A 2011 NBC News survey found that 83% of a sample of 4528 American teens said they owned an MP3 player such as an iPod. Assume that the sample was an SRS. Find the 90% confidence interval (± ± 0.0001) for the proportion of all American teens who own an MP3 player. The 90% confidence interval is from to
Solution :
Given that,
Point estimate = sample proportion =
= 0.83
1 -
= 1 - 0.83 = 0.17
Z/2
= 1.645
Margin of error = E = Z
/ 2 *
((
* (1 -
)) / n)
= 1.645 * (((0.83
* 0.17) / 4528)
Margin of error = E = 0.009
A 90% confidence interval for population proportion p is ,
- E < p <
+ E
0.83 - 0.009 < p < 0.83 + 0.009
0.821 < p < 0.839
The 90% confidence interval is from 0.821 to 0.839
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