In a study of 301 Hispanic women living in San Antonia, Texas, one variable of interest was the percentage of subjects with impaired fasting glucose (IFG). IFG refers to a metabolic stage intermediate between normal glucose homeostasis and diabetes. In the study, 24 women were classified in the IFG stage. The article cites population estimates for IFG among Hispanic women in Texas as 6.3%. Is there sufficient evidence to indicate the population of Hispanic women in San Antonio has a prevalence of IFG higher than 6.3%? Use 95% confidence level and the p -value method to form your conclusion. The computed z score was 1.193. Find the p-value.
In a study of 301 Hispanic women living in San Antonia, Texas, one variable of interest was the percentage of subjects with impaired fasting glucose (IFG). IFG refers to a metabolic stage intermediate between normal glucose homeostasis and diabetes. In the study, 24 women were classified in the IFG stage. The article cites population estimates for IFG among Hispanic women in Texas as 6.3%. Is there sufficient evidence to indicate the population of Hispanic women in San Antonio has a prevalence of IFG higher than 6.3%? Use 95% confidence level and the p -value method to form your conclusion. The computed z score was 1.193. Find the p-value.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter4: Equations Of Linear Functions
Section4.5: Correlation And Causation
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In a study of 301 Hispanic women living in San Antonia, Texas, one variable of interest was the percentage of subjects with impaired fasting glucose (IFG). IFG refers to a metabolic stage intermediate between normal glucose homeostasis and diabetes. In the study, 24 women were classified in the IFG stage. The article cites population estimates for IFG among Hispanic women in Texas as 6.3%. Is there sufficient evidence to indicate the population of Hispanic women in San Antonio has a prevalence of IFG higher than 6.3%? Use 95% confidence level and the p -value method to form your conclusion.
The computed z score was 1.193. Find the p-value.
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