divide that by 100. 59 00:03:16.170 --> 00:03:21.170 So once we do the divisions and then we do the addition 60 00:03:21.990 --> 00:03:24.750 and then once we get the square root, 61 00:03:24.750 --> 00:03:29.190 our final answer here is 0.232. 62 00:03:29.190 --> 00:03:34.153 Now we have to take this answer and now multiply it by 1.96. 63 00:03:35.250 --> 00:03:37.803 And once we do that, we get 0.455. 64 00:03:39.570 --> 00:03:44.570 Now 0.455 is going to be subtracted and added 65 00:03:45.120 --> 00:03:50.100 to the natural log of 1.81 and that is 0.59. 66 00:03:51.240 --> 00:03:54.660 So first we are going to take 0.455 and deduct it from 0.59 67 00:04:00.480 --> 00:04:05.383 and then we are going to add 0.455 to 0.59. 68 00:04:06.810 --> 00:04:09.150 Once we do the subtraction and addition, 69 00:04:09.150 --> 00:04:14.150 we get 0.135 and 1.045. 70 00:04:16.200 --> 00:04:20.340 The 95% confidence interval for the natural log 71 00:04:20.340 --> 00:04:25.340 of RR is 0.135 and 1.045. 72 00:04:27.960 --> 00:04:31.050 To generate the confidence interval for the RR, 73 00:04:31.050 --> 00:04:35.970 we take the anti-log of the lower and the upper limits. 74 00:04:35.970 --> 00:04:39.150 And when we do that, again, by using our calculator, 75 00:04:39.150 --> 00:04:44.150 we get 1.14 and 2.84. 76 00:04:44.310 --> 00:04:46.830 So as you can see, the confidence interval here 77 00:04:46.830 --> 00:04:48.570 does not include a one. 78 00:04:48.570 --> 00:04:51.660 Therefore, we can state with 95% confidence 79 00:04:51.660 --> 00:04:54.720 that a statistically significant difference exists 80 00:04:54.720 --> 00:04:56.253 between the two groups. 81 00:04:57.120 --> 00:04:58.520 Thank you for your attention 82 00:04:59.400 --> 00:05:01.263 and I will see you in the next video.