A research group conducted an extensive survey of 3053 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?" 1537 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
Sample proportion = 1537 / 3053 = 0.503
90% confidence interval for p is
- Z * sqrt( ( 1 - ) / n) < p < + Z * sqrt( ( 1 - ) / n)
0.503 - 1.645 * sqrt( 0.503 * 0.497 / 3053) < p < 0.503 + 1.645 * sqrt( 0.503 * 0.497 / 3053)
0.488 < p < 0.518
90% CI is ( 0.488 , 0.518)
Lower limit = 0.488
Upper limit = 0.518
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