Question

Reading proficiency: An educator wants to construct a 98% confidence interval for the proportion of elementary school children in Colorado who are proficient in reading.

(a) The results of a recent statewide test suggested that the proportion is 0.61 Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.03?

Answer #1

Solution :

Given that,

= 0.61

1 - = 1 - 0.61 = 0.39

margin of error = E = 3% = 0.03

At 98% confidence level the z is,

= 1 - 98%

= 1 - 0.98 = 0.02

/2 = 0.01

Z_{/2}
= 2.326 ( Using z table )

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

= (2.326 / 0.03)^{2} * 0.61 * 0.39

=1430.11

Sample size = 1431

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