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In 2004, the General Social Survey (which uses a method similar to simple random sampling) asked, "Do you consider yourself athletic?" For this question, 255 people said that they did out of 2373 randomly selected people. What is the 95% confidence interval for the proportion of all Americans who consider themselves athletic?

Respuesta :

Diatonic254

(0.0974, 0.1226)

Step-by-step explanation:

We first find the sample proportion;

255/ 2373

0.11

The Z value for a 95% confidence interval is 1.960

Therefore;

0.11 +- 1.96 * √ (0.11(1 – 0.11))/2373

0.11 +- 1.96 * 0.00642

0.11 +- 0.0126

= 0.1226

= 0.0974

(0.0974, 0.1226)

Ver imagen Diatonic254

The 95% confidence interval for the proportion of all Americans who consider themselves athletic is (0.0974, 0.1226)

Calculation of the confidence interval:

Since  255 people said that they did out of 2373 randomly selected people.

So, the sample proportion;

[tex]= 255\div 2373[/tex]

= 0.11

Since Z value for a 95% confidence interval is 1.960

So,

[tex]= 0.11 \pm1.96 \times \sqrt (0.11(1 – 0.11))\div 2373\\\\= 0.11 \pm 1.96 \times 0.00642\\\\= 0.11 \pm 0.0126[/tex]

So,

= 0.1226 or 0.0974

Therefore, The 95% confidence interval for the proportion of all Americans who consider themselves athletic is (0.0974, 0.1226)

learn more about proportion here: https://brainly.com/question/12328649