Question

You are a researcher studying the lifespan of a certain species of bacteria. From a previous study, it was found that the standard deviation was 5.6 hours. You would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.45 hours at a 98% level of confidence. What sample size should you gather to achieve a 0.45 hour margin of error?

Answer 1

Answer:

The sample size is  $n =  841$

Step-by-step explanation:

From the question we are that

    The standard deviation is  $\sigma = 5.6 \ hours$

     The margin of error is  $E = 0.45$

From the question we are told the confidence level is  98% , hence the level of significance is    

      $\alpha = (100 - 98) \%$

=>   $\alpha = 0.02$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } =  2.33$

Generally the sample size is mathematically represented as  

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

=>   $n = [2.33 } *  5.6}{0.45} ] ^2$

=>   $n =  841$