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In: Statistics and Probability

A researcher wishes to estimate the number of households with two computers. How large a sample...

A researcher wishes to estimate the number of households with two computers. How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 4%? A previous study indicates that the proportion of households with two computers is 25%.

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Expert Solution

Solution :

Given that,

= 0.25

1 - = 1 - 0.25= 0.75

margin of error = E = 4% = 0.04

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 / 0.04)2 * 0.25 * 0.75

=634.01

Sample size = 634

orchestra answered 2 years ago
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