Question

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

Answer 1

Answer:

The sample size is  $n = 426$

Step-by-step 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

      $\sigma = \sqrt{\sigma^2 }$

=>   $\sigma = \sqrt{1.44 }$

=>   $\sigma = 1.2$

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  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 2.58$

Generally the sample size is mathematically represented as

    $n = [\frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ] ^2$

=> $n = [\frac{2.58 * 1.2 }{0.15} ] ^2$

=>  $n = 426$