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

The sample size is
`n =  841`

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
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  2.33`

Generally the sample size is mathematically represented as

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

=>
`n = [2.33 } *  5.6}{0.45} ] ^2`

=>
`n =  841`