Question

Only 10.82, 10.83, 10.84. Thank you!

Pulmonary Disease Asthma is an important health problem for inner-city chil- dren, frequently resulting in hospital admission if symptoms become exacerbated. It is well known that compliance of children with asthma medication is often poor. Also, many household allergens (e.g., roaches) worsen asthma symp- toms. A study is proposed in which children will be rand- omized to either an active intervention where a community health worker comes to the home and educates the children and parents as to approaches to reduce the risk of asthma symptoms or a control intervention where households will receive the same information in print but no home visits will be performed. It is expected that 30% of the children in the active group vs. 10% of the children in the control group will have an improvement in asthma symptoms. 10.81 How many subjects should be recruited in each group (same number per group) to have a 90% chance of detecting a significant difference using a two-sided test with a = .05? 10.82 Suppose that 50 households are randomized per group. How much power would the study have under the above assumptions? The results of the study were as follows: 14 of the 50 active intervention children had an improvement in symptoms com- pared with 6 of the 50 control intervention children. 10.83 What test can be used to compare the results in the active and control groups? 10.84 Perform the test in Problem 10.83, and report a p-value (two-tailed).

Answer 1

Answen 1) Asthama es an import and health problem, for inner esty, frequently resulting in hospital. admission. ito symptoms o become "excer batedo - here households will recieve the same information in Print but no home visits will be performed. . it is expected that 30% of the children in the active group vs. 10% of the children in the controll group. .. . . . to have a got chance detecting a significant difference with d=0.5

1. The level of significance is 0:05 here the power is 90% here, " Picos q, 2017 Pg 20.1 .92-019 Q10.05 1-8-0.9 2 for 15.1.. level of significance and two tail test is 1.96 z for 80% Powen & two tail testis 1:28. by calculating Peq P = (PitPg) By substituting the values. 408 - 10:33 +0:1) 2. 9-1- 20.2. By substituting the value.

- 2 w here by calculating the sample size as folowo. no Thai (3 14]+ $7.9489,12,] - By subuliditing all values Jalo-360871-9.6) + JC0-32017) 4 otkroba (1028) By solving 10.3.0.112 3.245 0.04. . *82. ². In each group to have 90% chance of detecting a significance - two sidled text with significance ice, ning = 82... leviel. 0.5 is: