In: Statistics and Probability
A research group conducted an extensive survey of 3180 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hour-long telephone interviews with a nationally representative sample. In response to the question, "What does success mean to you?" 1614 responded, "Personal satisfaction from doing a good job." Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. Find a 90% confidence interval for p. (Round your answers to three decimal places.)
lower limit | |
upper limit |
Solution :
Given that,
n = 3180
x = 1614
Point estimate = sample proportion = = x / n = 1614/3180=0.5075
1 - = 1-0.5075=0.4925
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 (((0.5075*0.4925) /3180 )
E = 0.015
A 90% confidence interval for population proportion p is ,
- E < p < + E
0.5075-0.015 < p < 0.5075+0.015
0.493< p < 0.523
The 90% confidence interval for the population proportion p is : lower limit=0.493,upper limit=0.523