Question

In: Statistics and Probability

In a study of cell phone use and brain hemispheric dominance, an internet survey was emailed...

In a study of cell phone use and brain hemispheric dominance, an internet survey was emailed to 2601 subjects randomly selected from an online group involved with ears. 975 surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys. a) What is the best point estimate of the population P c) Construct the confidence level b) Identify the value of the margin of error E. c) Construct the confidence level.

Solutions

Expert Solution

Solution:

Given:

n = 2601

x = 975

thus

Construct a 90% confidence interval for the proportion of returned surveys.

Part a) What is the best point estimate of the population P

the best point estimate of the population P is sample proportion

that is:

Part  b) Identify the value of the margin of error E.

Zc is z critical value for c = 90% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.90) / 2 = 1.90 / 2 = 0.9500

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Zc = 1.645

thus

Part c) Construct the confidence level.


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