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A business student is interested in estimating the 90% confidence interval for the proportion of students who bring laptops to campus. He wishes a precise estimate and is willing to draw a large sample that will keep the sample proportion within seven percentage points of the population proportion. What is the minimum sample size required by this student, given that no prior estimate of the population proportion is available

Respuesta :

Answer:

n = 106

Step-by-step explanation:

Given  

p = 0.5 and 1-p = q = 0.5

Margin of error = 0.08  

Confidence level = 90%  

Z score for 90% confidence level =     1.65

As we know -  

Margin of error = z * Sqrt (pq/n)

Substituting the given value, we get –  

0.08 = 1.65 * Sqrt (0.5*0.5/n)

Squaring both the sides and solving, we get  

n = 1.65^2*0.5^2/0.08^2

n = 106.34 = 106

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