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, 286girls aged 10to 15years were followed to determine the effects of observation only (129patients), an underarm plastic brace (111patients), 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

Expert verified

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 H0. 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

01

Step 1. Given information.

Consider the given question,

02

Step 2. Consider the null and alternative hypotheses.

According to the null and alternative hypotheses,

H0denotes race distributions among the four U.S. regions are homogeneous.

H1 denotes race distributions among the four U.S. regions are non-homogeneous.

Specific level of significance is α=0.05

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

03

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 is28.13

04

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

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

df=r-1·c-1=3-1·2-1=2

For α=0.05, chi-square table reveals that the critical value is x20.05=5.991.

Hence, we can see that the value of the test statistic falls in the rejection region. Thus, we reject H0. 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.

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Most popular questions from this chapter

12.53 AIDS Cases. Refer to Exercise 12.47. For AIDS cases in the United States in 2011, solve the following problems:

RegionWhiteBlackOtherTotal
Northeast1,1002,493
5,177
Northwest1,137
5043221
South2,7617,848
12,867
West
76417664,230
Total

605225,435

a. Find and interpret the conditional distributions of region by race.

b. Find and interpret the marginal distribution of region.

c. Are the variables "region" and "race" associated? Explain your answer.

d. What percentage of AIDS cases were in the South?

e. What percentage of AIDS cases among whites were in the South?

f. Without doing further calculations, respond true or false to the distributions of race by region are not identical.

g. Find and interpret the marginal distribution of race and the conditional distributions of race by region.

The t-table has entries for areas of 0.10,0.05,0.025,0.01and0.005. In contrast, the χ2-table has entries for those areas and for 0.995,0.99,0.975,0.95and 0.90.Explain why the t-values corresponding to these additional areas can be obtained from the existing t-table but must be provided explicitly in the χ2-table.

Suppose that bivariate data are to be grouped into a contingency table. Determine the number of cells that the contingency table will have if the numbers of possible values for the two variables are

a. two and three.
b. four and three.
c. m and n.

Girls and Boys. One probability model for child gender is that a boy or a girl is equally likely to be born. If that model is correct, then, for a two-child family, the probabilities are 0.25,0.50, and 0.25 for two girls, one girl and one boy, and two boys, respectively. W. Stansfield and M. Carlton examined data collected in the National Health Interview Study on two-child families in the article "The Most Widely Published Gender Problem in Human Genetics" (Human Biology, Vol. 81, No. 1, pp. 3-11). Of 42,888 families with exactly two children, 9,523 had two girls, 22,031 had one girl and one boy, and 11,334 had two boys.

a) At the 1% significance level, do the data provide sufficient evidence to conclude that the distribution of genders in two-children families differs from the distribution predicted by the model described?

b) In view of your result from part (a), what conclusion can you draw?

Suppose that you have bivariate data for an entire population.

a. How would you decide whether an association exists between the two variables under consideration?

b. Assuming that you make no calculation mistakes, could your conclusion be in error? Explain your answer.

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