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In a study of cell phone use and brain hemispheric​ dominance, an Internet survey was​ e-mailed...

In a study of cell phone use and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 2446 subjects randomly selected from an online group involved with ears. 1083 surveys were returned. Construct a 99​% confidence interval for the proportion of returned survey.

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Expert Solution

Solution :

Given that,

n = 2446

x = 1083

Point estimate = sample proportion = = x / n = 0.443

1 - = 0.557

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

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

= 2.576 * (((0.443 * 0.557) / 2446)

= 0.026

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.443 - 0.026 < p < 0.443 + 0.026

0.417 < p < 0.469

The 99% confidence interval is : (0.417 , 0.469)

milcah answered 9 months ago
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