Question

A
random sample of 150 people shows that 30 are left-handed. Form a
95% confidence interval for the true proportion of
left-handers

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 30 / 150 = 0.200

1 - = 1 - 0.200 = 0.8

Z_{/2}
= 1.96

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

= 1.96 * (((0.200 * 0.8) / 150)

Margin of error = E = 0.064

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.200 - 0.064 < p < 0.200 + 0.064

0.136 < p < 0.264

The 95% confidence interval for the population proportion p is :
**0.136 , 0.264**

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