Help with number #4 please ### Simulation Example: Denise's Class Attendance**Scenario:** Denise has a tendency to oversleep, making her late for her 8 a.m. class 10% of the time. We will simulate 15 days to assess this behavior.#### 1. Simulation SetupSelect the correct digit assignment for the simulation of Denise's attendance:- **a)** 0 to 8 = on time, 9 = late to class- **b)** 00 to 49 = on time, 50 to 99 = late to class- **c)** 0 to 4 = on time, 5 to 9 = late to class- **d)** 00 to 90 = on time, 91 to 99 = late to class**Correct Answer:** a) 0 to 8 = on time, 9 = late to class.#### 2. Conducting the SimulationUsing the random digit assignment method, starting from line 104 of the random digit table (Attached Appendix A), simulate the 15 class days.| Day | Random Number | Was she on time or late? ||-----|---------------|--------------------------|| 1st | 0 | On time || 2nd | 3 | On time || 3rd | 4 | On time || 4th | 0 | On time || 5th | 0 | On time || 6th | 9 | Late || 7th | 2 | On time || 8th | 6 | On time || 9th | 3 | On time || 10th| 6 | On time || 11th| 1 | On time || 12th| 5 | On time || 13th| 4 | On time || 14th| 0 | On time || 15th| 8 | On time |**Notes:**- Line 104: Number 52711, first day number is '5'- Line 105: Number 95929, gives '9'- Line 106: Number 68417, gives '6'#### 3. ResultsCalculate the percentage of days Denise was late: **Table A: Random Digits**This table presents a series of random digits organized into lines for use in various statistical and educational applications. Each line is labeled with a line number for reference. Below are the lines of random digits from 101 to 150:- **Line 101**: 19223 95034 05756 28713 96409 12531 42544 82853- **Line 102**: 73676 47150 99400 01927 27754 42648 82425 36290- **Line 103**: 45467 71709 77558 00095 32363 29485 82264 59626- **Line 104**: 52711 38889 93074 91481 60779 53791 17297 50335- **Line 105**: 35902 41047 64811 51659 58989 35079 50211 47837- **Line 106**: 68417 35013 15529 72765 81766 55300 94383 14889- **Line 107**: 82739 40502 20807 49938 24543 91249 36689 04228- **Line 108**: 60940 72024 17868 24983 61790 90686 83485 41979- **Line 109**: 80009 49998 31682 83832 23647 93942 42006 78530- **Line 110**: 56450 63646 69536 24499 23496 12338 27649 93059- **Line 111**: 81486 89487 60513 71197 19352 73089 84869 41585- **Line 112**: 45889 70753 78814 03699 74035 56002 02938 70015- **Line 113**: 92110 98904 17336 45091 73986 14528 50490 61181- **Line 114**: 65684 61573 48724 71921 27990 21465 41446 99882

We are given that, Denise has a bad habit of sleeping through her alarm in the morning. She oversleeps and is late for her 8 am class 10% of the time. Let us denote p be the proportion of times Denise late for her class. Among 15 days observed Denise was late for just one day. Then the sample proportion (p^) is given by, p^=115=0.0667. The null (H0) and alternative (H1) hypothesis in this are as follows; H0 : p = 0.10 Vs H1 : p < 0.10 Since the alternative hypothesis contains sign "<" (smaller than), the test is the left tailed test.The formula for calculating test statistic is given by, Here in this example, we have n =15, p = 0.10 and p^ =0.0667. By plugging these values in the formula we get, The p-value is given by, P( Z < Z0) = P(Z < -0.43). This can be found using the excel command, "NORMSDIST(z)". Hence we get, p-value = 0.3336.Here we consider significance level (α) is, 0.05. Hence, the decision rule based on p-value is given by: "If p-value ≤ α then we reject null hypothesis otherwise we fail to reject null hypothesis." Here we get p-value = 0.3336 > α = 0.05, hence we fail to reject null hypothesis. Conclusion: There is not sufficient evidence to support the claim that fewer than 10% of the times Denise late for her class. Just because we take the small sample size (observed number of days) we get there is a difference between sample proportion and true population proportion.