A recent survey of 349 people ages 18 to 29 found that 86% of them own a smartphone. Find the 99% confidence interval of the population proportion.
Solution :
Given that,
n = 349
Point estimate = sample proportion = = 0.86
1 - = 1 - 0.86 = 0.14
At 99% confidence level
= 1 - 99%
=1 - 0.99 =0.01
/2
= 0.005
Z/2
= Z0.005 = 2.576
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 2.576 (((0.86 * 0.14) / 349)
= 0.048
A 99% confidence interval for population proportion p is ,
± E
= 0.86 ± 0.048
= ( 0.812, 0.908 )
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