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Beth wants to determine a 99 percent confidence interval for the true proportion pp of high school students in the area who attend their home basketball games. Out of nn randomly selected students she finds that that exactly half attend their home basketball games. About how large would nn have to be to get a margin of error less than 0.01 for pp

Sagot :

JeanaShupp

Answer: 16590

Step-by-step explanation:

Let p be the population proportion of high school students in the area who attend their home basketball games.

As per given,

prior p = 0.5

Margin of error E= 0.01

Criticfor z-value for 99% confidence = 2.576

Sample size will be computed as

[tex]n=p(1-p)(\frac{z^*}{E})^2\\\\ n= 0.5(1-0.5)(\frac{2.576}{0.01})^2\\\\=0.25(257.6)^2\\\\=0.25 (66357.76)\approx16590[/tex]

Hence, required sample size = 16590

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