In: Math
A research group conducted an extensive survey of 3144 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?" 1508 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 = 3144
x = 1508
= x / n = 1508 / 3144 = 0.480
1 - = 1 - 0.480 = 0.52
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.1
/ 2 = 0.1 / 2 = 0.05
Z/2 = Z0.025 = 1.645
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.645 * (((0.480*0.52) / 3144 )
= 0.015
A 90% confidence interval for population proportion p is ,
- E < P < + E
0.480 - 0.015 < p < 0.480 + 0.015
0.465 < p < 0.495
The 90% confidence interval for the population proportion p is : (0.465 , 0.495)
lower limit = 0.465
upper limit = 0.495