Question

A sample size must be determined for estimating a population mean given that the confidence level is ​% and the desired margin of error is . The largest value in the population is thought to be and the smallest value is thought to be . Calculate the sample size required to estimate the population mean using a generously large sample size.​ (Hint: Use the​ range/6 option.)

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

 $n =290$

b

 $n =129$

c

 The correct option is  D

Step-by-step explanation:

From the question we are told that

    The margin of error is   $E = 0.23$

     The largest value is  k =  15

     The smallest value is  u = 7

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

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

   $Z_{\frac{\alpha }{2} } =  1.96$

Generally population  standard deviation estimate is mathematically represented as

      $\sigma = \frac{Range }{4}$

Here the Range is mathematically represented as

       $Range = k - u$

=>    $Range = 15 - 7$

=>    $Range = 8$

=>   $\sigma = \frac{8 }{4}$

=>   $\sigma = 2$

Generally the sample size is mathematically represented as

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

=>   $n = [1.96  *  2 }{0.23} ] ^2$

=>   $n =290$

Generally to obtain a  conservatively small sample size the population standard deviation estimate becomes

=>   $\sigma = \frac{Range }{6}$

=>   $\sigma = \frac{8 }{6}$

=>   $\sigma =1.333$

Generally the conservatively small sample size is mathematically evaluated as

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

=>   $n = [1.96  *  1.333 }{0.23} ] ^2$

=>   $n =129$