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uysha [10]
2 years ago
6

In a survey of 7200 T.V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want

95% confidence in our estimate of the percent of T.V. viewers who watch network news programs.
Computers and Technology
1 answer:
olga_2 [115]2 years ago
6 0

Answer:

Margin of Error=M.E= ± 0.0113

Explanation:

Margin of Error= M.E= ?

Probability that watched network news programs = p = 0.4

α= 95%

Margin of Error =M.E= zₐ/₂√p(1-p)/n

Margin of Error=M.E= ±1.96 √0.4(1-0.4)/7200

Margin of Error=M. E = ±1.96√0.24/7200

Margin of Error=M. E- ±1.96* 0.005773

Margin of Error=M.E= ±0.0113

The Margin of Error is the estimate of how much error is possible as a result of random sampling.

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#include <iostream>

using namespace std;

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Explanation:

The header files for input and output are imported.

#include <iostream>

using namespace std;

All the variables are taken as float except labour charge per hour and number of rooms.

The user is asked to input the number of rooms to be painted. An array holds the square feet in each room to be painted.

cout<<"Enter the number of rooms to be painted "<<endl;

cin>>rooms;  

for(int i=0; i <= rooms; i++)

{

cout<<"Enter the square feet in room "<<endl;

cin>>feetPerRoom[i];  

totalsqft += feetPerRoom[i];

}  

The above code asks for square feet in each room and calculates the total square feet to be painted simultaneously.

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All the calculated values are displayed in the mentioned order.

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