Question

Question 17 1 pts A researcher wishes to estimate the number of households with two computers. How large a sample is needed in order to be 99% confident that the sample proportion will not differ from the true proportion by more than 2%? A previous study indicates that the proportion of households with two computers is 18% 2447 2004 2984 5

Answer 1

SOLUTION-

CONDIFENCE LEVEL(c)= 99%

MARGIN OF ERROR(m)= 0.02

PROPORTION OF HOUSEHOLD WITH TWO COMPUTERS($\hat{p}$)= 0.18

CRITICAL VALUE FOR TWO TAIL WITH SIGNIFICANCE 1% IS $Z_c$ = 2.576

THE SAMPLE SIZE IS DETERMINED APPROXIMATELY AS FOLLOWS,

$n= \frac{Z_c^{2}}{m^{2}}*\hat{p}(1-\hat{p})$

OR, $n=( \frac{2.576}{0.02})^{2}*0.18*(1-0.18)$

OR,

n2447

ANSWER- 2447

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