An insurance company checks police records on 596 accidents selected at random and notes that teenagers were at the wheel at 87 of them. if we want to construct a 95% confidence interval for the proportion of all auto accidents that involve teenage drivers, what would be the standard error of the sample proportion of successes?
Solution :
Given that,
n = 596
x = 87
Point estimate = sample proportion = = x / n = 87 / 596 = 0.146
1 - = 1 - 0.146 = 0.854
standar error = = [p ( 1 - p ) / n] = [(0.146 * 0.854) / 596 ] = 0.0145
At 95% confidence level
= 1 - 95%
=1 - 0.95 =0.05
/2
= 0.025
Z/2
= Z0.025 = 1.960
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.96 (((0.146 * 0.854) / 596 )
= 0.028
A 95% confidence interval for population proportion p is ,
± E
= 0.146 ± 0.028
= ( 0.118, 0.174 )
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