Question

It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A sample of 240 companies showed that 105 of them provide such facilities on site.

Construct a 96% confidence interval for the percentage of all such companies that provide such facilities on site. What is the margin of error for this estimate?

Round your answers to one decimal place.

Confidence interval: _____% to _____%

Margin of error: ______%

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 105 / 240 = 0.438

1 - = 0.562

Z_/2 = 2.054

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.054 * (((0.438 * 0.562) / 240)

= 0.066

Margin of error = E = 6.6%

A 96% confidence interval for population proportion p is ,

- E < p < + E

0.438 - 0.066 < p < 0.438 + 0.066

0.372 < p < 0.504

confidence interval = 37.2% to 50.4%