Question

Assume that all​ grade-point averages are to be standardized on a scale between 0 and 4. How many​ grade-point averages must be obtained so that the sample mean is within 0.011 of the population​mean? Assume that a 99​% confidence level is desired. If using the range rule of​ thumb, σ can be estimated as range/4 = 4-0/4 = 1. (a) what is the required sample size? (b) Does the sample size seem practical? Explain why.

Answer 1

a) The required sample size is 54926

b) The sample size is not practical because it is too large to consider

Explanation:

Given:

Sample mean = Margin of error, E = 0.011

Confidence level = 99%

Za= 100%-99%=1% => 0.01

Standard deviation, s.d =
`(4-0)/(4) = 1`

a) For required sample size, n:

`n= [((Z_a*s.d)/(2))/(E)]^2`

`but (Z_a)/(2) = (0.01)/(2) = 0.005`

From the normal distribution table,

NORMSDIST(0.005)

= 2.5758

The required sample size will now be:

`n = [(2.578*1)/(0.011)]^2`

= [234.3636]²

= 54,926.314 => 54926

Sample size is approximatelty 54926

b) The sample size is not practical because it is too large to consider. It will be very hard to collect a sample data of almost 54926 subjects

Answer 2

a) The required sample size n= 55011,57025

b) The sample size is not practical, because that size is extremely large and would be very costly to collect, normally the size have to be significant but no to large to avoid cost unnecessary

Explanation:

Range

`σ = (4-0)/(4) = 1`

`E = 0,011 \\c= 99 %\\`⇒ 0,99

Sample size can be determinate using equation:

`n=((Z_(\alpha /2 * σ) )/(E))^(2)`

Using confidence level desired 99%
`Z_(\alpha /2)` = 2,58

Table used :

Confidence
`Z_(\alpha/2 )`

90 % 1,60

95 % 1,96

99 % 2,58

99,9 % 3,291

So replacing:

`n=((2,58 * 1 )/(0,011))^(2)`

`n= 234,5454545^(2) \\<strong>n=55011,57025`