Question

Find the sample size needed to estimate, within margin of error 1.5%, the percentage of people in a certain country that are left-handed. Use a 90% confidence level and assume about 11% of people in the country are left-handed.

Answer #1

Solution :

Given that,

= 0.11

1 - = 1 - 0.11 = 0.91

margin of error = E = 1.5% = 0.015

At 90% confidence level z

= 1 - 90%

= 1 - 0.90 =0.10

/2
= 0.05

Z/2
= Z0.05 = 1.645 ( Using z table ( see the 0.05 value in standard
normal (z) table corresponding z value is 1.645 )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.645 / 0.015)2 * 0.11 * 0.91

=1204

Sample size = 1204

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