Gamers are described as people who regularly play video games. A random sample of 80 people...

Gamers are described as people who regularly play video games. A random sample of 80 people were selected and found 50 to be mail. The 99% confidence interval for the proportion based on the sample is

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Expert Solution

Solution :

Given that,

n = 80

x = 50

Point estimate = sample proportion = = x / n = 50/80=0.625

1 - = 1- 0.625 =0.375

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

Margin of error = E = Z/2 * ( $\sqrt$ ((( * (1 - )) / n)

= 2.576* ( $\sqrt$ ((0.625*0.375) /80 )

E = 0.139

A 99% confidence interval for proportion p is ,

- E < p < + E

0.625-0.139 < p < 0.625+0.139

0.486< p < 0.764

The 99% confidence interval for the proportion p is : (0.486 ,0.764)

orchestra answered 1 year ago

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