Answer: i) M.E = 1.96×100/√484
M.E = 8.91
ii) M.E = 1.96×100/√1764
M.E = 4.67
Therefore, the margin of error decreases as the sample size increases.
Step-by-step explanation:
Margin of error in statistics can be defined as a small amount that is allowed for in case of miscalculation or change of circumstances.
For a statistical data margin of error can be expressed as;
M.E = zr/√n
Given that;
Standard deviation r = 100
Confidence interval = 95%
sample size n1 = 484, n2 = 1764
Z(at 95% confidence interval) = 1.96
i) M.E = 1.96×100/√484
M.E = 8.91
ii) M.E = 1.96×100/√1764
M.E = 4.67
Therefore, the margin of error decreases as the sample size increases.
The margin of error at 95% when sample size n= 484 and 1764 are 8.91 and 4.67 respectively.
From the information, we need to understand that we can use the z-critical value for estimating the margin of error since the population of the sample sizes are too large.
Using the tables, at the z-critical level of 95%, [tex]\mathbf{z_c = 1.960}[/tex]
The margin of error (M.E) can be determined by using the formula:
[tex]\mathbf{M.E =z_c \times \dfrac{\sigma}{\sqrt{n}}}[/tex]
where;
[tex]\mathbf{M.E =1.96 \times \dfrac{100}{\sqrt{484}}}[/tex]
M.E = 8.91
[tex]\mathbf{M.E =1.96 \times \dfrac{100}{\sqrt{1764}}}[/tex]
M.E = 4.67
Therefore, we can conclude that the margin of error at 95% when sample size n= 484 and 1764 are 8.91 and 4.67 respectively.
Learn more about the margin of error here:
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