A random sample of 200 individuals working in a large city indicated that 58 are dissatisfied with their working conditions. Based upon this, compute a 90% confidence interval for the proportion of all individuals in this city who are dissatisfied with their working conditions. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.)


Solution :
Given that,
Point estimate = sample proportion = = x / n = 58 / 200 = 0.290
1  = 1  0.290 = 0.710
Z/2 = Z0.05 = 1.645
Margin of error = E = Z / 2 * (( * (1  )) / n)
= 1.645 * (((0.290 * 0.710) / 200)
= 0.053
A 90% confidence interval for population proportion p is ,
 E < p < + E
0.290  0.053 < p < 0.290 + 0.053
0.24 < p < 0.34
The lower limit of the 90% confidence interval is 0.24
The Upper limit of the 90% confidence interval is 0.34
Get Answers For Free
Most questions answered within 1 hours.