WEBVTT 1 00:00:09.120 --> 00:00:11.310 Hello students, and welcome 2 00:00:11.310 --> 00:00:15.076 to Biostat ER Chapter 8: Example 2. 3 00:00:15.076 --> 00:00:18.630 In this example, we will learn how to calculate sample size 4 00:00:18.630 --> 00:00:21.240 for one sample continuous outcome 5 00:00:21.240 --> 00:00:23.250 to determine confidence interval, 6 00:00:23.250 --> 00:00:27.003 and we will utilize the data from example one. 7 00:00:28.050 --> 00:00:30.420 The problem statement hence is going to be the same, 8 00:00:30.420 --> 00:00:32.340 but I will read it again. 9 00:00:32.340 --> 00:00:35.640 The issue here is that we have collected some data, 10 00:00:35.640 --> 00:00:38.640 and we need to know if our sample mean is different 11 00:00:38.640 --> 00:00:41.850 from some established mean, which is 100. 12 00:00:41.850 --> 00:00:45.240 We want a hypothesis test with 80% power, 13 00:00:45.240 --> 00:00:46.830 and we know that we will fail 14 00:00:46.830 --> 00:00:48.930 to reject the null hypothesis 15 00:00:48.930 --> 00:00:51.570 within a margin of error of five. 16 00:00:51.570 --> 00:00:54.282 Therefore, if our mean is between 95 17 00:00:54.282 --> 00:00:56.700 and 105, we know that we will fail 18 00:00:56.700 --> 00:00:59.040 to reject the null hypothesis. 19 00:00:59.040 --> 00:01:01.080 From publications for similar studies, 20 00:01:01.080 --> 00:01:05.763 we know that the standard deviation is between 8.5 and 9.5. 21 00:01:06.750 --> 00:01:10.080 So the question is how many subjects are required 22 00:01:10.080 --> 00:01:13.212 to determine a confidence interval that will have a margin 23 00:01:13.212 --> 00:01:15.813 of error no greater than five? 24 00:01:16.756 --> 00:01:21.000 So again, for your convenience, 25 00:01:21.000 --> 00:01:23.436 I have summarized the information here, 26 00:01:23.436 --> 00:01:26.130 as you can take a look at it very quickly. 27 00:01:26.130 --> 00:01:28.020 And I want to also emphasize 28 00:01:28.020 --> 00:01:31.650 that the Z value here is again going to be 1.96. 29 00:01:31.650 --> 00:01:33.840 I've also inserted the formula here, 30 00:01:33.840 --> 00:01:37.006 and it is N equal to Z times sigma divided 31 00:01:37.006 --> 00:01:39.153 by E and the whole square. 32 00:01:41.160 --> 00:01:42.540 Now the standard deviation, 33 00:01:42.540 --> 00:01:45.990 which was obtained from previous literature is a range, 34 00:01:45.990 --> 00:01:49.230 and we want to use the highest value 35 00:01:49.230 --> 00:01:51.300 as it'll give us the most variation 36 00:01:51.300 --> 00:01:55.200 and ensure that we have enough subjects in the study. 37 00:01:55.200 --> 00:01:57.990 We are here erroring on the side of caution, 38 00:01:57.990 --> 00:02:02.580 and we want to ensure that we are enrolling enough people. 39 00:02:02.580 --> 00:02:06.375 So once we insert all the values, so here Z is going 40 00:02:06.375 --> 00:02:10.410 to be 1.96, sigma is going to be 9.5, 41 00:02:10.410 --> 00:02:11.820 the E is going to be five. 42 00:02:11.820 --> 00:02:14.910 So we insert all the values and we perform the algebra, 43 00:02:14.910 --> 00:02:16.710 we get 13.9. 44 00:02:16.710 --> 00:02:19.020 And as always, we are going to round it up 45 00:02:19.020 --> 00:02:20.403 and that's going to be 14. 46 00:02:21.265 --> 00:02:22.923 So 14 subjects here are required.