In: Statistics and Probability
In a survey of 8000 ?women, 5431 say they change their nail polish once a week. Construct a 90 % confidence interval for the population proportion of women who change their nail polish once a week. A 90?% confidence interval for the population proportion is ?( nothing?, nothing?). ?(Round to three decimal places as? needed.)
Solution :
Given that,
n = 8000
x = 5431
= x / n = 5431 / 8000 = 0.679
1 - = 1 - 0.679 = 0.321
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
Z/2 = Z0.05 = 1.645
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.645 * (((0.679 * 0.321) / 8000)
= 0.009
A 90% confidence interval for population proportion p is ,
- E < P < + E
0.679 - 0.009 < p < 0.679 + 0.009
0.670 < p < 0.687
(0.670 , 0.687)