One estimate of the proportion of children with autism in the United States is 1 in 100 (Source: http://www.cbsnews.com/stories/2009/10/05/health/main5363192.shtml). Suppose you are interested in the rate of autism among current school-aged children in Utah. You collect a sample of 400 children between the ages of 5 and 18 and find that three have had a previous diagnosis of an autism disorder. You plan to calculate a 95% confidence interval estimator of the proportion of school-aged children in Utah who have ever had a diagnosis of an autism disorder. Which of the following is the most likely reason you would use a Wilson estimator to calculate the confidence interval estimator? It is uncomfortable to define having been diagnosed with autism as a success. It is possible that if even the actual proportion in Utah is 1%, your sample may only have very few children who have had a previous diagnosis of an autism disorder. It is an easier way to calculate the confidence interval estimator. A confidence interval estimator based on the Wilson estimator does not use the normal distribution. Let p be the proportion of school-aged children in Utah who have ever had a diagnosis of an autism disorder. Using the Wilson estimate, the 95% confidence interval estimate of p is ± . (Hint: For a 95% confidence interval estimate, use zα/2zα/2 = 1.96.)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter4: Equations Of Linear Functions

Section: Chapter Questions

Problem 8SGR

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One estimate of the proportion of children with autism in the United States is 1 in 100 (Source: http://www.cbsnews.com/stories/2009/10/05/health/main5363192.shtml). Suppose you are interested in the rate of autism among current school-aged children in Utah. You collect a sample of 400 children between the ages of 5 and 18 and find that three have had a previous diagnosis of an autism disorder. You plan to calculate a 95% confidence interval estimator of the proportion of school-aged children in Utah who have ever had a diagnosis of an autism disorder.

Which of the following is the most likely reason you would use a Wilson estimator to calculate the confidence interval estimator?

It is uncomfortable to define having been diagnosed with autism as a success.

It is possible that if even the actual proportion in Utah is 1%, your sample may only have very few children who have had a previous diagnosis of an autism disorder.

It is an easier way to calculate the confidence interval estimator.

A confidence interval estimator based on the Wilson estimator does not use the normal distribution.

Let p be the proportion of school-aged children in Utah who have ever had a diagnosis of an autism disorder. Using the Wilson estimate, the 95% confidence interval estimate of p is ± . (Hint: For a 95% confidence interval estimate, use zα/2zα/2 = 1.96.)

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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