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 such companies showed that 99 of them provide such facilities on site.

Construct a 90 % 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: Enter your answer; confidence interval, lower bound % to Enter your answer; confidence interval, upper bound %

Margin of error: Enter your answer; Margin of error

Answer 1

Solution :

Given,

n = 240 ....... Sample size

x = 99 .......no. of successes in the sample

Let denotes the sample proportion.

     = x/n = 99/240 = 0.413  

Our aim is to construct 90% confidence interval.

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05 and 1- /2 = 0.950

= 1.645 (use z table)

Now , the margin of error is given by

E = *  

= 1.645 * [ 0.413 *(1 - 0.413)/240]

= 0.052

Now the confidence interval is given by

( - E)   ( + E)

( 0.413 - 0.052 )   ( 0.413 + 0.052 )

0.361   0.465

Required 90% Confidence Interval is ( 0.361 , 0.465 )

The confidence interval lower bound = 36.1%

The confidence interval upper bound = 46.5%

The margin of error = 0.052