Question

A random sample of 125 people shows that 25 are left-handed.
Form a 99% confidence interval for the true proportion of
left-handers.

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 25 / 125 = 0.20

1 - = 1 - 0.20 = 0.80

Z/2
= Z_{0.005} = 2.576

Margin of error = E = Z_{
/ 2} * ((
* (1 -
))
/ n)

= 2.576 (((0.20 * 0.80) / 125)

= 0.092

A 99% confidence interval for population proportion p is ,

± E

= 0.20 ± 0.092

= ( 0.108, 0.292 )

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