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* (
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
orchestra answered 1 year agoRelated Solutions
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