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

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
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.96`

Generally population standard deviation estimate is mathematically represented as

`\sigma = (Range )/(4)`

Here the Range is mathematically represented as

`Range = k - u`

=>
`Range = 15 - 7`

=>
`Range = 8`

=>
`\sigma = (8 )/(4)`

=>
`\sigma = 2`

Generally the sample size is mathematically represented as

`n = [\frac{Z_{(\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 = (Range )/(6)`

=>
`\sigma = (8 )/(6)`

=>
`\sigma =1.333`

Generally the conservatively small sample size is mathematically evaluated as

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

=>
`n = [1.96  *  1.333 }{0.23} ] ^2`

=>
`n =129`