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

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.