A research group conducted an extensive survey of 2944 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?" 1600 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,
Point estimate = sample proportion = = x / n = 1600 / 2944 = 0.543
1 - = 1 - 0.543 = 0.457
Z/2 = 1.645
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.645 * (((0.543 * 0.457) / 2944)
Margin of error = E = 0.015
A 90% confidence interval for population proportion p is ,
- E < p < + E
0.543 - 0.015 < p < 0.543 + 0.015
0.528 < p < 0.558
lower limit: 0.528
upper limit: 0.558
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