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bija089 [108]
3 years ago
15

A sample size must be determined for estimating a population mean given that the confidence level is ​% and the desired margin o

f 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.)
Mathematics
1 answer:
Sauron [17]3 years ago
6 0

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  

 

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