Question
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%.
Solutions
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 agoRelated Solutions
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