In: Statistics and Probability
Of 94 adults selected randomly from one town, 62 have health insurance. Find a 90% confidence interval for the proportion of all adults in the town who have health insurance.
Solution :
Given that,
n = 94
x = 62
Point estimate = sample proportion = = x / n =62/94=0.660
1 - = 0.34
At 90% confidence level
= 1 - 90%
= 1 - 0.90 =0.10
/2
= 0.05
Z/2
= Z0.05 = 1.645 ( Using z table )
Margin of error = E Z/2 *(( * (1 - )) / n)
= 1.645 *((0.66*0.34) /94 )
E = 0.080
A 90% confidence interval for population proportion p is ,
- E < p < + E
0.660-0.080 < p < 0.660+0.080
(0.580,0.740)