## Suppose you are reading a study conducted in the year 2000 about adolescent girls with anorexia nervosa in the United States. The authors re

Question

Suppose you are reading a study conducted in the year 2000 about adolescent girls with anorexia nervosa in the United States. The authors report the following frequency data on regions where the 933 adolescent girls with anorexia nervosa in their random sample reside:
Observed Frequencies Region West South Midwest Northeast 229 369 197 138 You wonder whether the differences in the frequencies are due to the disease being more prevalent in certain areas of the country, whether the study simply included more girls from certain regions, or whether the distribution matches the geographic distribution of US population. You obtain the following data from the 2000 census: Percent Distribution of the U.s. Population by Region Region West South Midwest Northeast 22.46% 35.62% 22.88% 19.04% [Source: Hobbs F., 8 stoops, N. (2002). Census 2000 special reports: Demographic Drends in the 20th century, US. Census Bureau.] You use a chi-square test for goodness of fit to see how well the sample of adolescent girls with anorexia nervosa fits the census data. What is the most appropriate null hypothesis?
a. The distribution of where the adolescent girls live is the same as that provided by the census data.
b. The distribution of where the adolescent girls live is different from that provided by the census data.
c. The distribution of where the adolescent girls live is equal across the four geographic regions.
d. The distribution of where the adolescent girls live is not equal across the four geographic regions.

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2 weeks 2021-09-11T00:01:27+00:00 1 Answer 0

Step-by-step explanation:

H0: All are as per the percent data

Ha: atleast one is not as in percent data

(Two tailed chi square test)

Observed O 229              369           197                138             933

Expected % 22.46      35.62         22.88         19.04

Expected e 209.5518     332.3346 213.4704  177.6432

(o-e)^2/e         1.804959     4.045175 1.270781         8.846853      15.96777

degrees of freedom = (r-1)(c-1)  = 3

Critical value = 7.81

Since chi square > critical value, reject null hypothesis.

Therefore you can conclude that proportions differ significantly from the percent.