Need help asap! Thanks!! ### Investigating the Effectiveness of External Hip Protectors in Preventing Hip Fractures in Elderly PeoplePhysicians conducted an experiment to investigate the effectiveness of external hip protectors in preventing hip fractures in elderly people. They randomly assigned some participants to receive hip protectors while others were assigned to be the control group (i.e., no protection). The incidence of hip fractures was then recorded for each person in both groups.### Hypothesis TestingWe aim to test the hypothesis that two different treatments (hip protectors and no protection) create no difference in their hip fracture risks. For this test, we will use a significance level of \( \alpha = 0.02 \).#### Null Hypothesis (H₀):There is no difference in the risk of hip fractures between the two groups (hip protectors and control).#### Alternative Hypothesis (H₁):There is a difference in the risk of hip fractures between the two groups.### Data CollectionThe results from the experiment are summarized in the table below:| | Hip Protector | Control ||----------------|---------------|---------|| Hip Fracture | 13 | 67 || No Hip Fracture| 640 | 1081 |### Steps to Calculate1. **Complete the 2x2 table with expected counts and group totals:** - Total in Hip Protector group = 13 (Hip Fracture) + 640 (No Hip Fracture) = 653 - Total in Control group = 67 (Hip Fracture) + 1081 (No Hip Fracture) = 1148 - Grand total = 653 + 1148 = 18012. **Compute the expected counts for each cell:** - Expected count for Hip Fracture in Hip Protector group = (Total Hip Fracture * Total Hip Protector) / Grand total - Expected count for No Hip Fracture in Hip Protector group = (Total No Hip Fracture * Total Hip Protector) / Grand total - Expected count for Hip Fracture in Control group = (Total Hip Fracture * Total Control) / Grand total - Expected count for No Hip Fracture in Control group = (Total No Hip Fracture * Total Control) / Grand total - **Note:** Calculations need to be done to fill these cells.3

The provided information is Hip protector Control Hip fracture 13 67 No hip fracture 640…

Physicians conducted an experiment to investigate the effectiveness of external hip protectors in preventing hip fractures in elderly people. They randomly assigned some people to get hip protectors and others to be the control group (no protection). They recorded the hip fractures status for each person in each treatment arm (treatment and control). Perform the appropriate test on the hypothesis that two different treatments create no difference in their hip fracture risks. Use a = 0.02. Clearly state the null and alternative hypothesis, calculate the test statistics value, find P-value, and state your conclusion. (hint: complete the 2 by 2 table with expected counts and group totals, first) Hip fracture No hip fracture Hip protector 13 640 Control 67 1081

Physicians conducted an experiment to investigate the effectiveness of external hip protectors in preventing hip fractures in elderly people. They randomly assigned some people to get hip protectors and others to be the control group (no protection). They recorded the hip fractures status for each person in each treatment arm (treatment and control). Perform the appropriate test on the hypothesis that two different treatments create no difference in their hip fracture risks. Use a = 0.02. Clearly state the null and alternative hypothesis, calculate the test statistics value, find P-value, and state your conclusion. (hint: complete the 2 by 2 table with expected counts and group totals, first) Hip fracture No hip fracture Hip protector 13 640 Control 67 1081

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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