Question

Scoliosis is a condition involving curvature of the spine. In a study by A. Nachemson and L. Peterson, reported in the Journal of Bone and Joing Surgery, $286$girls aged $10$to $15$years were followed to determine the effects of observation only ($129$patients), an underarm plastic brace ($111$patients), and nighttime surface electrical stimulation ($46$ patients). A treatment was deemed to have failed if the curvature of the spine increased by 6 on two successive examinations. The following table summarizes the results obtained by the researchers.

At the $5\%$significance level, do the data provide sufficient evidence to conclude that a difference in failure rate exists among the three types of treatments?

Short answer

We know, chi-square is $28.13,df=2$.

We can say that the value of the test statistic falls in the rejection region. Thus, we reject $H_0$. The test results are significant.

At $5\%$significance level, the data provide sufficient evidence to conclude that a difference in failure rate exists among the three types of treatments.

Step-by-step solution

Step 1. Given information.

Consider the given question,

Step 2. Consider the null and alternative hypotheses.

According to the null and alternative hypotheses,

$H_0$denotes race distributions among the four U.S. regions are homogeneous.

$H_1$ denotes race distributions among the four U.S. regions are non-homogeneous.

Specific level of significance is $\alpha =0.05$

The given table contains the expected frequencies corresponding to the observed frequencies.

Step 3. Calculate the chi-square.

All of the expected frequencies are greater than $1$, we can verify using the table.

At most $20\%$of the expected frequencies are less than $5$.

Hence, we can say all the assumptions are satisfied.

On calculating the chi-square,

The test statistics is$28.13$

Step 4. Plot a graph to reveal the critical value.

The row variable has $4$values and the column variable has $3$values. Hence, $r-3,c=2$.

$$\begin{gathered} df=\left(r-1\right)·\left(c-1\right) \\ =\left(3-1\right)·\left(2-1\right) \\ =2 \end{gathered}$$

For $\alpha =0.05$, chi-square table reveals that the critical value is ${x^2}_{0.05}=5.991$.

Hence, we can see that the value of the test statistic falls in the rejection region. Thus, we reject $H_0$. The test results are significant.

At the $5\%$significance level, the data provide sufficient evidence to conclude that a difference in failure rate exists among the three types of treatments.