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A software developer wants to know how many new computer games people buy each year. Assume a previous study found the variance to be 1.44. She thinks the mean is 6 computer games per year. What is the minimum sample size required to ensure that the estimate has an error of at most 0.15 at the 99% level of confidence

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Answer:

The sample size is
n = 426

Explanation:

From the question we are told that

The population variance is
\sigma^2 = 1.44

The mean is
\= x = 6

The margin of error is E = 0.15

Generally the standard deviation is mathematically represented as


\sigma = √(\sigma^2 )

=>
\sigma = √(1.44 )

=>
\sigma = 1.2

From the question we are told the confidence level is 99% , hence the level of significance is


\alpha = (100 - 99 ) \%

=>
\alpha = 0.01

Generally from the normal distribution table the critical value of
(\alpha )/(2) is


Z_{(\alpha )/(2) } =  2.58

Generally the sample size is mathematically represented as


n = [\frac{Z_{(\alpha )/(2) } *  \sigma }{E} ] ^2

=>
n = [(2.58 * 1.2 )/(0.15) ] ^2

=>
n = 426

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