Question

please help me

Scenario 2: Strategies to combat Scoliosis Scoliosis is a condition involving curvature of the spine. Recent research compared the effects of two treatments commonly prescribed to combat scoliosis symptoms from advancing (i.e., the spine increasing in curvature). The observational study followed 286 girls aged 10 to 15 who had been diagnosed with adolescent scoliosis and were either prescribed (Group 1) no treatment, (Group 2) an underarm plastic brace, or (Group 3) nighttime electrical stimulation. Six months later, the patient's 'outcome' was recorded as either a success or failure, with 'failure' representing cases where the curvature of the spine had progressed by 6 degrees or more. Results are shown below: Success (no . Failure Treatment progression in curvature Total curvature) progressed) Underarm brace Electrical stimulation Control group Total 06 286 180 Question 10 (1 point) Saved Suppose a chi-squared test of independence is conducted to assess whether there is an association between the two variables Treatment and Outcome. Correctly identity which of the hypotheses below is the null and alternative hypothesis by selecting the appropriate option

Answer 1

Answer 11) 0

If the null hypothesis is true, then the observed and expected frequencies will be equal in value. As a result, the χ2 statistic will be equal to zero.

Answer 12) 69.8601

Expected Value of a Cell = (Row Total)*(Column Total)/Total No. of Observations

Row Total for Underarm brace treatment = 111

Column Total for success = 180

Total no. of observations = 286

Expected Value = (111*180)/286

Expected Value = 69.8601

Answer 13) 2.8620

The formula for contribution = (Observed value - Expected value )^2/Expected value

In this case, Observed value = 84

Expected value = 69.8601 (Refer to Answer 12)

Contirbution = (84 - 69.8601)^2/69.8601

Contirbution = 2.8620

Answer 14) 0.0018

Given χ2 = 12.6266

Degrees of Freedom = (r-1)*(c-1) = (3-1)*(2-1) = 2

P-value corresponding to χ2 = 12.6266 and df = 2 is 0.0018 (Obtained using p-value calculator. Screenshot below)

12.626 Chi-square score: DR Significance Level: 0.01 0.05 0.10 The P-value is .001813. The result is significant at p<.05. Calculate

*** Dear Student, We can answer 4 sub-parts of a question per post. Please post remaining sub-part separately***