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 98% confidence interval for the percentage of all such companies that provide such facilities on site. What is the margin of error for this estimate?

Confidence interval is 0.7 ≤ p ≤ 0.8

The margin of error is 7.1 %

Step-by-step explanation:

We have to calculate a 98% confidence interval for the proportion.

The sample proportion is p=0.75.

The standard error of the proportion is:

Here we have, the proportion or point estimate given by

= 150/200 = 0.75

Sample size, n = 200

The formula for confidence interval, CI, given a proportion, is;

At 98% z = ±2.326348

Plugging in the values of, , z and n we get;

CI = 0.6787704 ≤ p ≤ 0.8212296

To one decimal place, we have

CI = 0.7 ≤ p ≤ 0.8

step-by-step explanation:

she was supposed to add the hours(38.5) and multiply the amount of money she earned(7.50) resulting in $288.75 .

correct me if i'm wrong and if it is i'm truly sorry.