WEBVTT 1 00:00:01.680 --> 00:00:02.513 Hello students. 2 00:00:02.513 --> 00:00:07.513 Welcome to Biostats ER, example one for chapter seven. 3 00:00:07.590 --> 00:00:08.580 In this example, 4 00:00:08.580 --> 00:00:10.110 we will learn how to use 5 00:00:10.110 --> 00:00:12.873 a one-sample continuous outcome formula. 6 00:00:14.220 --> 00:00:17.910 The problem here is from our textbook, problem 20. 7 00:00:17.910 --> 00:00:19.297 The problem states, 8 00:00:19.297 --> 00:00:24.217 "The mean lifetime for cardiac stents is 8.9 years. 9 00:00:24.217 --> 00:00:28.297 "A medical device company has implemented some improvements 10 00:00:28.297 --> 00:00:30.247 "in the manufacturing process, 11 00:00:30.247 --> 00:00:35.197 "and hypothesizes that the lifetime is now longer. 12 00:00:35.197 --> 00:00:37.747 "A study of 14 new devices reveals 13 00:00:37.747 --> 00:00:40.657 "a mean lifetime of 9.7 years 14 00:00:40.657 --> 00:00:44.437 "with a standard deviation of 3.4 years. 15 00:00:44.437 --> 00:00:46.177 "Is there statistical evidence 16 00:00:46.177 --> 00:00:49.237 "of a prolonged lifetime of the stents?" 17 00:00:49.237 --> 00:00:54.150 "Run the test at a 5% level of significance." 18 00:00:54.150 --> 00:00:57.757 Now, before I move forward with solving the problem, 19 00:00:57.757 --> 00:00:59.490 again, I would like to let you know 20 00:00:59.490 --> 00:01:01.890 that for this Biostats ER example, 21 00:01:01.890 --> 00:01:03.720 the solution is typed, 22 00:01:03.720 --> 00:01:06.330 because again, as I mentioned previously, 23 00:01:06.330 --> 00:01:07.707 once I heard from a student, 24 00:01:07.707 --> 00:01:11.913 that all of my Biostats ER examples are solved by hand. 25 00:01:12.840 --> 00:01:15.930 Hence, I have again typed the solution here, 26 00:01:15.930 --> 00:01:18.273 but I'm going to go through it step-by-step. 27 00:01:19.260 --> 00:01:20.670 So the first step here 28 00:01:20.670 --> 00:01:23.303 is for us to set up the hypothesis, 29 00:01:26.790 --> 00:01:29.283 and determine the level of significance. 30 00:01:30.690 --> 00:01:33.706 So here, the null hypothesis states, 31 00:01:33.706 --> 00:01:38.706 that the mean lifetime for cardiac stents 32 00:01:43.020 --> 00:01:45.540 is equal to 8.9 years, 33 00:01:45.540 --> 00:01:47.040 and the alternative states 34 00:01:47.040 --> 00:01:49.913 that it is greater than 8.9 years, 35 00:01:49.913 --> 00:01:54.913 and the significance level is 0.05 as stated in the problem. 36 00:01:58.140 --> 00:02:01.710 Now we have to select the appropriate test statistics. 37 00:02:01.710 --> 00:02:04.770 Now, given our sample size is 40, 38 00:02:04.770 --> 00:02:06.690 we can select a Z test. 39 00:02:06.690 --> 00:02:08.130 But then the question is, 40 00:02:08.130 --> 00:02:10.863 which type of Z test are we going to use? 41 00:02:14.280 --> 00:02:16.680 Now here we have to look at the problem again, 42 00:02:16.680 --> 00:02:18.630 and as we can see that this is 43 00:02:18.630 --> 00:02:21.843 a one-sample continuous outcome, 44 00:02:23.850 --> 00:02:25.980 because we are comparing our mean 45 00:02:25.980 --> 00:02:30.180 from a single sample to an expected value. 46 00:02:30.180 --> 00:02:34.530 And our outcome is continuous as we are measuring years. 47 00:02:34.530 --> 00:02:38.237 Hence, we will select the correct Z test, 48 00:02:42.390 --> 00:02:43.893 which is provided here. 49 00:02:45.180 --> 00:02:47.940 Then we have to set up the decision rule, 50 00:02:47.940 --> 00:02:50.910 and given that alpha is 0.05, 51 00:02:50.910 --> 00:02:54.180 we will reject null if our Z calculated 52 00:02:54.180 --> 00:02:58.623 is greater than or equal to 1.645. 53 00:02:59.520 --> 00:03:03.840 Now, in step four we have to compute the test statistics. 54 00:03:03.840 --> 00:03:07.380 The X bar here is 9.7, 55 00:03:07.380 --> 00:03:09.900 and the mu naught is 8.9. 56 00:03:09.900 --> 00:03:12.540 The standard deviation is 3.4, 57 00:03:12.540 --> 00:03:13.860 and it is divided by the 58 00:03:13.860 --> 00:03:17.250 square root of the sample size, which is 40. 59 00:03:17.250 --> 00:03:19.413 So after we do all the algebra, 60 00:03:20.460 --> 00:03:23.403 our Z calculated is 1.49. 61 00:03:24.390 --> 00:03:27.090 Now, we have to draw the conclusion. 62 00:03:27.090 --> 00:03:28.680 So the conclusion here, 63 00:03:28.680 --> 00:03:31.590 we realize that we fail to reject the null 64 00:03:31.590 --> 00:03:36.590 because 1.49 is less than 1.645. 65 00:03:36.840 --> 00:03:40.904 So, we do not have statistically significant evidence 66 00:03:40.904 --> 00:03:45.540 at alpha of 0.05 to show 67 00:03:45.540 --> 00:03:49.173 that the lifetime is longer than 8.9 years. 68 00:03:56.040 --> 00:04:00.120 I hope this was helpful, 69 00:04:00.120 --> 00:04:01.530 and if you have any questions, 70 00:04:01.530 --> 00:04:03.240 please let me know, 71 00:04:03.240 --> 00:04:05.163 and I'll see you in the next video. 72 00:04:06.000 --> 00:04:06.833 Thank you.