Question

In: Statistics and Probability

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

In the study of cell phone use and brain hemispheric dominance, an internet survey was emailed to 2524 subjects randomly selected from an online group involved with ears. 930 surveys were returned. Construct a 99% confidence interval for the proportion of returned surveys.

A. Find the best point estimate of the population proportion p

B. Identify the value of the margin of error E.

C. Construct the confidence interval

Solutions

Expert Solution

Solution :

Given that,

n = 2524

x = 930

Point estimate = sample proportion = = x / n = 930/2524=0.368

1 -   = 1- 0.368 =0.632

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

  Margin of error = E = Z / 2    * (((( * (1 - )) / n)

= 2.576* (((0.368*0.632) /2524 )

E = 0.0247

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.368 -0.0247 < p <0.368+ 0.0247

0.3433< p < 0.3927

The 99% confidence interval for the population proportion p is : 0.3433,0.3927


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