Question

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level. In an earlier study, the population proportion was estimated to be 0.21 . How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 99% confidence level with an error of at most 0.02? Round your answer up to the next integer.

Answer #1

Solution :

Given that,

= 0.21

1 - = 1 - 0.21 = 0.79

margin of error = E = 0.02

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_{/2}
= 2.58 ( Using z table ( see the 0.005 value in standard normal (z)
table corresponding z value is 2.58 )

Sample size = n = (Z_{/2}
/ E)^{2} *
* (1 -
)

= (2.58 / 0.02)^{2} * 0.21 * 0.79

=2761

Sample size = 2761

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