Question

# A random sample of 200 individuals working in a large city indicated that 58 are dissatisfied...

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.)

 What is the lower limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval?

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

#### Earn Coins

Coins can be redeemed for fabulous gifts.