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Which of the following [tex]$z$[/tex]-values, standard deviations, and sample sizes produce:

[tex]\[ M E = \frac{Z \cdot S}{\sqrt{n}} \][/tex]

A. [tex]\( z = 2.14; s = 4; n = 9 \)[/tex]

B. [tex]\( z = 2.14; s = 4; n = 81 \)[/tex]

C. [tex]\( z = 2.14; s = 16; n = 9 \)[/tex]

D. [tex]\( z = 2.14; s = 16; n = 81 \)[/tex]

Sagot :

To find the margin of error (ME) for each set of given values of [tex]\( z \)[/tex], [tex]\( s \)[/tex], and [tex]\( n \)[/tex], you use the formula:

[tex]\[ ME = \frac{z \cdot s}{\sqrt{n}} \][/tex]

Here are the calculations for each set of given values:

1. For [tex]\( z = 2.14 \)[/tex], [tex]\( s = 4 \)[/tex], [tex]\( n = 9 \)[/tex]:

[tex]\[ ME = \frac{2.14 \cdot 4}{\sqrt{9}} = \frac{2.14 \cdot 4}{3} = \frac{8.56}{3} = 2.8533333333333335 \][/tex]

2. For [tex]\( z = 2.14 \)[/tex], [tex]\( s = 4 \)[/tex], [tex]\( n = 81 \)[/tex]:

[tex]\[ ME = \frac{2.14 \cdot 4}{\sqrt{81}} = \frac{2.14 \cdot 4}{9} = \frac{8.56}{9} = 0.9511111111111111 \][/tex]

3. For [tex]\( z = 2.14 \)[/tex], [tex]\( s = 16 \)[/tex], [tex]\( n = 9 \)[/tex]:

[tex]\[ ME = \frac{2.14 \cdot 16}{\sqrt{9}} = \frac{2.14 \cdot 16}{3} = \frac{34.24}{3} = 11.413333333333334 \][/tex]

4. For [tex]\( z = 2.14 \)[/tex], [tex]\( s = 16 \)[/tex], [tex]\( n = 81 \)[/tex]:

[tex]\[ ME = \frac{2.14 \cdot 16}{\sqrt{81}} = \frac{2.14 \cdot 16}{9} = \frac{34.24}{9} = 3.8044444444444445 \][/tex]

So the margins of error for the given values are:

1. [tex]\( z = 2.14 \)[/tex], [tex]\( s = 4 \)[/tex], [tex]\( n = 9 \)[/tex]: [tex]\(\mathbf{2.8533333333333335}\)[/tex]
2. [tex]\( z = 2.14 \)[/tex], [tex]\( s = 4 \)[/tex], [tex]\( n = 81 \)[/tex]: [tex]\(\mathbf{0.9511111111111111}\)[/tex]
3. [tex]\( z = 2.14 \)[/tex], [tex]\( s = 16 \)[/tex], [tex]\( n = 9 \)[/tex]: [tex]\(\mathbf{11.413333333333334}\)[/tex]
4. [tex]\( z = 2.14 \)[/tex], [tex]\( s = 16 \)[/tex], [tex]\( n = 81 \)[/tex]: [tex]\(\mathbf{3.8044444444444445}\)[/tex]