Use the sample data and confidence level given below to complete parts (a) through (d). In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2437 subjects randomly selected from an online group involved with ears. 1180 surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys. Click the icon to view a table of z scores. a) Find the best point estimate of the population proportion p. (Round to three decimal places as needed.) b) Identify the value of the margin of error E. E= (Round to three decimal places as needed.) c) Construct the confidence interval.
Use the sample data and confidence level given below to complete parts (a) through (d). In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2437 subjects randomly selected from an online group involved with ears. 1180 surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys. Click the icon to view a table of z scores. a) Find the best point estimate of the population proportion p. (Round to three decimal places as needed.) b) Identify the value of the margin of error E. E= (Round to three decimal places as needed.) c) Construct the confidence interval.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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
Transcribed Image Text:**Educational Exercise: Analyzing Survey Data**
Use the sample data and confidence level given below to complete parts (a) through (d).
In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2437 subjects randomly selected from an online group involved with ears. 1180 surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys.

Transcribed Image Text:**Standard Normal (z) Distribution Table**
This table displays the cumulative probabilities of the standard normal distribution. The values represent the probability that a standard normal random variable Z is less than or equal to the given z-value.
The table consists of two sections:
- **Rows:** Each row represents the integer and first decimal of z-values from 0.0 to 2.4.
- **Columns:** Each column represents the second decimal place of z-values from .00 to .09.
For example, to find the cumulative probability of Z for a z-value of 1.65:
1. Locate the row for 1.6.
2. Move across to the column under .05.
3. The intersection gives you the cumulative probability: 0.9505.
This means that there is a 95.05% probability that a standard normal random variable will be less than or equal to 1.65.
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